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\author{massiveballs}
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  /Title (chemistry.pdf)
  /Creator (Cheatography)
  /Author (massiveballs)
  /Subject (Chemistry Cheat Sheet)
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    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Chemistry Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{massiveballs} via \textcolor{DarkBackground}{\uline{cheatography.com/202526/cs/42987/}}}
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}massiveballs \\
  \uline{cheatography.com/massiveballs} \\
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Not Yet Published.\\
   Updated 12th April, 2024.\\
   Page {\thepage} of \pageref{LastPage}.
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  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
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\begin{multicols*}{2}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Average atomic mass of isotopes}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Atomic mass x abundance of isotope for each isotope, then add together \newline % Row Count 2 (+ 2)
Example: \newline % Row Count 3 (+ 1)
Iron has 4 isotopes.  \newline % Row Count 4 (+ 1)
Fe-54, 56, 57, and 58 \newline % Row Count 5 (+ 1)
To calculate the average atomic mass, first get the abundance \% and atomic mass of each isotope. \newline % Row Count 7 (+ 2)
Fe-54 \newline % Row Count 8 (+ 1)
abundance \% = 5.845\% \newline % Row Count 9 (+ 1)
Atomic mass = 53. 9396 \newline % Row Count 10 (+ 1)
Convert abundance to decimal format by dividing by 100 \newline % Row Count 12 (+ 2)
= 0.05845 \newline % Row Count 13 (+ 1)
Now, multiply the atomic mass with the \% as a decimal \newline % Row Count 15 (+ 2)
5.845 x 0.05845 = 0.34164025  \newline % Row Count 16 (+ 1)
Do the same for the remaining isotopes, then add the final numbers together \newline % Row Count 18 (+ 2)
= 55.845 \newline % Row Count 19 (+ 1)
Highest abundance \% is the most commonly found isotope btw% Row Count 21 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{max obtainable mass}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Maximum mass that can be obtained \newline % Row Count 1 (+ 1)
Laughing gas" or nitrous oxide, N 2O, is prepared by the thermal decomposition of ammonium nitrate: NH4NO3 (s)  N 2O (g) + 2 H2O (l) (a) What is the maximum mass in grams of N2O (g) that can be obtained from 1.53  10 2 g of ammonium nitrate? \newline % Row Count 6 (+ 5)
Calculate moles of ammonium nitrate \newline % Row Count 7 (+ 1)
-	Mass is 153g, molar mass is 80g/mol, divide mass by molar mass \newline % Row Count 9 (+ 2)
-	Moles = 153/80 = 1.91 moles \newline % Row Count 10 (+ 1)
According to equation, 1 mole of NH4NO3 yields 1 mole of N20 \newline % Row Count 12 (+ 2)
Number of moles of N2O produced will also be 1.91 \newline % Row Count 13 (+ 1)
Convert moles of N2O to grams \newline % Row Count 14 (+ 1)
Multiply N2O molar mass (44.02) by number of moles \newline % Row Count 16 (+ 2)
Mass of N2O = 1.91 x 44.02g/mol = 84.1 grams \newline % Row Count 17 (+ 1)
If the percentage yield was 76\%, what mass (grams) of N 2O was actually produced? \newline % Row Count 19 (+ 2)
Calculate maximum theoretical yield – 84.1 grams \newline % Row Count 21 (+ 2)
Apply percentage yield to find actual yield \newline % Row Count 22 (+ 1)
Given the percentage yield as 76\%, we multiply it by the maximum theoretical yied \newline % Row Count 24 (+ 2)
Actual yield = 84.1 g x 76/100 = 63.9g% Row Count 25 (+ 1)
} \tn 
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\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{\% (w/w)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Percentage by weight, indicating the mass of the solute per 100 grams of total product (solution) \newline % Row Count 2 (+ 2)
Tells us the proportion of the solute to the entire product \newline % Row Count 4 (+ 2)
Number of grams of NaOh present in a 375g can of oven cleaner labelled 4.2\% (w/w) NaOH \newline % Row Count 6 (+ 2)
Calculate mass of NaOH per gram \newline % Row Count 7 (+ 1)
Given percentage of NaOH in oven cleaner Is 4.2\% w/w \newline % Row Count 9 (+ 2)
To convert \% to grams/gram, divide it by 100 \newline % Row Count 10 (+ 1)
-	4.2/100 = 0.042g \newline % Row Count 11 (+ 1)
Determine total mass of NaOH in the can \newline % Row Count 12 (+ 1)
Total mass is given as 375g \newline % Row Count 13 (+ 1)
To find mass of NaOH in the can, you multiply the mass of NaOH per gram of product by the total mass of the can \newline % Row Count 16 (+ 3)
-	NaOH mass in can = 0.042 g/g x 375g \newline % Row Count 17 (+ 1)
-	= 15.75g% Row Count 18 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Evaporation/ \%Mass}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Calculate concentration of a solid as a percentage by mass (\% mass) \newline % Row Count 2 (+ 2)
\% of mass calculated by dividing the mass of the solute by the total mass of the solution and then multiply by 100 \newline % Row Count 5 (+ 3)
Calculate the concentration of the solid mass per unit volume g/L \newline % Row Count 7 (+ 2)
This concentration represents the amount of solid material (solute) per unit volume of the solution \newline % Row Count 9 (+ 2)
A 1.62 kg (1.71 L) sample of creek water was evaporated to dryness, leaving 6.33 g of solid material. \newline % Row Count 12 (+ 3)
Calculate the concentration of this solid as percentage by mass and calculate the concentration of this solid mass per unit volume g/l \newline % Row Count 15 (+ 3)
Concentration of the solid as a percentage by mass (\% mass) \newline % Row Count 17 (+ 2)
The mass of the solid material left after evaporating is 6.33g \newline % Row Count 19 (+ 2)
The total mass of the solution is 1.62kg = 1620g \newline % Row Count 20 (+ 1)
\% mass = (6.33g/1620g)  x 100\% - 0.391\% \newline % Row Count 21 (+ 1)
Calculate the concentration \newline % Row Count 22 (+ 1)
The volume of creek water is 1.71L \newline % Row Count 23 (+ 1)
Concentration = mass of solute / volume of solution \newline % Row Count 25 (+ 2)
= 6.33g/1.71L = 3.70 g/L% Row Count 26 (+ 1)
} \tn 
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\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{concentration in dissolved solution}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Figure out number of moles first \newline % Row Count 1 (+ 1)
Molarity = moles per litre (mol.L\textasciicircum{}-1\textasciicircum{}) \newline % Row Count 2 (+ 1)
work out number of moles first, then divide by the volume \newline % Row Count 4 (+ 2)
Molar mass of Na = 22.99 \newline % Row Count 5 (+ 1)
MM of Cl = 35.45 \newline % Row Count 6 (+ 1)
divide by the given volume (6g) \newline % Row Count 7 (+ 1)
6/(22.99 + 35.35 moles) \newline % Row Count 8 (+ 1)
=6g/58.44 mol \newline % Row Count 9 (+ 1)
= 0.103 gmol \textasciicircum{}-1\textasciicircum{} \newline % Row Count 10 (+ 1)
volume in mL, convert to L \newline % Row Count 11 (+ 1)
750ml/1000 = 0.75L \newline % Row Count 12 (+ 1)
concentration is the mols/volume \newline % Row Count 13 (+ 1)
0.103/0.75 \newline % Row Count 14 (+ 1)
= 0.14M% Row Count 15 (+ 1)
} \tn 
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\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Moles corresponding to molecules}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{convert number of molecules to moles by dividing given number of molecules by Avo number \newline % Row Count 2 (+ 2)
Number of moles = number of molecules/6.022 x 10\textasciicircum{}23\textasciicircum{} \newline % Row Count 4 (+ 2)
Moles of CH4 corresponding to 1.56 x 10\textasciicircum{}20\textasciicircum{} molecules \newline % Row Count 6 (+ 2)
N(CH4) = 1.56 x 10\textasciicircum{}20 molecules/6.022 x 10\textasciicircum{}23\textasciicircum{} \newline % Row Count 7 (+ 1)
= 25.9 x 10\textasciicircum{}-4\textasciicircum{}% Row Count 8 (+ 1)
} \tn 
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\end{tabularx}
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\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Mass of product}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{{\bf{If 100.0 g of Al is added to 0.11 mol of O2, how many grams of Al 2O3 (s) will be produced?}} \newline % Row Count 2 (+ 2)
Write balanced equation \newline % Row Count 3 (+ 1)
3Al + 3O2 -{}-\textgreater{} 2Al2O3 \newline % Row Count 4 (+ 1)
- 4 moles of Al react with 3 moles of oxygen to make 2 moles of aluminum oxide \newline % Row Count 6 (+ 2)
Moles of Al using its molar mass \newline % Row Count 7 (+ 1)
= moles of Al = mass of Al/Molar mass of Al \newline % Row Count 8 (+ 1)
= 100g/27 g/mol = 3.70 mol \newline % Row Count 9 (+ 1)
Mols of oxygen are already given, they are 0.11 mol \newline % Row Count 11 (+ 2)
Determine limiting reactants \newline % Row Count 12 (+ 1)
we assume 1 is in excess \newline % Row Count 13 (+ 1)
- We use stochiometry to determine how much of the O2 would be needed to react completely with the excess reactant. if there is more of the other reactant than the calculated amount, then it is in excess; otherwise it is the limiting reactant. \newline % Row Count 18 (+ 5)
- to react all alumium, it would require 3/4 x moles of Al \newline % Row Count 20 (+ 2)
- 3/4 x 3.70 = 2.78 mol of O2 \newline % Row Count 21 (+ 1)
Assuming O2 is in excess, it would require 4/3 x moles of O2 \newline % Row Count 23 (+ 2)
= 4/3 x 0.11 = 15 mol Al \newline % Row Count 24 (+ 1)
Since theres less than 2.78 mol O2 to react with the Al, and more than 0.15 mol Al to react with the O2, O2 is the limiting reactant \newline % Row Count 27 (+ 3)
Calculate moles of Al2O3 \newline % Row Count 28 (+ 1)
from the equation, we see that 4 moles of Al react with 3 moles of O to produce 2 moles of Al2O3. \newline % Row Count 30 (+ 2)
} \tn 
\end{tabularx}
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\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Mass of product (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Since O is limtiing reactant, we use its moles to find the moles of Al2O3 \newline % Row Count 2 (+ 2)
moles of Al2O3 = 2/3 moles of O2 \newline % Row Count 3 (+ 1)
= 2/3 x 011 = 0.073 \newline % Row Count 4 (+ 1)
Mass of Al2O3 produces \newline % Row Count 5 (+ 1)
= Moles of Al2O3 x molar mass of Al2O3 \newline % Row Count 6 (+ 1)
= 0.073 x 102.0 = 7.48 grams% Row Count 7 (+ 1)
} \tn 
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\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{lewis structure}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Count total number of valence electrons and add them together \newline % Row Count 2 (+ 2)
- for exanmple, NCl3, nitrogen has 5, chlorine has 7. since there are 3 Cl atoms, thats 3x7 = 21 valence electrons. \newline % Row Count 5 (+ 3)
- 5 + 21 = 26 valence electrons for the entire molecules \newline % Row Count 7 (+ 2)
Choose the central atom \newline % Row Count 8 (+ 1)
- the central atom is the less electronegative one, as it can make more bonds \newline % Row Count 10 (+ 2)
- NCl3, nitrogen is the central atom \newline % Row Count 11 (+ 1)
Connect the atoms with single bonds \newline % Row Count 12 (+ 1)
- Connect the central nitrogen atom to each of the 3 chlorine atoms using single bonds \newline % Row Count 14 (+ 2)
- this uses up 3 pairs of electrons, 26-3 = 23 electrons remaining \newline % Row Count 16 (+ 2)
distribute the remaining electrons \newline % Row Count 17 (+ 1)
- place lone pairs on the outer atoms first, and then fill the remaning electrons around the central atoms \newline % Row Count 20 (+ 3)
- each lone pair is represented by 2 electrons \newline % Row Count 21 (+ 1)
check for formal charges after \newline % Row Count 22 (+ 1)
- a formal charge occurs when an atom doesnt have the expected number of valence electrons \newline % Row Count 24 (+ 2)
- calculate the formal charge for each atom by subtracting the number of lone pair electrons and half the number of bonding electrons from the number of valence electrons the atom brings \newline % Row Count 28 (+ 4)
If any atoms have formal charges, try to minimize them by moving lone pairs or changing the arrangement of bonds \newline % Row Count 31 (+ 3)
} \tn 
\end{tabularx}
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\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{lewis structure (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{- the goal is to have the lowest formal charges possible while still satisfying the octet rule \newline % Row Count 2 (+ 2)
N - 1 lone pair (2 electrons) \newline % Row Count 3 (+ 1)
Each Cl - 3 lone pairs (6 electrons)% Row Count 4 (+ 1)
} \tn 
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\end{tabularx}
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\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{mass of production in reaction}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{KCN (aq) + HCl (aq) -{}-\textgreater{} KCl (aq) + HCN (g) \newline % Row Count 1 (+ 1)
{\bf{If 0.250 g KCN reacts with excess acid, calculate the mass (in g) of HCN produced?}} \newline % Row Count 3 (+ 2)
Determine molar masses of the substances involved \newline % Row Count 4 (+ 1)
- Molar mass of KCN = molar mass of K + molar mass of C + molar mass of N \newline % Row Count 6 (+ 2)
- Molar mass of HCN =  molar mass of H + C + N \newline % Row Count 7 (+ 1)
conver the mass of KCN to moles \newline % Row Count 8 (+ 1)
- number of moles = mass/molar mass \newline % Row Count 9 (+ 1)
apply stoichiometry to find the molar ratio of KCN and HCN to find the number of moles of HCN produced when a certain number of moles of KCN reacts \newline % Row Count 12 (+ 3)
Convert moles of HCN to mass \newline % Row Count 13 (+ 1)
- mass = number of moles x molar mass \newline % Row Count 14 (+ 1)
Moles of KCN = 0.250/65 \newline % Row Count 15 (+ 1)
= 0.003846 \newline % Row Count 16 (+ 1)
the ratio is 1:1, so the number of HCN moles is the same as the number of KCN moles \newline % Row Count 18 (+ 2)
To find the mass of HCN, multiply the number of moles of HCN by it's molar mass \newline % Row Count 20 (+ 2)
Moles of HCN x Molar mass of HCN \newline % Row Count 21 (+ 1)
= 0.003846 mol x 27g/mol \newline % Row Count 22 (+ 1)
= 0.104 g \newline % Row Count 23 (+ 1)
mass of HCN produced when 0.250 g of KCN reacts with excess acid is 0.104% Row Count 25 (+ 2)
} \tn 
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\end{tabularx}
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\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{concentration of a solution}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{{\bf{"Calculate the concentration (mol/L) of a saline solution of 9 g table salt in 1 L water."}} \newline % Row Count 2 (+ 2)
Determine the molar mass of NaCl \newline % Row Count 3 (+ 1)
- sum of the atomic mass of Na and Cl \newline % Row Count 4 (+ 1)
- MM Na + MM Cl \newline % Row Count 5 (+ 1)
= 22.99 g/mol + 35.45 g/mol \newline % Row Count 6 (+ 1)
= 58.44 g/mol \newline % Row Count 7 (+ 1)
Convert the mass to moles \newline % Row Count 8 (+ 1)
- the given mass of table salt is 9g \newline % Row Count 9 (+ 1)
- number of moles of Nacl \newline % Row Count 10 (+ 1)
= Mass of NaCl/ molar mass of NaCl \newline % Row Count 11 (+ 1)
= 9g/58.44g/mol \newline % Row Count 12 (+ 1)
= 0/154 mol \newline % Row Count 13 (+ 1)
Now calculate the morality/concentration of the NaCl in the saline solution \newline % Row Count 15 (+ 2)
Concentration (mol/L) = number of mols of solute/ volume of solution (L) \newline % Row Count 17 (+ 2)
= 0/154mol/1L \newline % Row Count 18 (+ 1)
= 0/154 mol/L \newline % Row Count 19 (+ 1)
The concentration of the saline solution, 9g of NaCl in 1L of water, is 0.154 mol/L \newline % Row Count 21 (+ 2)
{\bf{Explain how you would proceed to make 50 ml of 0.030 M solution out of the above stock saline solution.}} \newline % Row Count 24 (+ 3)
Calculate volume of stock solution needed \newline % Row Count 25 (+ 1)
use the dilution formula to calculate the volume of the stock solution (0.154 M) to prepare the desired result: \newline % Row Count 28 (+ 3)
C1V1 = C2V2 \newline % Row Count 29 (+ 1)
- C1 = concentration of stock solution (0.514 M) \newline % Row Count 30 (+ 1)
} \tn 
\end{tabularx}
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\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{concentration of a solution (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{- V1 = volume of stock needed (?) \newline % Row Count 1 (+ 1)
- C2 = desired concentration of final solution (0.030 M) \newline % Row Count 3 (+ 2)
- V2 = Volume of final solution (50mL) \newline % Row Count 4 (+ 1)
Solve for V1 \newline % Row Count 5 (+ 1)
V1 = C2 X V2 / C1 \newline % Row Count 6 (+ 1)
= (0.030 mol/L) x (0.050L) / 0.514 mol/L \newline % Row Count 7 (+ 1)
= 0.0015 / 0.154 L \newline % Row Count 8 (+ 1)
convert to mL \newline % Row Count 9 (+ 1)
= 0.0097 X 1000 mL/L \newline % Row Count 10 (+ 1)
= V1 = 9.7mL \newline % Row Count 11 (+ 1)
You need 9.7 mL of the solution to prepare 50 mL of the desired solution.% Row Count 13 (+ 2)
} \tn 
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\end{tabularx}
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\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{water solubility and ions}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Determing which compounds will form ions when dissolved in water. \newline % Row Count 2 (+ 2)
Identify if compound is ionic, polar covalent or non-polar covalent \newline % Row Count 4 (+ 2)
- ionic: consist of metal and a nonmetal or a metal and a polyatomic ion \newline % Row Count 6 (+ 2)
- polar: significant different in electronegativity btwn bonded atoms, resulting in partial pos/neg charges \newline % Row Count 9 (+ 3)
- non polar: similar electronegativity, resulting in equal sharing of electrons and no signfiicant charge seperations \newline % Row Count 12 (+ 3)
Solubility rules: \newline % Row Count 13 (+ 1)
- Most nitrate (NO3-), acetate (C2H3O2-), and chlorate (ClO3-) salts are soluble \newline % Row Count 15 (+ 2)
- most alkali metal (group 1) and ammonium salts (NH4+) salts are soluble \newline % Row Count 17 (+ 2)
- most chloride (Cl-), bromide (Br-). and iodide (I-) salts are soluble, except for those of silver, lead(ii) and mercury(i) \newline % Row Count 20 (+ 3)
- most sulfate (SO4\textasciicircum{}2-\textasciicircum{}) salts are soluble, except for those of calcium, stronium, barium. lead(ii), and some silver salts \newline % Row Count 23 (+ 3)
{\bf{Dissociation in water}} \newline % Row Count 24 (+ 1)
ionic compounds with ions are soluble according to the rules, it will dissociate into ions when dissolved in water \newline % Row Count 27 (+ 3)
if polar or nonpolar, it wont \newline % Row Count 28 (+ 1)
Some compounds that fully dissociate into ions in solution are considered strong electrolytes \newline % Row Count 30 (+ 2)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{water solubility and ions (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{- these compounds conduct electricity well in solution due to the presence of free ions \newline % Row Count 2 (+ 2)
- strong acids, bases and soluble ionic compounds \newline % Row Count 3 (+ 1)
Some dissociation behaviour of a compound can vary, some partially dissociate into ions while others fully dissociate \newline % Row Count 6 (+ 3)
- weak acids and bases partially dissociate into ions in solution, theyre waek electrolytes \newline % Row Count 8 (+ 2)
- non-electrolytes dont dissociate into ions when dissolved in water \newline % Row Count 10 (+ 2)
{\bf{Example}} \newline % Row Count 11 (+ 1)
Li2SO4 \newline % Row Count 12 (+ 1)
Lithium sulfate, containing lithium (Li) and sulfate (SO4\textasciicircum{}2-\textasciicircum{}) ions. \newline % Row Count 14 (+ 2)
Li salts, such as lithium sulfate, are generalyl soluble in water as lithium compounds are highly soluble \newline % Row Count 17 (+ 3)
most sulfate salts are soluble \newline % Row Count 18 (+ 1)
based on this, lithium sulfate will dissociate into its constituent ions \newline % Row Count 20 (+ 2)
- Li+ and SO4\textasciicircum{}-2\textasciicircum{}  \newline % Row Count 21 (+ 1)
Li2SO4 -{}-\textgreater{} 2Li+ + SO4\textasciicircum{}2-\textasciicircum{}% Row Count 22 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{volume of acid to neutralise}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{volume 0.355 M perchloric acid to neautralise 15.5mL of a 0.179 calcium hydroxide solution \newline % Row Count 2 (+ 2)
Write a balanced equation for the reaction \newline % Row Count 3 (+ 1)
- perchloric acid (HClO4)  and calcium hydroxide (Ca(OH)2) \newline % Row Count 5 (+ 2)
- 2HClO4 + Ca(OH)2 -{}-\textgreater{} Ca(ClO4)2 + Ca(ClO4)2 +2H2O \newline % Row Count 7 (+ 2)
Determine the stoichiometry \newline % Row Count 8 (+ 1)
From the equation, you can see that 2 moles of HClO4 react with 1 mole of Ca(OH)2, meaning the stoichiometric ratio is 2:1 \newline % Row Count 11 (+ 3)
Calculate the moles of Ca(OH)2 \newline % Row Count 12 (+ 1)
Use the given concentration and volume of the calcium hydroxide solution to calculate the number of moles present. \newline % Row Count 15 (+ 3)
- Moles of Ca(OH)2 = concentration x volume \newline % Row Count 16 (+ 1)
= 0.179 M x 15.5mL/1000L \newline % Row Count 17 (+ 1)
= 0.179 M x 0.0155L  \newline % Row Count 18 (+ 1)
= 0.00277 Mol \newline % Row Count 19 (+ 1)
find the moles of HClO4 \newline % Row Count 20 (+ 1)
 since the ratio between them is 2:1, the number of moles of perchloric acid needed is twice the number of calcium hydroxide \newline % Row Count 23 (+ 3)
- moles of HClO4  \newline % Row Count 24 (+ 1)
= 2 x 0.00277 \newline % Row Count 25 (+ 1)
= 0.00554 Mol \newline % Row Count 26 (+ 1)
now use the concentration of perchloric acid solution to find the volume of it thats needed \newline % Row Count 28 (+ 2)
Volume = moles/concentration \newline % Row Count 29 (+ 1)
= 0.00554mol/0.355M \newline % Row Count 30 (+ 1)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{volume of acid to neutralise (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{= 0.00554/0.355 L \newline % Row Count 1 (+ 1)
= 0.0156 litres \newline % Row Count 2 (+ 1)
So approx. 0.0156 litres or 15.6 mL of 0.355 M of perchloric acid needed to neutralise 15.5mL of 0.179 M calcium hydroxide solution% Row Count 5 (+ 3)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Molecular formula from \% composition}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Follow previous steps to get the empirical formula, then find the molar mass of it by adding up the atomic masses of all atoms in the formula \newline % Row Count 3 (+ 3)
Determine the molecular formula mass \newline % Row Count 4 (+ 1)
You need to know the molecular weight/molar mass of the compound, if not known, you have to determine it experimentally \newline % Row Count 7 (+ 3)
Calculate the 'multiplier' or 'factor' \newline % Row Count 8 (+ 1)
Divide the molecular weight of the compound by the empirical formula mass to get the 'multiplier' that represents how many times the empirical formula must be multiplied to get the molecular formula \newline % Row Count 12 (+ 4)
Multiply the subscripts in the empirical formula by the 'multiplier' \newline % Row Count 14 (+ 2)
{\bf{Example:}} \newline % Row Count 15 (+ 1)
Compound: 40\% carbon, 6,7\% hydrogen, and 53.3\% oxygen, with a molecular weight of 180 g/mol \newline % Row Count 17 (+ 2)
the empirical formula is CH2O \newline % Row Count 18 (+ 1)
to find the mass of it: \newline % Row Count 19 (+ 1)
Mass of C = 12.01 g/mol \newline % Row Count 20 (+ 1)
Mass of H = 1.01 g/mol \newline % Row Count 21 (+ 1)
Mass of O = 16 g/mol \newline % Row Count 22 (+ 1)
The empirical formula mass: \newline % Row Count 23 (+ 1)
(12.01 x 1) + (1.01 x 2) + (16.00 x 1) \newline % Row Count 24 (+ 1)
= 30.03g/mol \newline % Row Count 25 (+ 1)
Calculate the 'multiplier'.  \newline % Row Count 26 (+ 1)
- Multiplier= Molecular weight/Empirical formula mass=180/30.03 \newline % Row Count 28 (+ 2)
= 6 \newline % Row Count 29 (+ 1)
Multiply the subscripts in the empirical formula by the multiplier \newline % Row Count 31 (+ 2)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Molecular formula from \% composition (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{C1H2O1 x 6 = C6H12O6 \newline % Row Count 1 (+ 1)
so the molecular formula is C6H12O6 \newline % Row Count 2 (+ 1)
By following these steps, you can determine the molecular formula of a compound from its percent composition and molecular weight.% Row Count 5 (+ 3)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Excess reactant}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Write balanced equation and determine limiting reactant \newline % Row Count 2 (+ 2)
- calculate moles of each reactant involved and compare these values to determine which is limiting and which is in excess \newline % Row Count 5 (+ 3)
- the limiting reactant is the one that produces the least amount of product    \newline % Row Count 7 (+ 2)
Calculate theoretical yield \newline % Row Count 8 (+ 1)
use the stoichiometry of the balanced equation \newline % Row Count 9 (+ 1)
- multiply the number of moles of the limiting reactant by the stochiometric coefficient of the product in the balanced equation to get the theoretical yield in moles. \newline % Row Count 13 (+ 4)
Determine the excess reactant \newline % Row Count 14 (+ 1)
This is the one that is not limiting \newline % Row Count 15 (+ 1)
- subtract the amount of excess reactant that reacts from the initial amount of excess reactant given \newline % Row Count 18 (+ 3)
now calculate the amount left after the reaction \newline % Row Count 19 (+ 1)
- subtract the amount of excess reactant that reacted from the initial amount of excess reactant given.  \newline % Row Count 22 (+ 3)
Example: \newline % Row Count 23 (+ 1)
Hydrogen gas and oxygen gas to form water \newline % Row Count 24 (+ 1)
balance equation: \newline % Row Count 25 (+ 1)
\seqsplit{2H2​(g)+O2​(g)→2H2​O(g)} \newline % Row Count 26 (+ 1)
- Suppose we have 10.0 grams of hydrogen gas and 20.0 grams of oxygen gas. \newline % Row Count 28 (+ 2)
Determine limiting reactant \newline % Row Count 29 (+ 1)
calculate moles of each reactant \newline % Row Count 30 (+ 1)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Excess reactant (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{- H2 molar mass = 2.02g/mol \newline % Row Count 1 (+ 1)
- O2 molar mass = 32 g/mol \newline % Row Count 2 (+ 1)
divide by the respective grams amount \newline % Row Count 3 (+ 1)
- H2 - 10.0g / 2.02g/mol = 4.95 \newline % Row Count 4 (+ 1)
- O2 - 20.9g / 32g/mol = 0.625 \newline % Row Count 5 (+ 1)
O2 has fewer moles, so its the limiting reactant, while H2 is the excess \newline % Row Count 7 (+ 2)
Calculate theoretical yield \newline % Row Count 8 (+ 1)
Balanced equation shows us that 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O. \newline % Row Count 10 (+ 2)
O2 is limiting, so the amount of H2O produced is determined by its quantity \newline % Row Count 12 (+ 2)
- the theoretical yield of H2O is 2 x 0.625 mol = 1.25 mol \newline % Row Count 14 (+ 2)
Since all the O2 is consumed, we don't need to calculate the amount of excess reactant (H2) that reacts. \newline % Row Count 17 (+ 3)
Since all O2 was consumed, and the initial moles of H2 was 4.95, all moles of H2 will react, leaving no excess H2 after reaction. \newline % Row Count 20 (+ 3)
{\bf{7g ammonia reacted with 10 g of oxygen, which reagent is in excess?}} \newline % Row Count 22 (+ 2)
Determine number of moles of each reactant \newline % Row Count 23 (+ 1)
N = mass (gs)/molar mass \newline % Row Count 24 (+ 1)
NH3/ammonia \newline % Row Count 25 (+ 1)
N(NH3) = 7g/17.034g/mol \newline % Row Count 26 (+ 1)
For O2/oxygen  \newline % Row Count 27 (+ 1)
N(O2) – 10g/32g/mol \newline % Row Count 28 (+ 1)
Compare the number of moles to find the limiting reagent \newline % Row Count 30 (+ 2)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Excess reactant (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Use the stoichiometric ratio 4:5 moles of NH3 react with 1 mole of O2 \newline % Row Count 2 (+ 2)
N(NH3) = 4/5 x n(O2) \newline % Row Count 3 (+ 1)
Calculate the number of moles of NH3 using the same ratio \newline % Row Count 5 (+ 2)
Identify the limiting reagent by comparing the calculated moles of NH3 and O2 \newline % Row Count 7 (+ 2)
If n(NH3) \textless{} n(O2), NH3 is limiting, and vice versa \newline % Row Count 9 (+ 2)
O2 is the limiting reagent, bc n(O2) = 0.411 moles, which is less than n(NH3) = 0.25 moles \newline % Row Count 11 (+ 2)
Calculate the mass of the product based o the limiting reagent \newline % Row Count 13 (+ 2)
Use the stochiometry to find the number of moles of the product NO, then use the molar mass of NO to find the mass of the product \newline % Row Count 16 (+ 3)
Mass(NO) = n(NO) X molar mass (NO) \newline % Row Count 17 (+ 1)
0.25 moles x 30g/mol \newline % Row Count 18 (+ 1)
= 7.5g, round to 8g% Row Count 19 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Moles of substance A to B}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Write balanced equation and identify given and unknown quantities \newline % Row Count 2 (+ 2)
- Given: number of moles of A you have \newline % Row Count 3 (+ 1)
- Unknown: The substance B you want to convert to \newline % Row Count 4 (+ 1)
Use molar ratios \newline % Row Count 5 (+ 1)
- From the balanced equation, identify the molar ratios between A and B \newline % Row Count 7 (+ 2)
- these ratios are determined by the coefficients in front of the substances \newline % Row Count 9 (+ 2)
- if the balanced equation is aA + bB -{}-\textgreater{} cC + dD, then the molar ratio of A to B b/a \newline % Row Count 11 (+ 2)
Use the molar ratio to calculate the number of moles of substance B that will form or react with the moles of substance A \newline % Row Count 14 (+ 3)
- Multiply the moles of A by the appropriate molar ratio \newline % Row Count 16 (+ 2)
Ensure all reactants and products are included in the stoichiometric calculation: \newline % Row Count 18 (+ 2)
{\bf{- Limiting and excess reactants: }} \newline % Row Count 19 (+ 1)
limiting: reactant completely consumed in reaction, limiting amount of product that can be formed \newline % Row Count 21 (+ 2)
excess: reactant not completely consumed and is left over after limiting reactant is consumed fully \newline % Row Count 23 (+ 2)
{\bf{- Theoretical yield}} \newline % Row Count 24 (+ 1)
Max. amount of product that can be obtained from given amount of reactants assuming all reactants are converted to products according to the stoichiometry of the balanced equation \newline % Row Count 28 (+ 4)
{\bf{-compare actual and theoretical yields}} \newline % Row Count 29 (+ 1)
actual yield: amount of product obtained from reaction in practice \newline % Row Count 31 (+ 2)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Moles of substance A to B (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{if its less than the theoretical yield, it indicates one or more reactants were limiting and the reaction didnt proceed to completion. \newline % Row Count 3 (+ 3)
if any reactant is limiting or in excess, adjust the stoichiometric calculations: \newline % Row Count 5 (+ 2)
limiting: calculate amount of product formed based on its quantity \newline % Row Count 7 (+ 2)
excess: determine amount left over after reaction is complete \newline % Row Count 9 (+ 2)
To conclude, indicate the number of B moles formed or reacted based n the given A moles% Row Count 11 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{grams to atoms}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{find molar mass and calculate number of moles by dividing number of grams by molar mass \newline % Row Count 2 (+ 2)
multiply the number of moles by avogrado's number \newline % Row Count 3 (+ 1)
Example: \newline % Row Count 4 (+ 1)
10 grams of Na to atoms \newline % Row Count 5 (+ 1)
molar mass of Na is approx. 22.99 g/mol \newline % Row Count 6 (+ 1)
divide 10 grams by molar mass \newline % Row Count 7 (+ 1)
= 10/22.99 \newline % Row Count 8 (+ 1)
= 0.435 moles \newline % Row Count 9 (+ 1)
Multiply number of moles by Avogrado's number \newline % Row Count 10 (+ 1)
0.434 x (6.022x10\textasciicircum{}23\textasciicircum{}) \newline % Row Count 11 (+ 1)
= 2.6 x 10\textasciicircum{}23\textasciicircum{} atoms in 10 grams of Na% Row Count 12 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Number of protons, neutrons and electrons}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Number of protons = atomic number \newline % Row Count 1 (+ 1)
Number of neutrons = mass number - atomic number \newline % Row Count 2 (+ 1)
Number of electrons = atomic number - the charge \newline % Row Count 3 (+ 1)
Example: \newline % Row Count 4 (+ 1)
Carbon - 12 has an atomic number of 6 and a mass number of 12 \newline % Row Count 6 (+ 2)
number of protons= 6 \newline % Row Count 7 (+ 1)
number of neutrons= 12-6 = 6 \newline % Row Count 8 (+ 1)
number of electrons = 6 - 0 (neutral atom) = 6 \newline % Row Count 9 (+ 1)
if it is an isotope/has a charge: \newline % Row Count 10 (+ 1)
Carbon - 13 \newline % Row Count 11 (+ 1)
number of protons = 6 \newline % Row Count 12 (+ 1)
number of neutrons = 13 - 6 = 7 \newline % Row Count 13 (+ 1)
number of electrons = 6 - +1 = 5 (C-13 has a +1 charge, so it loses 1 electron)% Row Count 15 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Naming ionic main metal compounds}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Some ionic compounds containing a transition metal require Roman numerals while some don't. The way we determine this is whether or not that transition metal can exhibit multiple oxidation states. \newline % Row Count 4 (+ 4)
Elements in groups 1, 2 and 13 commonly have 1 oxidation state: \newline % Row Count 6 (+ 2)
- Sodium, Na, is usually Na\textasciicircum{}+\textasciicircum{} with a +1 charge, calcium, Ca, is usually Ca\textasciicircum{}2+\textasciicircum{} \newline % Row Count 8 (+ 2)
- They don't require the use of roman numerals \newline % Row Count 9 (+ 1)
For these kinds of ionic compounds, we simply use the same naming system as the other ionic compounds.% Row Count 12 (+ 3)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Ionic compounds chemical formula}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Determine if the elements in the compound typically form ions and which charge it forms. \newline % Row Count 2 (+ 2)
Then, balance the charges \newline % Row Count 3 (+ 1)
- since ionic compounds are electrically negative, the total positive charge from the cations must balance the total negative charge from the anions. To achieve this, you may need to adjust the number of ions present in the compound. \newline % Row Count 8 (+ 5)
when writing the formula, simplify the subscripts by dividing them by their greatest common number, but don't change the ratio between them. \newline % Row Count 11 (+ 3)
if there is more than one possible ionization state, use Roman numerals. \newline % Row Count 13 (+ 2)
Example:  \newline % Row Count 14 (+ 1)
Sodium chloride (NaCl) \newline % Row Count 15 (+ 1)
- Sodium, Na, has a +1 charge = Na\textasciicircum{}+\textasciicircum{} \newline % Row Count 16 (+ 1)
- Chlorine, Cl, has a -1 charge = Cl\textasciicircum{}-\textasciicircum{} \newline % Row Count 17 (+ 1)
- Since they have equal opposite charges, no need for adjustment/balancing \newline % Row Count 19 (+ 2)
Aluminium chloride \newline % Row Count 20 (+ 1)
- Al has a charge of \textasciicircum{}+3\textasciicircum{}, while chlorine has a charge of \textasciicircum{}-1\textasciicircum{} \newline % Row Count 22 (+ 2)
- So, we just need to swap the subscripts with the number of the other element present:  \newline % Row Count 24 (+ 2)
- Al\textasciicircum{}+3\textasciicircum{} + Cl\textasciicircum{}-1\textasciicircum{} -{}-\textgreater{} Al1 Cl3 (we can omit the 1 from Al coz its already a given) \newline % Row Count 26 (+ 2)
- Now, the total Al charge is +3, and the total Cl charge is -3, making them balanced \newline % Row Count 28 (+ 2)
Aluminum chloride = AlCl3% Row Count 29 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Grams to molecules}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Find total molar mass of all elements in substance. \newline % Row Count 2 (+ 2)
Use the converstion factor Avogadro's number - 6.022 x 10\textasciicircum{}23\textasciicircum{} molecules/mol \newline % Row Count 4 (+ 2)
Calculate number of moles \newline % Row Count 5 (+ 1)
- divide number of grams by molar mass \newline % Row Count 6 (+ 1)
Convert moles to molecules by multiplying the number of moles by Avogadro's number \newline % Row Count 8 (+ 2)
{\bf{10 g H2O to molecules}} \newline % Row Count 9 (+ 1)
Molar mass = 18.016g/mol \newline % Row Count 10 (+ 1)
Divide 10 grams by molar mass \newline % Row Count 11 (+ 1)
=10 grams/18.016g/mol = 0.555 \newline % Row Count 12 (+ 1)
Convert to molecules \newline % Row Count 13 (+ 1)
0.555 x (6.022 x 10\textasciicircum{}23\textasciicircum{}) \newline % Row Count 14 (+ 1)
= 3.34 x 10\textasciicircum{}23\textasciicircum{} molecules of H2O% Row Count 15 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Naming transition metals}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{For binary compounds (those with only two elements), the naming convention follows a specific set of rules. \newline % Row Count 3 (+ 3)
However, compounds involving a metal and a non-metal, like iron chloride, the non-metal is typically ended with an 'ide'. \newline % Row Count 6 (+ 3)
Non-metals typically form a negatively charged ion (anion), which are named with an 'ide'. \newline % Row Count 8 (+ 2)
This is why both FeCl2 and FeCl3 are referred to as "chloride." \newline % Row Count 10 (+ 2)
Chlorine forms the chloride ion when it gains an electron, becoming C\textasciicircum{}l\textasciicircum{}-. \newline % Row Count 12 (+ 2)
In the case of iron chloride, the 'ide' ending used in both Fecl2 and Fecl3 doesn't refer to the number of chlorine present, but the nature of the charge formed by the chlorine, which is an anion \newline % Row Count 16 (+ 4)
The use of 'ate' and 'ite' are typically used when the non-metal in a compound is oxygen \newline % Row Count 18 (+ 2)
- Chlorate (ClO\textasciicircum{}3\textasciicircum{}-) and chlorite (ClO\textasciicircum{}2\textasciicircum{}-) have suffixes indicating the number of oxygen present/the specific arrangement of atoms around oxygen \newline % Row Count 21 (+ 3)
This naming convention doesn't directly apply to compounds involving chlorine and other elements like iron. \newline % Row Count 24 (+ 3)
Instead, they are named according to their oxidation states. When an element can have multiple oxidation states, it's indicated using Roman numerals. \newline % Row Count 27 (+ 3)
So, to name transition metals, you must first figure out the oxidation state of the compound, and indicate this using roman numerals \newline % Row Count 30 (+ 3)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Naming transition metals (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Example: \newline % Row Count 1 (+ 1)
FeCl\textasciicircum{}2\textasciicircum{} \newline % Row Count 2 (+ 1)
- The iron here has a 2+ oxidation state, meaning it's lost 2 electrons \newline % Row Count 4 (+ 2)
- Therefore, it's named Iron (ii) Chloride \newline % Row Count 5 (+ 1)
FeCl\textasciicircum{}3\textasciicircum{} \newline % Row Count 6 (+ 1)
- In this case, iron has lost 3 electrons making it +3 \newline % Row Count 8 (+ 2)
- So it's named Iron (iii) chloride \newline % Row Count 9 (+ 1)
Another example CuS - Copper sulfide \newline % Row Count 10 (+ 1)
To figure out if it's i-iii, we figure out the charge of copper by first figuring out the charge of sulfur \newline % Row Count 13 (+ 3)
- Sulfur is a chalogen found in group 6a, and they form \textasciicircum{}2-\textasciicircum{} anions \newline % Row Count 15 (+ 2)
- Copper can form either +1 or +2 ions, so in this case it would have to be a +2 charge to balance out the charge of sulfur. \newline % Row Count 18 (+ 3)
- {\emph{if the charge of copper was +1, then 2 copper atoms would be needed to balance the compound, but given that there is only 1 in CuS, we can determine that the charge would be +2}} \newline % Row Count 22 (+ 4)
Since Cus is a neutral compound, the total positive charge of copper must balance the total negative charge of sulfur by determining coppers charge, which we already determined was -2. since the compound is neutral, the total sum of charges must = 0 \newline % Row Count 27 (+ 5)
We denote the charge of copper as 'x' \newline % Row Count 28 (+ 1)
- X+(-2) = 0x \newline % Row Count 29 (+ 1)
    We denote the charge of copper as "x." Since the compound is neutral, the sum of the charges of copper and sulfur must equal zero: \newline % Row Count 32 (+ 3)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Naming transition metals (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{- x + (−2) = 0 \newline % Row Count 1 (+ 1)
- solving for x, we find that x = +2, meaning copper must have a +2 charge \newline % Row Count 3 (+ 2)
Therefore, in naming CuS, we put the number of the charge in roman numerals \newline % Row Count 5 (+ 2)
- CuS = Copper(ii) sulfide. \newline % Row Count 6 (+ 1)
In the case of Cu2S, we'd follow the same steps \newline % Row Count 7 (+ 1)
- S has a charge of -2, we must balance the copper \newline % Row Count 9 (+ 2)
- 2x + (-2) = 0 \newline % Row Count 10 (+ 1)
- solving for x, we find x = +1 because 1 x +1 + (-2) = 0 \newline % Row Count 12 (+ 2)
Therefore, copper has a +1 charge \newline % Row Count 13 (+ 1)
so Cu2S = Copper (i) sulfide% Row Count 14 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Grams to moles}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Find the atomic mass of each element and multiply it by the number of that element in 1 molecule of the substance. \newline % Row Count 3 (+ 3)
Add the masses of all the atoms to find the molar mass. \newline % Row Count 5 (+ 2)
Multiply the number of grams by the reciprocal molar mass of the substance. \newline % Row Count 7 (+ 2)
Example: \newline % Row Count 8 (+ 1)
50 grams of H2O to moles \newline % Row Count 9 (+ 1)
- Molar mass of H2O is sum of the masses of H and O \newline % Row Count 11 (+ 2)
- (H) 1.008g/mol x 2 + (O) 16g/mol \newline % Row Count 12 (+ 1)
= 18.016 \newline % Row Count 13 (+ 1)
    Determine the substance: We want to convert grams of water (H2O) to moles. \newline % Row Count 15 (+ 2)
- The reciprocal of the molar mass of H2O is 1/18.016g/mol \newline % Row Count 17 (+ 2)
- multiply 50 grams by 1/18.016g/mol \newline % Row Count 18 (+ 1)
= 2.78 moles% Row Count 19 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Find amount of mols in gs of substance}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Find molar mass of substance and divide it by the amount of grams \newline % Row Count 2 (+ 2)
Example: \newline % Row Count 3 (+ 1)
Moles in 1 gram H2O \newline % Row Count 4 (+ 1)
- molar mass = 18.016g.mol \newline % Row Count 5 (+ 1)
- Divide by 1  \newline % Row Count 6 (+ 1)
- 1/18.016 = 0.0555 \newline % Row Count 7 (+ 1)
Moles in 2 grams H2O \newline % Row Count 8 (+ 1)
- multiply reciprocal of H2O by 2 \newline % Row Count 9 (+ 1)
= 2 x (1/18.016) \newline % Row Count 10 (+ 1)
= 0.1109 moles in 2 grams of H2O% Row Count 11 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Naming polyatomc ions}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Polyatomic ions are ions that contain more than 1 atom, monatomic ions only contain 1 atom \newline % Row Count 2 (+ 2)
NO3-, NO4- = polyatomic \newline % Row Count 3 (+ 1)
N3- = monatomic \newline % Row Count 4 (+ 1)
Suffixes: \newline % Row Count 5 (+ 1)
'ate' or 'ite' - typically polyatomic ion that contains 1 oxygen atom \newline % Row Count 7 (+ 2)
'ide' - typically monoatomic ion that lacks oxygen \newline % Row Count 9 (+ 2)
so, N3- is Nitr{\bf{ide}} \newline % Row Count 10 (+ 1)
and NO3- is Nitr{\bf{ate}} \newline % Row Count 11 (+ 1)
Example: \newline % Row Count 12 (+ 1)
Cl- Chloride (no oxygen - ide) \newline % Row Count 13 (+ 1)
ClO- Hypochloride (1 more oxygen - 'hypo' prefix) \newline % Row Count 14 (+ 1)
ClO2- Chlorite (1 more oxygen - ide) \newline % Row Count 15 (+ 1)
ClO3- Chlorate (1 more oxygen - ate) \newline % Row Count 16 (+ 1)
ClO4- Perchlorate (1 more oxygen - 'per' prefix)% Row Count 17 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Dilution p2}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{{\bf{Water needed to dilute 100ml of 1.0 M solution of Nacl to a 0.25 solution}} \newline % Row Count 2 (+ 2)
Initial volume and concentration \newline % Row Count 3 (+ 1)
100ml and 1.0 M respectively \newline % Row Count 4 (+ 1)
Final concentration = 0.25 M (as desired) \newline % Row Count 5 (+ 1)
Calculate amount of solute (initial) \newline % Row Count 6 (+ 1)
Initial volume x initial concentration \newline % Row Count 7 (+ 1)
= 100ml x 1.0M = 100 moles \newline % Row Count 8 (+ 1)
Calculate final volume needed \newline % Row Count 9 (+ 1)
- we want to dilute solution to a final concentration of 0.25 M, so we use this formula \newline % Row Count 11 (+ 2)
Final concentration = amount of solute (initial)/final volume \newline % Row Count 13 (+ 2)
rearrange formula \newline % Row Count 14 (+ 1)
Final volume = amount of solute (initial)/final concentration \newline % Row Count 16 (+ 2)
= 100 moles/0.25 = 400ml \newline % Row Count 17 (+ 1)
To get the amount of water needed, we calculate the difference btwn the final volume and the initial volume \newline % Row Count 20 (+ 3)
Final volume - initial volume \newline % Row Count 21 (+ 1)
= 400ml - 100ml = 300ml \newline % Row Count 22 (+ 1)
So you add 300ml of water% Row Count 23 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Covalent/molecular compound chemical formula}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{{\bf{Sulfur Dioxide}} \newline % Row Count 1 (+ 1)
No prefix in front of 'sulfur' means there is only one in the compound \newline % Row Count 3 (+ 2)
'di' prefix infront of 'oxygen' means there are 2 in the compound \newline % Row Count 5 (+ 2)
So, the formula would be SO2 \newline % Row Count 6 (+ 1)
{\bf{Dinitrogen pentoxide}} \newline % Row Count 7 (+ 1)
'di' in front of nitrogen means there are two in the compound \newline % Row Count 9 (+ 2)
'penta' in front of oxygen means there are 5 in the compound \newline % Row Count 11 (+ 2)
So, the formula would be N2O5% Row Count 12 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Grams to make solution}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Number of moles can be calculated using the formula \newline % Row Count 2 (+ 2)
n = c x v \newline % Row Count 3 (+ 1)
c - concentration in mol/L, V - volume in L \newline % Row Count 4 (+ 1)
mass = n x molar mass \newline % Row Count 5 (+ 1)
How many grams of solid Mg(NO3)2 are required to make 2.5 L of a 1.5 M Mg(NO3)2 solution? \newline % Row Count 7 (+ 2)
V = 2.5, the C = 1.5 mol/L, we want to find the mass of solid Mg(NO3)2 needed to make this solution \newline % Row Count 9 (+ 2)
N = 1.5mol/L x 2.5L \newline % Row Count 10 (+ 1)
= 3.75 moles \newline % Row Count 11 (+ 1)
Now calculate the mass \newline % Row Count 12 (+ 1)
3.75 moles x 148.31 g/mol \newline % Row Count 13 (+ 1)
= 556.1625 grams, rounded to 556 grams% Row Count 14 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Naming: ionic compounds}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{A compound is a substance with more than 1 element \newline % Row Count 2 (+ 2)
NaCl is Sodium chloride - two different elements present, making it a compound \newline % Row Count 4 (+ 2)
An ionic compound is one that is composed of ions \newline % Row Count 5 (+ 1)
NaCl is composed of Na+ and Cl-, an anion and a cation respectively \newline % Row Count 7 (+ 2)
When naming an anionic compound, the ending of the last element is changed to 'ide' \newline % Row Count 9 (+ 2)
NaCl - Sodium + chlorine = sodium chlor{\emph{ide}} \newline % Row Count 10 (+ 1)
Name the first element, and end the second element in 'ide' \newline % Row Count 12 (+ 2)
Example: \newline % Row Count 13 (+ 1)
AlP contains Aluminium and phosphorous = Aluminium phosphide \newline % Row Count 15 (+ 2)
Polyatomic ionic compounds follow the same rules, look at the section for them% Row Count 17 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Atoms in grams}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Calculate molar mass of compounds then calculate the number of moles \newline % Row Count 2 (+ 2)
n = mass/molar mass \newline % Row Count 3 (+ 1)
multiply with avo number \newline % Row Count 4 (+ 1)
sodium atoms in 1kg of Na2SO4 \newline % Row Count 5 (+ 1)
molar mass Na = 22.99 \newline % Row Count 6 (+ 1)
S = 32.07 \newline % Row Count 7 (+ 1)
O = 16 \newline % Row Count 8 (+ 1)
2 x 22.99 + 32.07 + 4 x 16 \newline % Row Count 9 (+ 1)
= 141.05 g/mol \newline % Row Count 10 (+ 1)
Moles of Na2SO4 \newline % Row Count 11 (+ 1)
= mass/molar mass = 1000/141.04 \newline % Row Count 12 (+ 1)
= 7.09 \newline % Row Count 13 (+ 1)
since there are 2 moles of Na for every 1 mole of Na2SO4, Na atoms are doubled \newline % Row Count 15 (+ 2)
2 x 7,09 = 14.18 \newline % Row Count 16 (+ 1)
14.18 x 6.022 x 10\textasciicircum{}23\textasciicircum{} \newline % Row Count 17 (+ 1)
Na atoms = 8.53 x 10\textasciicircum{}24\textasciicircum{}% Row Count 18 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Amount of molecules in gs}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Divide mass (g) by the molar mass of the molecule, then multiply it by avos number (6.022 x 10\textasciicircum{}23\textasciicircum{}) \newline % Row Count 2 (+ 2)
Oxygen molecules in 6 grams of oxygen \newline % Row Count 3 (+ 1)
n = mass (g)/molar mass (g/mol\textasciicircum{}-1\textasciicircum{}) \newline % Row Count 4 (+ 1)
molar mass of O2 = 16 x 2 = 32 \newline % Row Count 5 (+ 1)
6/32 = 0.187 \newline % Row Count 6 (+ 1)
0.187 x 6.022 x 10\textasciicircum{}23\textasciicircum{} molecules/mol% Row Count 7 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Product formed in reaction}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Balance the equation, and determine the number of moles for each reactant \newline % Row Count 2 (+ 2)
N = mass of reactant/molar mass \newline % Row Count 3 (+ 1)
Barium peroxide reacts with hydrochloric acid to form peroxide, H2O2:  \newline % Row Count 5 (+ 2)
BaO2(s) + 2HCl(aq) -{}-\textgreater{} H2O2(aq) + BaCl2(aq).  \newline % Row Count 6 (+ 1)
If 2.01 g of barium peroxide is reacted with 0.75 g of acid (HCl) how much peroxide will be produced? \newline % Row Count 9 (+ 3)
Determine moles for each reactant \newline % Row Count 10 (+ 1)
BaO2 barium peroxide \newline % Row Count 11 (+ 1)
-	N(BaO2) = 2.01g/169.3 g/mol \newline % Row Count 12 (+ 1)
Hydrochloric acid HCL \newline % Row Count 13 (+ 1)
-	N(HCl) = 0.75g/36.458 g/mol \newline % Row Count 14 (+ 1)
Compare the number of moles \newline % Row Count 15 (+ 1)
From the equation, $\frac{1}{2}$ moles of BaO2 react with 1 mole of Hcl \newline % Row Count 17 (+ 2)
N(BaO2) = $\frac{1}{2}$ x n(HCl) \newline % Row Count 18 (+ 1)
Calculate the number of moles of BaO2 using the same ratio \newline % Row Count 20 (+ 2)
Compare the moles, whichever is smaller is the limiting reagent \newline % Row Count 22 (+ 2)
Hcl = 0.0199 moles \newline % Row Count 23 (+ 1)
BaO2 = 0.0103 moles \newline % Row Count 24 (+ 1)
-	Hcl limiting reagent \newline % Row Count 25 (+ 1)
Calculate product mass based on limiting reagent/HCl using stochiometry to find number of moles \newline % Row Count 27 (+ 2)
In the equation, the stoichiometric coefficients represent the mole ratio btwn reactants and products \newline % Row Count 30 (+ 3)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Product formed in reaction (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Hcl is limiting, and 2 moles of HCl react to produce 1 mole of H2O2 \newline % Row Count 2 (+ 2)
Therefore, n(H2O2) = $\frac{1}{2}$ n(HCl) \newline % Row Count 3 (+ 1)
Molar mass of H2O2 is 34.014gmol \newline % Row Count 4 (+ 1)
To convert moles to grams, use its molar mass \newline % Row Count 5 (+ 1)
-	Number of moles x molar mass \newline % Row Count 6 (+ 1)
-	0.0103 moles x 34.014g/mol% Row Count 7 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Naming covalent/molecular prefixes}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Mono - 1 \newline % Row Count 1 (+ 1)
Die - 2 \newline % Row Count 2 (+ 1)
Tri - 3 \newline % Row Count 3 (+ 1)
Tetra - 4 \newline % Row Count 4 (+ 1)
Penta - 5 \newline % Row Count 5 (+ 1)
Hexa - 6 \newline % Row Count 6 (+ 1)
Hepta - 7 \newline % Row Count 7 (+ 1)
Octo - 8 \newline % Row Count 8 (+ 1)
Nona - 9 \newline % Row Count 9 (+ 1)
Deca - 10 \newline % Row Count 10 (+ 1)
CO = Carbon monoxide \newline % Row Count 11 (+ 1)
- 1st element has subscript of 1, but doesn't need prefix 'mono' \newline % Row Count 13 (+ 2)
- 2nd element has subscript of 1 (1 oxygen atom), so prefix 'mono' is used \newline % Row Count 15 (+ 2)
CO2 = Carbon dioxide \newline % Row Count 16 (+ 1)
- 2nd element has subscript of 2 (2 oxygen atoms), so 'di' is used \newline % Row Count 18 (+ 2)
NO2 = Nitrogen Dioxide \newline % Row Count 19 (+ 1)
N2O5 = Dinitrogen pentoxide \newline % Row Count 20 (+ 1)
- 1st element has subscript of 2, so 'die' is used \newline % Row Count 22 (+ 2)
- 2nd element has subscript of 5, so 'penta' is used% Row Count 24 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{molecular shape and molecule polarity}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{To predict molecular shape and whetehr a molecule is polar; \newline % Row Count 2 (+ 2)
identify central atom (which can form most bonds/least electronegative) \newline % Row Count 4 (+ 2)
determine the electron geometry around the central atom by considering both bonding and non-bonding electron pairs \newline % Row Count 7 (+ 3)
- NCL3, N has 1 lone pair and forms 3 single bonds with the Cl \newline % Row Count 9 (+ 2)
- this gives it a tetrahedral electron geomotry \newline % Row Count 10 (+ 1)
Determien the molecular shape by considering only the positions of bonded atoms \newline % Row Count 12 (+ 2)
- in NCl3, the lone pair of electrons on N repels the bonding pair, causing the molecule to adopt a trigonal pyramidal shape \newline % Row Count 15 (+ 3)
Determine polarity - consider the electronegativity of the atoms and the molecular shape \newline % Row Count 17 (+ 2)
- NCl3, N is less electronegative than Cl, meaning the bonds btwn N and Cl are polar \newline % Row Count 19 (+ 2)
- the chlorine atom carries a partial neg charge, and the N carries a partial pos charge \newline % Row Count 21 (+ 2)
While the arrangment of atoms in Ncl3 is symmetrical, symmetry alone doesnt determine polarity \newline % Row Count 23 (+ 2)
the distribution of electron density due to the lone pair of N still results in a net dipole moment (overall polarity of the molecule), making the molecule polar% Row Count 27 (+ 4)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Moles produced and molecules to react}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{2CO + O2 -{}-\textgreater{} 2CO2 \newline % Row Count 1 (+ 1)
{\bf{If 5.23 moles of CO react with excess oxygen, how many moles of CO2 are produced?}} \newline % Row Count 3 (+ 2)
Moles of CO = 5.23 mol \newline % Row Count 4 (+ 1)
need to find moles of CO2 produced \newline % Row Count 5 (+ 1)
From equation, 2 moles of CO react with 1 mole of O2 to make 2 moles of CO2, meaning ratio is 2:2 -{}-\textgreater{} 1:1 \newline % Row Count 8 (+ 3)
Using the given moles of CO and the ratio, calculate moles of CO2 produced \newline % Row Count 10 (+ 2)
Moles of CO2 = (given moles of CO) X moles of CO2/ moles of CO) \newline % Row Count 12 (+ 2)
= 5.23 mol x 2 mol CO2/2 mol CO \newline % Row Count 13 (+ 1)
= 5.23 x 1 \newline % Row Count 14 (+ 1)
moles of co2 = 5.23 mol \newline % Row Count 15 (+ 1)
when 5.23 moles of CO react, 5.23 moles of CO2 are produced \newline % Row Count 17 (+ 2)
{\bf{How many oxygen molecules were required to react all the CO?}} \newline % Row Count 19 (+ 2)
The equation shows us that 2 moles of CO react with 1 mole of O2 to produce 2 moles of CO2, so the ratio between CO and O2 is 2:1 \newline % Row Count 22 (+ 3)
Moles of O2 = (given moles of CO) x moles of O2/moles of CO \newline % Row Count 24 (+ 2)
= 5.23 x 1 mol O2/2 mol CO \newline % Row Count 25 (+ 1)
= 5.23/2 \newline % Row Count 26 (+ 1)
= Moles of O2 = 2.615 \newline % Row Count 27 (+ 1)
Convert to molecules \newline % Row Count 28 (+ 1)
1 mole of substance contains 6.022 x 10\textasciicircum{}23\textasciicircum{} molecules, so we multiply 2.615 by this number \newline % Row Count 30 (+ 2)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Moles produced and molecules to react (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{= 1.572 x 10\textasciicircum{}24\textasciicircum{} \newline % Row Count 1 (+ 1)
exponant increases by 1 (23-24) bc when we convert moles to molecules, we are multiplying by avogrados number - 6.022 x 10\textasciicircum{}23\textasciicircum{} molecules per mole. \newline % Row Count 4 (+ 3)
if we have 2 moles of a substance, we have 2 x 6.022 x 10\textasciicircum{}23\textasciicircum{} molecules, which is 1.2044 x 10\textasciicircum{}24\textasciicircum{} molecules \newline % Row Count 7 (+ 3)
- each additional mole increases the number of molecules in Avo's number \newline % Row Count 9 (+ 2)
when we increase the number of moles by 1, the exponant of 10 increases by 1 in the scientific notation represeantation of the number of molecules% Row Count 12 (+ 3)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Electronic configuration}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Understand the subshells \newline % Row Count 1 (+ 1)
Electronic configuration describes the distribution of electrons in the atomic orbitals of an atom \newline % Row Count 3 (+ 2)
each subshell is labelled with the principal quantum number (n) and the orbital type (s,p,d,f) \newline % Row Count 5 (+ 2)
1s2, 2s2, 2p4 \newline % Row Count 6 (+ 1)
- n is 1 for the 1s subshell, 2 is for the 2s and 2p subshell \newline % Row Count 8 (+ 2)
- the value of n represents the energy level of the orbital \newline % Row Count 10 (+ 2)
the superscript after each subshell represents the number of electrons in that subshell \newline % Row Count 12 (+ 2)
- 1s2, 1s subshell is filled with 2 electrons \newline % Row Count 13 (+ 1)
- 2s2, 2s subshell is filled with 2 electrons \newline % Row Count 14 (+ 1)
- 2p4, 2p subshell filled with 4 electrns \newline % Row Count 15 (+ 1)
To determine the element that corresponds to the electron configuration, you use the periodic table \newline % Row Count 17 (+ 2)
findthe element with the atomic number that matches the sum of the superscripts in the electron configuration \newline % Row Count 20 (+ 3)
{\bf{Elements and ions in their ground state}} \newline % Row Count 21 (+ 1)
identify the atomic number (z) of an element \newline % Row Count 22 (+ 1)
- carbon has a z of 6 - 6 protons and 6 electrons in its neutral state \newline % Row Count 24 (+ 2)
determine the shell occupied by the valence electrons \newline % Row Count 26 (+ 2)
- C has an atomic number of 6, so we fill the electrons into the available order of increasing energy \newline % Row Count 29 (+ 3)
use the Aufbau principle \newline % Row Count 30 (+ 1)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Electronic configuration (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{- electrons fill the lowest energy orbitals before moving to the higehr energy ones \newline % Row Count 2 (+ 2)
- fill the 1s, then the 2s, and then the 2p orbitals \newline % Row Count 4 (+ 2)
For carbon (z=6), the electronic configuration is 1s2 2s2 2sp2 \newline % Row Count 6 (+ 2)
- 2 electrons in 2s orbital, 2 electrons in the 2s orbital and 2 in the 2p orbital% Row Count 8 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Moles}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{1 mole = 6.022 x 10\textasciicircum{}23\textasciicircum{} \newline % Row Count 1 (+ 1)
1 mol of carbon atoms = 6.022 x 10\textasciicircum{}23\textasciicircum{} atoms of Carbon \newline % Row Count 3 (+ 2)
1 mol of CO2 = 6.022 x 10\textasciicircum{}23\textasciicircum{} \newline % Row Count 4 (+ 1)
2 mol of carbon = 2 x 6.022 x 10\textasciicircum{}23\textasciicircum{} \newline % Row Count 5 (+ 1)
4 mol of C \newline % Row Count 6 (+ 1)
4 mol C/1 x 6x10\textasciicircum{}23\textasciicircum{} C atoms/1 mol C \newline % Row Count 7 (+ 1)
= 4 x 6 = 24 \newline % Row Count 8 (+ 1)
= 24 x 10\textasciicircum{}23\textasciicircum{} \newline % Row Count 9 (+ 1)
Move the decimals to the left \textless{}-{}- = \textasciicircum{}23\textasciicircum{} goes up by as many places as you moved to the left \newline % Row Count 11 (+ 2)
Move decimals to the right -{}-\textgreater{} it goes down% Row Count 12 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Moles}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{The mass number of an element also represents the 'molar mass' \newline % Row Count 2 (+ 2)
- 1 mol of an element has a mass of (its mass number) \newline % Row Count 4 (+ 2)
- Nitrogen's mass number is 14 = 1 mol of N has a mass of 14g; 14g of N contains 6.022 x 10\textasciicircum{}23\textasciicircum{} atoms \newline % Row Count 7 (+ 3)
- Mole is proportional to it's 'molar mass'  \newline % Row Count 8 (+ 1)
- 2 mol of N = 28g N \newline % Row Count 9 (+ 1)
Figuring out the molar mass of a compound can be done by identifying the molar mass of each element present and adding them together \newline % Row Count 12 (+ 3)
{\bf{O3}} \newline % Row Count 13 (+ 1)
- 1 oxygen atom has an mass number of 16, so 6 of them would be 3x16 = 48 \newline % Row Count 15 (+ 2)
- Molar mass of ozone/O3 = 48g/mol \newline % Row Count 16 (+ 1)
{\bf{CO2}} \newline % Row Count 17 (+ 1)
- Carbon mass number is 12.01, Oxygen mass number is 16, so 16x2 = 32 \newline % Row Count 19 (+ 2)
- 32 + 12.01 = 44.01g/mol \newline % Row Count 20 (+ 1)
{\bf{Calcium phosphate}} \newline % Row Count 21 (+ 1)
- 3 calcium atoms, 2 phosphate groups with 1 P atom and 4 oxygen atoms \newline % Row Count 23 (+ 2)
- first, balance the formula \newline % Row Count 24 (+ 1)
- Calcium has 3 atoms \newline % Row Count 25 (+ 1)
- 1 P atom in each phosphate group, since there are 2 groups, that's 2 P in total \newline % Row Count 27 (+ 2)
- Each group has 4 O atoms, so 4x2 = 8 O atoms in total \newline % Row Count 29 (+ 2)
= Ca3(PO4)2 \newline % Row Count 30 (+ 1)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Moles (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{- Molar mass of C = 40.08 x 3 = 80.16 \newline % Row Count 1 (+ 1)
- Molar mass of P = 30.97 x 2 = 61.94 \newline % Row Count 2 (+ 1)
- Molar mass of O = 16 x 4 = 64 \newline % Row Count 3 (+ 1)
Then you add them together \newline % Row Count 4 (+ 1)
Molar mass of Calcium phosphate/Ca3(PO4)2 = 310.18g/mol% Row Count 6 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{oxidation state of an atom in a compound}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{{\bf{General rules}} \newline % Row Count 1 (+ 1)
- Oxidation state of an atom in its elemental form is always 0 \newline % Row Count 3 (+ 2)
- For monatomic ions, the oxidation state is equal to the charge of the ion \newline % Row Count 5 (+ 2)
- The sum of the oxidation state of all atoms in a neutral compound is 0, and it equals the charge of the compound if it's an ion \newline % Row Count 8 (+ 3)
- In compounds, some elements have fixed oxidation states (Group 1 metals always have an oxidation state of +1, group 2 metals always have an oxidation state of +2, oxygen s usually -2) \newline % Row Count 12 (+ 4)
Start with the elements that have a fixed oxidation state/are in elemental form, and assign their states based on rules above \newline % Row Count 15 (+ 3)
If an element has variable oxidation states, use these rules \newline % Row Count 17 (+ 2)
- Assign the oxidation state of oxygen as -2, unless it's in a peroxide or when combines with fluroine, where it has a positive state \newline % Row Count 20 (+ 3)
- Hydrogen usually has a +1 state, except when bonded with metals where it's -1 \newline % Row Count 22 (+ 2)
- Group 1 metals are +1, group 2 metals are +2, group 13 are +3 in compounds \newline % Row Count 24 (+ 2)
- in compounds, flourine is always -1 \newline % Row Count 25 (+ 1)
- the sum of oxidation states in neautral compounds is 0 \newline % Row Count 27 (+ 2)
After assigning states to each atom, check that the sum of the states is equal to the total charge of the compound or ion, and if its neautral it should be 0 \newline % Row Count 31 (+ 4)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{oxidation state of an atom in a compound (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{{\bf{Example}} \newline % Row Count 1 (+ 1)
NH3 \newline % Row Count 2 (+ 1)
H usually is +1, O is usually -2 \newline % Row Count 3 (+ 1)
The sum of the oxidation states should equal 0 as its neutral \newline % Row Count 5 (+ 2)
H is typically +1, so all 3 atoms in NH3 will equal a +3 charge. \newline % Row Count 7 (+ 2)
NH3 is neutral, so the sum of oxidation states must equal 0, so Nitrogen must have a state that offsets the total charge from the H atoms - +3 \newline % Row Count 10 (+ 3)
N's state can be calcualted by substracting the sum of oxidation states of H from 0 \newline % Row Count 12 (+ 2)
Oxidation state of N is x \newline % Row Count 13 (+ 1)
- x + 3(+1) = 0 \newline % Row Count 14 (+ 1)
- x + 3 = 0 \newline % Row Count 15 (+ 1)
- x = -3 \newline % Row Count 16 (+ 1)
Nitrogen's oxidation state therefore is -3 \newline % Row Count 17 (+ 1)
{\bf{example}} \newline % Row Count 18 (+ 1)
Nitrate ion has a -1 charge, and O has a -2 charge \newline % Row Count 20 (+ 2)
lets say the nitrogen state is x \newline % Row Count 21 (+ 1)
- 3 O atoms, each -2 - 3(-2) = -6 \newline % Row Count 22 (+ 1)
The sum of oxidation states must equal the charge of the ion, -1, so nitrogen must have an oxidation state that offsets the total charge of the oxygen atoms -6. \newline % Row Count 26 (+ 4)
x + -6 = -1 \newline % Row Count 27 (+ 1)
x = -1 + 6 \newline % Row Count 28 (+ 1)
x = +5 \newline % Row Count 29 (+ 1)
so the oxidation state of NH3 is -1, and the nitrate ion has a state of +5, while the O atom has a state of -2% Row Count 32 (+ 3)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Hydrated ionic compound empirical formula}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Identify the ionic compound and the number of water molecules in it \newline % Row Count 2 (+ 2)
Determine the masses of each component separately, including the mass of the andhydrous salt (w/o water) and the mass of the water molecules. \newline % Row Count 5 (+ 3)
These can be find from the given total mass of the compound. \newline % Row Count 7 (+ 2)
Calculate the molar mass \newline % Row Count 8 (+ 1)
Determine molar mass of the andhydrous salt by summing the molar masses of each element in the compound \newline % Row Count 11 (+ 3)
To find the molar mass of the anhydrous salt, sum each of the molar masses in the compound \newline % Row Count 13 (+ 2)
- molar mass of compound = molar mass of each atom added together  \newline % Row Count 15 (+ 2)
Then, determine the molar mass of H2O \newline % Row Count 16 (+ 1)
Calculate the moles \newline % Row Count 17 (+ 1)
Use the masses and the molar masses to find the number of moles in each component \newline % Row Count 19 (+ 2)
For the anhydrous salt, use this formula: \newline % Row Count 20 (+ 1)
- Moles of A.salt = mass of A.salt/molar mass of A. salt \newline % Row Count 22 (+ 2)
for the water: \newline % Row Count 23 (+ 1)
- Moles of water = mass of water/molar mass of water \newline % Row Count 25 (+ 2)
Determine the simplest ratio \newline % Row Count 26 (+ 1)
divide the number of moles of each component by the smallest number of moles calculated, to get the simplest ratio of ions to water molecules \newline % Row Count 29 (+ 3)
Round to whole numbers if not already whole numbers \newline % Row Count 31 (+ 2)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Hydrated ionic compound empirical formula (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Write the empirical formula using the whole number ratios \newline % Row Count 2 (+ 2)
- the subscripts in the formula represent the number of ions or water molecules in one formula unit of the compound \newline % Row Count 5 (+ 3)
{\bf{Example}} \newline % Row Count 6 (+ 1)
Copper (II) sulfate pentahydrate - CuSO4 + 5H2O \newline % Row Count 7 (+ 1)
- made of copper (II) sulfate - CuSO4 - and 5 H2O/water molecules. \newline % Row Count 9 (+ 2)
Lets say we have 250 grams, to determine the mass of the A.salt (CuSO4) and the mass of water by weighing the sample \newline % Row Count 12 (+ 3)
Calculate molar masses \newline % Row Count 13 (+ 1)
CuSO4 molar mass = 159.55 g/mol \newline % Row Count 14 (+ 1)
H2O molar mass = 18.02 x 5 = 90.10 g/mol \newline % Row Count 15 (+ 1)
calculate moles \newline % Row Count 16 (+ 1)
CuSO4 = 100/159.55 = 0.627 mol \newline % Row Count 17 (+ 1)
5 H2O = 50/90.10 = 0.554 mol \newline % Row Count 18 (+ 1)
Determine the simplest ratio \newline % Row Count 19 (+ 1)
divide the number of moles of each component by the smallest number of moles calculated \newline % Row Count 21 (+ 2)
-   0.627/0.554 \newline % Row Count 22 (+ 1)
= 1.13 \newline % Row Count 23 (+ 1)
Round to nearest whole number \newline % Row Count 24 (+ 1)
= 1 \newline % Row Count 25 (+ 1)
The ratio is therefore 1:5, so the empirical formula is CuSO4 + 5H2O% Row Count 27 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Empircal formula from \% composition}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Convert \% to grams \newline % Row Count 1 (+ 1)
Start by assuming you have 100g sample of the compound, and convert the percentages of each element to grams \newline % Row Count 4 (+ 3)
- if a compound contains 40\% carbon, it means there are 40/100 x 100g =40g of carbon in 100g of the compound \newline % Row Count 7 (+ 3)
 Convert grams to moles \newline % Row Count 8 (+ 1)
use the molar mass of each element to convert to moles \newline % Row Count 10 (+ 2)
- Moles = grams/molar mass \newline % Row Count 11 (+ 1)
Determine simplest ratio \newline % Row Count 12 (+ 1)
Divide number of moles of each element by the smallest number of moles calculated \newline % Row Count 14 (+ 2)
- if the ratios obtained aren't whole numbers, then round them to the nearest whole number. \newline % Row Count 16 (+ 2)
- if the numbers are close to whole numbers, you can multiply all ratios by the same number to make them whole \newline % Row Count 19 (+ 3)
Write formula \newline % Row Count 20 (+ 1)
use the whole number ratios to write the empirical formula of the compound. \newline % Row Count 22 (+ 2)
the subscripts in the formula represent the number of atoms of each element in one molecule of the compounds \newline % Row Count 25 (+ 3)
{\bf{Example}} \newline % Row Count 26 (+ 1)
Compound with 40\% carbon, 6.7\% hydrogen, and 53.3\% oxygen by mass \newline % Row Count 28 (+ 2)
Convert \% to grams \newline % Row Count 29 (+ 1)
C = 40.0g, H= 6.7g, O= 53.3g \newline % Row Count 30 (+ 1)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Empircal formula from \% composition (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Convert G to moles \newline % Row Count 1 (+ 1)
Using the molar mass \newline % Row Count 2 (+ 1)
- moles = mass (grams)/molar mass (grams/mol) \newline % Row Count 3 (+ 1)
- C: 40.0/12.01 = 3.33 moles \newline % Row Count 4 (+ 1)
- H: 6.7/1.01 = 6.67 moles \newline % Row Count 5 (+ 1)
- O: 53.3/16 = 3.33 moles \newline % Row Count 6 (+ 1)
Determine the simplest ratio \newline % Row Count 7 (+ 1)
Divide the number of moles of each element by the smallest number of mles \newline % Row Count 9 (+ 2)
The ratio of C:H:O = 1:2:1 \newline % Row Count 10 (+ 1)
They are close enough to whole numbers, and therefore don't need to be rounded. \newline % Row Count 12 (+ 2)
The empirical formula would therefore be: \newline % Row Count 13 (+ 1)
CH2O \newline % Row Count 14 (+ 1)
{\bf{Example}} \newline % Row Count 15 (+ 1)
Analyses of a compound found it to contain, by mass: 63.68\% C, 12.38\% N, 9.80\% H and 14.14\% O. Calculate the empirical formula for this compound \newline % Row Count 18 (+ 3)
Carbon: 63.68/12.01 = 5.30  \newline % Row Count 19 (+ 1)
N: 12.38/14.01 = 0.884  \newline % Row Count 20 (+ 1)
H: 9.80/1.008 = 9.72  \newline % Row Count 21 (+ 1)
O: 14.14/16 = 0.900 \newline % Row Count 22 (+ 1)
Divide number of moles each by the smallest number, which is 0.884 from Nitrogen \newline % Row Count 24 (+ 2)
5.30 and 8.884 and 9.72 and 0.900 all divided by 0.884 \newline % Row Count 26 (+ 2)
= 6, 1, 11, 1 after simplifying  \newline % Row Count 27 (+ 1)
= empirical formula becomes C6NH11O1 \newline % Row Count 28 (+ 1)
Divide mass percentage of each element by its molar mass to find number of moles \newline % Row Count 30 (+ 2)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Empircal formula from \% composition (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Determine simplest whole-number ratio of moles by dividing each number of moles by the smallest number of moles \newline % Row Count 3 (+ 3)
Write the empirical formula \newline % Row Count 4 (+ 1)
Analyses of a compound found it to contain, by mass: 63.68\% C, 12.38\% N, 9.80\% H and 14.14\% O. Calculate the empirical formula for this compound \newline % Row Count 7 (+ 3)
Carbon: 63.68/12.01 = 5.30  \newline % Row Count 8 (+ 1)
N: 12.38/14.01 = 0.884  \newline % Row Count 9 (+ 1)
H: 9.80/1.008 = 9.72  \newline % Row Count 10 (+ 1)
O: 14.14/16 = 0.900 \newline % Row Count 11 (+ 1)
Divide number of moles each by the smallest number, which is 0.884 from Nitrogen \newline % Row Count 13 (+ 2)
5.30 and 8.884 and 9.72 and 0.900 all divided by 0.884 \newline % Row Count 15 (+ 2)
= 6, 1, 11, 1 after simplifying  \newline % Row Count 16 (+ 1)
= empirical formula becomes C6NH11O1% Row Count 17 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Mass of excess reactant}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Write balanced equation, find limiting reactant, calculate theoretical yield of the product, and determine the excess reactant. \newline % Row Count 3 (+ 3)
Now, calculate the amount left over by subtracting the amount of excess reactant that reacted from the initial amount of excess reactant given. \newline % Row Count 6 (+ 3)
Convert to mass \newline % Row Count 7 (+ 1)
Use the molar mass of the reactant to convert the amount of excess reactant left from moles to grams.% Row Count 10 (+ 3)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Dilution}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{First determine initial volume and concentration of the solution before dilution \newline % Row Count 2 (+ 2)
Then determine the final volume of the solution after dilution by adding the initial volume of the solution to the volume of the solvent added during dilution \newline % Row Count 6 (+ 4)
Caclualte the initial amount of solute by using the initial volume and concentration to calculate the amount of solute (substance being diluted) present in the solution before dilution. \newline % Row Count 10 (+ 4)
- Amount of solute (initial) = initial volume x initial concentration  \newline % Row Count 12 (+ 2)
Now determine the final concentration \newline % Row Count 13 (+ 1)
- Final concentration = amount of solute (initial)/final volume \newline % Row Count 15 (+ 2)
{\bf{if we dilute 100ml of a 1.0 M (concentration) of KCl to 400ml with 300ml of water, what will the final concentration be?}} \newline % Row Count 18 (+ 3)
Initial volume = 100ml \newline % Row Count 19 (+ 1)
initial concentration = 1.0 M \newline % Row Count 20 (+ 1)
Final volume = initial volume + volume of solvent added \newline % Row Count 22 (+ 2)
= 100ml + 400ml = 500ml \newline % Row Count 23 (+ 1)
Use the initial volume and concentration to calculate the amount of solute (KCl) present in the solution before dilution \newline % Row Count 26 (+ 3)
- initial solute amount = initial volume x initial concentration  \newline % Row Count 28 (+ 2)
= 100ml x 1.0 M = 100 moles \newline % Row Count 29 (+ 1)
Final concentration  \newline % Row Count 30 (+ 1)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Dilution (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{= amount of solute (initial)/final volume \newline % Row Count 1 (+ 1)
= 100 moles/500ml = 0.2 M% Row Count 2 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{A grams - B grams}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Balance equation and identify given and unknown qualities. \newline % Row Count 2 (+ 2)
Convert grams of A to moles \newline % Row Count 3 (+ 1)
use the given mass of A and it's molar mass \newline % Row Count 4 (+ 1)
- moles = mass (grams)/molar mass (grams/mol) \newline % Row Count 5 (+ 1)
Use the molar ratios \newline % Row Count 6 (+ 1)
Convert moles of A to B by using the molar ratio \newline % Row Count 7 (+ 1)
- multiply A moles by appropriate molar ratio \newline % Row Count 8 (+ 1)
Convert moles of B to grams \newline % Row Count 9 (+ 1)
- mass (grams) = moles x molar mass (grams/mol) \newline % Row Count 10 (+ 1)
Conclude by showing the mass of B formed or reacted based on the given mass of A% Row Count 12 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Stoichiometry: Moles Formed in Reaction}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Write balanced chemical equation \newline % Row Count 1 (+ 1)
Identify the given and unknown qualities \newline % Row Count 2 (+ 1)
- given: initial amount of moles \newline % Row Count 3 (+ 1)
- unknown: moles that will be formed \newline % Row Count 4 (+ 1)
Use the molar ratio to determine the answer \newline % Row Count 5 (+ 1)
Example: \newline % Row Count 6 (+ 1)
How many moles of SO3 will form when 3.4 moles of sulfur dioxide react with excess oxygen gas? \newline % Row Count 8 (+ 2)
Write the balanced equation \newline % Row Count 9 (+ 1)
Balance the equation for the reaction between SO2 and O2 to form sulfur trioxide \newline % Row Count 11 (+ 2)
- SO2 + O2 -{}-\textgreater{} SO3 \newline % Row Count 12 (+ 1)
- 2SO2 + O2 -{}-\textgreater{} 2SO3 \newline % Row Count 13 (+ 1)
Given: 3.4 moles of SO2 \newline % Row Count 14 (+ 1)
Unknown: moles of SO3 formed \newline % Row Count 15 (+ 1)
Use the molar ratio: \newline % Row Count 16 (+ 1)
The balanced equation shows us that 2 mole of SO2 react to form 2 moles of SO3, which means the molar ratio between them is 1:1. \newline % Row Count 19 (+ 3)
Therefore, if 3.4 moles of SO2 react, 3.4 moles of SO3 will form \newline % Row Count 21 (+ 2)
The reaction proceeds to completion, meaning all of the reactant (SO2) is consumed and converted into products - there no limiting factors that would prevent the complete consumption of the reactant. \newline % Row Count 25 (+ 4)
In the reaction btwn SO2 and O2 to form SO3, if SO2 is provided in excess, it implies theres enough oxygen to completely react with the SO2 presented. Excess oxygen ensures that all the SO2 molecules will find oxygen molecules to react with, thus the reaction can occur till all the SO2 is consumed.  \newline % Row Count 32 (+ 7)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Stoichiometry: Moles Formed in Reaction (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Because the reaction proceeds to completion and all the SO2 is consumed, the number of SO3 moles formed will be equal to the number of SO2 moles initially present. \newline % Row Count 4 (+ 4)
Therefore, the number of moles of SO3 formed will also be 3.4 moles. \newline % Row Count 6 (+ 2)
- 3.4 moles of SO3 will form when 3.4 moles of SO2 react with excess O2.% Row Count 8 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Dilution p3}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{{\bf{Dilute 250 mL of a 0.100 M solution from a 2.00 M solution}} \newline % Row Count 2 (+ 2)
Initial volume = ? \newline % Row Count 3 (+ 1)
Initial concentration = 2.00 \newline % Row Count 4 (+ 1)
Final volume = volume of diluted solution (250ml) \newline % Row Count 5 (+ 1)
Final concentration = 0.100m \newline % Row Count 6 (+ 1)
 Calculate amount of initial volume \newline % Row Count 7 (+ 1)
- since its unknown, we cant directly calculate the amount of solute from it, but we know that the amount of solute in the stock solution is equal to the amount of solute in the diluted solution after dilution (since no solute is added or removed during dilution) \newline % Row Count 13 (+ 6)
So we can calculate the amount of solute using the final concentration and volume of the diluted solution. \newline % Row Count 16 (+ 3)
Amount of initial solute = final concentration x final volume \newline % Row Count 18 (+ 2)
= 0.100 M x 250mL = 25 moles \newline % Row Count 19 (+ 1)
To find out how much of the stock solution (2.00M) we need to dilute to obtain the desired amount of solute (25 moles), we use this formula \newline % Row Count 22 (+ 3)
-  Final volume = Amount of solute (initial) / Initial concentration \newline % Row Count 24 (+ 2)
 = 25 moles / 2.00 M = 12.5 mL \newline % Row Count 25 (+ 1)
Calculate amount of water needed \newline % Row Count 26 (+ 1)
this is the difference btwn final volume of the stock solution and the volume of the diluted solution \newline % Row Count 29 (+ 3)
Water needed = Final volume of the stock solution - Final volume of the diluted solution \newline % Row Count 31 (+ 2)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Dilution p3 (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{= 12.5 mL - 250 mL = -237.5 mL \newline % Row Count 1 (+ 1)
The amount of water is negative so we dont need to add additional water to the stock solution for the desired concentration of 0.100M \newline % Row Count 4 (+ 3)
We just remove 12.5mL of the stock solution and then add water to make up the remaining volume to reach 250mL, giving us the desired diluted solution with a concentration of 0.100M \newline % Row Count 8 (+ 4)
Formula for dilution is C1V1 = C2V2, C1 and V2 are initial concentration and volume, and C2 V2 are final concentration and volume \newline % Row Count 11 (+ 3)
Rearrange the formula to solve for v1 \newline % Row Count 12 (+ 1)
V1 = C2V2/C1 \newline % Row Count 13 (+ 1)
A laboratory technician is required to prepare 1.00 m3 of 0.100 M H2SO4. What volume of 10 M H2SO4 is required? \newline % Row Count 16 (+ 3)
C1 = 10M, V2=1.00m\textasciicircum{}3, C2 = 0.100M \newline % Row Count 17 (+ 1)
V1 = 0.100Mx1.00m\textasciicircum{}3/10M \newline % Row Count 18 (+ 1)
= 10L% Row Count 19 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}


% That's all folks
\end{multicols*}

\end{document}