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\author{Jerstellar}
\pdfinfo{
  /Title (genchem-q1-module.pdf)
  /Creator (Cheatography)
  /Author (Jerstellar)
  /Subject (GenChem q1 module Cheat Sheet)
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\noindent
\begin{multicols}{3}
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        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{GenChem q1 module Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{Jerstellar} via \textcolor{DarkBackground}{\uline{cheatography.com/204102/cs/43494/}}}
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\noindent
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}Jerstellar \\
  \uline{cheatography.com/jerstellar} \\
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Not Yet Published.\\
   Updated 28th July, 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}{Module 1 - {\bf{Matter and its Properties}}}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Matter - has mass and occupies space.% Row Count 1 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{p{0.76 cm} x{4.028 cm} x{2.812 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{3}{x{8.4cm}}{\bf\textcolor{white}{3 States of Matter}}  \tn
% Row 0
\SetRowColor{LightBackground}
State & Definition & Examples \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
Solid & rigid; has a fixed shape and volume & ice cube, diamond, iron bar \tn 
% Row Count 4 (+ 2)
% Row 2
\SetRowColor{LightBackground}
\seqsplit{Liquid} & has a definite volume but takes the shape of its container & gasoline, water, blood \tn 
% Row Count 7 (+ 3)
% Row 3
\SetRowColor{white}
Gas & has no fixed volume or shape; takes the shape of its container & air, helium, oxygen \tn 
% Row Count 10 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}---}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Phase Changes of Matter}}  \tn
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\hhline{>{\arrayrulecolor{DarkBackground}}-}
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\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{p{0.76 cm} x{3.42 cm} x{3.42 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{3}{x{8.4cm}}{\bf\textcolor{white}{Elements and Compounds}}  \tn
% Row 0
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\seqsplit{Elements} & cannot be broken down into other substances by chemical means & iron, aluminum, oxygen, and hydrogen \tn 
% Row Count 4 (+ 4)
% Row 1
\SetRowColor{white}
\seqsplit{Compound} & substances that have the same composition no matter where we find them; can be broken down into elements & Water (H20), Salt (NaCl), Ammonia (NH3) \tn 
% Row Count 10 (+ 6)
\hhline{>{\arrayrulecolor{DarkBackground}}---}
\end{tabularx}
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\begin{tabularx}{8.4cm}{x{2.56 cm} x{5.44 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Physical and Chemical Properties and Changes}}  \tn
% Row 0
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Physical Properties & odor, color, volume, state (gas, liquid, or solid), density, melting point, boiling point \tn 
% Row Count 4 (+ 4)
% Row 1
\SetRowColor{white}
Chemical Properties & burning, digestion, fermentation, rusting, electrolysis \tn 
% Row Count 7 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{x{0.836 cm} x{3.42 cm} x{3.344 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{3}{x{8.4cm}}{\bf\textcolor{white}{Other Properties}}  \tn
% Row 0
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\seqsplit{Extensive} & changes when the amount of material changes & mass, length, volume, shape \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
\seqsplit{Intrinsive} & does not depend on the size of the material & temperature, odor, color, hardness, density \tn 
% Row Count 6 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}---}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{x{1.14 cm} x{3.192 cm} x{3.268 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{3}{x{8.4cm}}{\bf\textcolor{white}{Mixture and Pure Substances}}  \tn
% Row 0
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\seqsplit{Mixture} & has variable composition &  \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
 & Homogenous & also called a solution; does not vary in composition from one region to another \tn 
% Row Count 7 (+ 5)
% Row 2
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 & Heterogenous & contains regions that have different properties from those of other regions \tn 
% Row Count 12 (+ 5)
% Row 3
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Pure \seqsplit{Substance} & always have the same composition; either elements or compounds &  \tn 
% Row Count 16 (+ 4)
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\begin{tabularx}{8.4cm}{p{0.76 cm} x{3.42 cm} x{3.42 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{3}{x{8.4cm}}{\bf\textcolor{white}{Types of bonds}}  \tn
% Row 0
\SetRowColor{LightBackground}
Ionic & when one atom shifts or transfers an electron to another atom; metals + nonmetals & Na\textasciicircum{}+\textasciicircum{} (1A) and Cl\textasciicircum{}-\textasciicircum{} (7A) creates a stable bond (octet rule) \tn 
% Row Count 5 (+ 5)
% Row 1
\SetRowColor{white}
\seqsplit{Covalent} & atoms share electrons; nonmetals & O\textasciicircum{}2-\textasciicircum{}(6A) and 2 atoms of H\textasciicircum{}+\textasciicircum{}(1A) = H₂O \tn 
% Row Count 8 (+ 3)
% Row 2
\SetRowColor{LightBackground}
\seqsplit{Metallic} & a metal shares an electron with another metal; positively charged ions in electrons &  \tn 
% Row Count 13 (+ 5)
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\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Module 2 - Isotopes, Compounds, Empirical Formula}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Atoms have a constant or fixed number of protons \newline % Row Count 1 (+ 1)
Atomic Number - gives the protons in the nucleus of an atom; represented as {\bf{Z}} \newline % Row Count 3 (+ 2)
Neutral Atom - number of protons is equal to the number of electrons \newline % Row Count 5 (+ 2)
{\bf{{\emph{Z}} = nuclear charge = number of protons = number of electrons in neutral form}} \newline % Row Count 7 (+ 2)
Mass Number - sum of the number of protons and neutrons; represented by {\bf{A}} \newline % Row Count 9 (+ 2)
An atom can be represented by the nuclear symbol \textasciicircum{}A\textasciicircum{}zE \newline % Row Count 11 (+ 2)
Nucleons - protons + neutrons% Row Count 12 (+ 1)
} \tn 
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\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{John Dalton's Atomic Theory}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{All atoms of an element have the same mass, although isotopes are atoms of the same element but has different numbers of protons \newline % Row Count 3 (+ 3)
Ex: All carbons atoms (Z=6) have 6 protons and electrons, but only 98.89\% of naturally occuring carbon atoms have 6 neutrons (A=12)% Row Count 6 (+ 3)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
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\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Chemical Compounds}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Radicals/Polyatomic Ions - stable groups which form chemical bonds as an intact unit. \newline % Row Count 2 (+ 2)
{\emph{The valence numbe is taken as one.}}  \newline % Row Count 3 (+ 1)
If a molecule contains {\bf{more than one radical}} (At least two unpaired electrons), the formula uses {\bf{parentheses}}. Calcium Phosphate - Ca₃(PO₄)₂% Row Count 7 (+ 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}{Some Polyatomic Ions}}  \tn
% Row 0
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\mymulticolumn{1}{x{8.4cm}}{Monovalent (1\textasciicircum{}-\textasciicircum{} )} \tn 
% Row Count 1 (+ 1)
% Row 1
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\mymulticolumn{1}{x{8.4cm}}{Ammonium} \tn 
% Row Count 2 (+ 1)
% Row 2
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\mymulticolumn{1}{x{8.4cm}}{Acetate} \tn 
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% Row 3
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\mymulticolumn{1}{x{8.4cm}}{Chlorite} \tn 
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\mymulticolumn{1}{x{8.4cm}}{Biculfate} \tn 
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\mymulticolumn{1}{x{8.4cm}}{Hydroxide} \tn 
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\mymulticolumn{1}{x{8.4cm}}{Nitrate} \tn 
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\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{p{3.04 cm} x{4.96 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Diatomic Molecules}}  \tn
% Row 0
\SetRowColor{LightBackground}
H₂ & hydrogen \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
N₂ & nitrogen \tn 
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% Row 2
\SetRowColor{LightBackground}
F₂ & fluorine \tn 
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% Row 3
\SetRowColor{white}
O₂ & oxygen \tn 
% Row Count 4 (+ 1)
% Row 4
\SetRowColor{LightBackground}
I₂ & iodine \tn 
% Row Count 5 (+ 1)
% Row 5
\SetRowColor{white}
Cl₂ & chlorine \tn 
% Row Count 6 (+ 1)
% Row 6
\SetRowColor{LightBackground}
Br₂ & bromine \tn 
% Row Count 7 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Criss-Cross Method}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{p{8.4cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/jerstellar_1717077834_Screenshot 2024-05-30 220412.png}}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{-Determine the charge or valence number of the elements \newline -Exchange their valence numbers \newline -Reducing by their gcf is possible}  \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}{Calculating Empirical Formula}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Percentage Composition - amounts of the elements for a given amount of compound \newline % Row Count 2 (+ 2)
Empirical Formula - simpest formula of any compound (smallest ratio of moles); derived from mass analysis \newline % Row Count 5 (+ 3)
- Determine the given number of moles in each element \newline % Row Count 7 (+ 2)
- Divide each by the smallest number of moles given \newline % Row Count 9 (+ 2)
- Multiply each by the smallest number that will turn them into whole numbers.% Row Count 11 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Example}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{p{8.4cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/jerstellar_1717081230_Screenshot 2024-05-30 230017.png}}} \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}{Calculating Empirical Formula with Molar mass}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Given} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{28.03\% Mg} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{21.6\% Si} \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{1.16\% H} \tn 
% Row Count 4 (+ 1)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{49.21\% O} \tn 
% Row Count 5 (+ 1)
% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Answer = Mg₃Si₂H₃O₈} \tn 
% Row Count 6 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{-Divide the given percent composition to the mass number of each element \newline -Divide each quotient by the smallest number among them \newline -Multiply the quotients by the smallest number that will make them whole}  \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}{For more examples:}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{\{\{popup="https://docs.google.com/document/d/1BfapJ3aGJSu2U5F9YLPPx0elXItld3m7hX4O\_azEmek/edit?usp=sharing"\}\}Chem Calculation Worksheet\{\{/popup\}\}} \tn 
% Row Count 3 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{x{2.016 cm} x{1.008 cm} x{2.016 cm} x{2.16 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{4}{x{8.4cm}}{\bf\textcolor{white}{Calculating Molecular Formula By Empirical Formula}}  \tn
% Row 0
\SetRowColor{LightBackground}
Empirical \seqsplit{Composition} & Mass \seqsplit{number} & Product (rounded off) & Product (emp* (mass/x)) \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
Mg₃ & \seqsplit{24.305} & 73 & 6 \tn 
% Row Count 5 (+ 2)
% Row 2
\SetRowColor{LightBackground}
Si₂ & \seqsplit{28.086} & 56 & 4 \tn 
% Row Count 7 (+ 2)
% Row 3
\SetRowColor{white}
H₃ & 1.008 & 3 & 6 \tn 
% Row Count 8 (+ 1)
% Row 4
\SetRowColor{LightBackground}
O₈ & \seqsplit{15.999} & 128 & 16 \tn 
% Row Count 10 (+ 2)
% Row 5
\SetRowColor{white}
 &  & Σ = 260 &  \tn 
% Row Count 11 (+ 1)
% Row 6
\SetRowColor{LightBackground}
\mymulticolumn{4}{x{8.4cm}}{Suppose the molar mass is 520.8; divide it by the summation (520.8/260 ≈ 2). Multiply 2 by the empirical compostion of each element. Answer = Mg₆Si₄H₆O₁₆} \tn 
% Row Count 15 (+ 4)
\hhline{>{\arrayrulecolor{DarkBackground}}----}
\SetRowColor{LightBackground}
\mymulticolumn{4}{x{8.4cm}}{-Get the summation summation of the product of each empirical composition to their mass number \newline -Divide the summation from the molar mass \newline -Multiply the quotient to the empirical composition of each element}  \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}{Module 3 - Molar Mass, Chem Reactions, Eq}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Mole(mol) - SI unit for determining molar mass; amount of substance that contains the same number of atoms in 12g of Carbon-12 \newline % Row Count 3 (+ 3)
{\emph{Avogadro's number - 6.02214076 × 10\textasciicircum{}23\textasciicircum{}}} \newline % Row Count 4 (+ 1)
Elements - mass in amu of 1 atom of an element is the same as the mass in grams of 1 mole of atoms of the element \newline % Row Count 7 (+ 3)
{\emph{Mass of S (32.07 amu) is equal to the mass of 1 mol (6.02214076 × 10\textasciicircum{}23\textasciicircum{}) of S (32.07 amu)}}% Row Count 9 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{x{1.44 cm} x{1.944 cm} x{2.16 cm} x{1.656 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{4}{x{8.4cm}}{\bf\textcolor{white}{Calculating Molecular Mass/Weight}}  \tn
% Row 0
\SetRowColor{LightBackground}
\seqsplit{Composition} & Number of Atoms & Mass Number (amu) & Product (amu) \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
H₂ & 2 & 1.008 & 2.02 \tn 
% Row Count 3 (+ 1)
% Row 2
\SetRowColor{LightBackground}
O & 1 & 16.00 & 16.00 \tn 
% Row Count 4 (+ 1)
% Row 3
\SetRowColor{white}
 &  &  & Σ = 18.02 \tn 
% Row Count 6 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}----}
\SetRowColor{LightBackground}
\mymulticolumn{4}{x{8.4cm}}{-Determine the number of atoms of each element then multiply to their corresponding mass number \newline -Get the summation of the products}  \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}{Writing and Balancing Chem Eq}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Law of Conservation of Mass - mass is neither created nor destroyed in a chemical reaction \newline % Row Count 2 (+ 2)
Antoine Lavoisier - French chemist; proponent \newline % Row Count 3 (+ 1)
Reactants - starting material in a chemical reaction \newline % Row Count 5 (+ 2)
Product - substance formed in a chemical reaction \newline % Row Count 6 (+ 1)
{\emph{Reactants → Products}} \newline % Row Count 7 (+ 1)
"to yield" or "to form" (→) \newline % Row Count 8 (+ 1)
"to react with" or "to combine with" (+)% Row Count 9 (+ 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}{Examples}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{\{\{ac\}\}CH₄ + O₂ → CO₂ + H₂O ⟹ CH₄ + {\bf{2}}O₂ → CO₂ + {\bf{2}}H₂O} \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{\{\{ac\}\}Al + BaO → Al₂O₃ + Ba ⟹ {\bf{2}}Al + {\bf{3}}BaO → Al₂O₃ + {\bf{3}}Ba} \tn 
% Row Count 4 (+ 2)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{\{\{ac\}\}Cl₂ + KBr → KCl + Br₂ ⟹ Cl₂ + {\bf{2}}KBr → {\bf{2}}KCl + Br₂} \tn 
% Row Count 6 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{x{2.432 cm} x{2.812 cm} x{2.356 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{3}{x{8.4cm}}{\bf\textcolor{white}{Types of Chemical Reactions}}  \tn
% Row 0
\SetRowColor{LightBackground}
Type & Definition & Example \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\seqsplit{Combination/Synthesis} & two or more reactants combine to form a single product & 2Mg + O₂ → 2MgO \tn 
% Row Count 5 (+ 4)
% Row 2
\SetRowColor{LightBackground}
\seqsplit{Decomposition} & one reactant breaks down into two or more products & CaCO₃ → CaO + CO₂ \tn 
% Row Count 9 (+ 4)
% Row 3
\SetRowColor{white}
Single \seqsplit{Displacement} & one element is substituted for another element in a compound & K + NaCl → KCl + Na \tn 
% Row Count 14 (+ 5)
% Row 4
\SetRowColor{LightBackground}
Double \seqsplit{Displacement/salt} metathesis & two substances react by exchanging ions to produce two new molecules & AgNo₃ + NaCl → AgCl + NaNo₃ \tn 
% Row Count 19 (+ 5)
\hhline{>{\arrayrulecolor{DarkBackground}}---}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Module 4 - Mass Relationships in Chem Reactions}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Stoichiometry - quantitative relationship between reactants and products in a chemical reaction \newline % Row Count 2 (+ 2)
Stoichiometric coefficient - added before an element, ion, or molecule to balance chemical reactions \newline % Row Count 5 (+ 3)
Mole method using mole-mole factor: \newline % Row Count 6 (+ 1)
\{\{ac\}\}2Na(s) + 2HCl(aq) → 2NaCl(aq) + H₂(g) \newline % Row Count 7 (+ 1)
2 moles Na 	≅ 2 moles NaCl; hence,% Row Count 8 (+ 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}{Calculating Amount of Product and Reactant}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{\{\{ac\}\}3Hg₂(g) + N₂(g) → 2NH₃(g) \newline % Row Count 1 (+ 1)
How many moles of {\bf{H₂}} are needed to produce {\bf{26.5 moles of NH₃}}? \newline % Row Count 3 (+ 2)
\textasciitilde{}\textasciitilde{}26.5 moles NH₃\textasciitilde{}\textasciitilde{} x (3 moles H₂)/\textasciitilde{}\textasciitilde{}(2 moles NH₃)\textasciitilde{}\textasciitilde{} = 39.8 moles of H₂ \newline % Row Count 5 (+ 2)
How many moles of {\bf{NH₃}} will be produced if {\bf{33.7 moles of N₂}} reacts completely with {\bf{H₂}} \newline % Row Count 8 (+ 3)
\textasciitilde{}\textasciitilde{}33.7 moles of N₂\textasciitilde{}\textasciitilde{} x (2 moles of NH₃)/\textasciitilde{}\textasciitilde{}(1 mole of N₂)\textasciitilde{}\textasciitilde{} = 67.4 moles of NH₃% Row Count 10 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{-In using mole-mole factor, the arrangement of fractions is done in a way that there is cancellation of similar units}  \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}{Calculating " " with Molar Mass}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{\{\{ac\}\}2LiOH(s) + CO₂(g) → Li₂CO₃(s) + H₂O(l) \newline % Row Count 2 (+ 2)
How many grams of {\bf{CO₂}} can be absorbed by {\bf{236.1 g of LiOH}}? \newline % Row Count 4 (+ 2)
\textasciitilde{}\textasciitilde{}236.1g LiOH\textasciitilde{}\textasciitilde{} x (1 mole of LiOH)/\textasciitilde{}\textasciitilde{}(23.95g LiOH)\textasciitilde{}\textasciitilde{} x \textasciitilde{}\textasciitilde{}(1 mole CO₂)\textasciitilde{}\textasciitilde{} /(2 moles LiOH) x (44.01g of CO₂)/ \textasciitilde{}\textasciitilde{}(1 mole CO₂)\textasciitilde{}\textasciitilde{} = 221.1g CO₂% Row Count 7 (+ 3)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{-Determine the mass of each element and add each to the given compounds \newline (Li=6.941, O=15.999, H=1.008, C=12.011) \newline -Since the given number has 4 significant figures, the numbers also have 4 significant figures}  \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}{Limiting and Excess Reagent}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{In chemical reactions, the amount of reactants isn't always stoichiometrically exact, so scientists use cheaper reactants (excess) \newline % Row Count 3 (+ 3)
Limiting Reagent - Reagent that is completely reacted or used up \newline % Row Count 5 (+ 2)
Excess Reagent - Reactant present with higher quantity than what is required to react in a limiting reagent% Row Count 8 (+ 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}{Example}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{\{\{ac\}\} 3H₂ + 2N₂ → 2NH₃ \newline % Row Count 1 (+ 1)
Suppose 6 moles of H₂ was mixed with 4 moles of N₂ . To determine which is the limiting reagent, the amount of NH₃ must be computed given the moles of H₂ and N₂ and the mole-mole factor of the equation% Row Count 6 (+ 5)
} \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}{Solution}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{p{8.4cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/jerstellar_1717342267_Screenshot 2024-06-02 230451.png}}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{-Simplify the number of moles by multiplying each final no. of moles of reagent by the proportion of the initial number of moles and given reagent \newline -The reagent with lesser number of moles of NH₃ is the limiting reagent and vise versa; in this case, H₂ is the limiting reagent and N₂ is the excess reagent}  \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}{Calculate the excess}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{-To determine how much of 4 moles of N₂ is in excess, use mole-mole factor of N₂ and H₂ \newline % Row Count 2 (+ 2)
\textasciitilde{}\textasciitilde{}6 moles of H₂\textasciitilde{}\textasciitilde{} x (1 mole N₂)/(\textasciitilde{}\textasciitilde{}3 moles H₂\textasciitilde{}\textasciitilde{}) = 2 moles N₂ in excess \newline % Row Count 4 (+ 2)
-The number of moles of N₂ required to react with 6 moles of H₂ is only 2, thus, {\bf{6 moles of N₂ has an excess of 4 moles}} \newline % Row Count 7 (+ 3)
6 moles of N₂ - 2 moles of N₂ in 6 moles of H₂ = 4 moles excess N₂% Row Count 9 (+ 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}{Limiting and Excess Reagent with Molar Mass}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{\{\{ac\}\} Mg₂Si + 4H₂O → 2Mg(OH)₂ + SiH₄ \newline % Row Count 1 (+ 1)
If we start with 50.0 g of each reactant, how much in grams SiH₄ can be formed? \newline % Row Count 3 (+ 2)
\textasciitilde{}\textasciitilde{}50g Mg₂Si\textasciitilde{}\textasciitilde{} x (1 mol Mg₂Si)/(\textasciitilde{}\textasciitilde{}76.7 g Mg₂Si\textasciitilde{}\textasciitilde{}) x (\textasciitilde{}\textasciitilde{}1 mol SiH₄\textasciitilde{}\textasciitilde{})/(1 mol Mg₂Si) x (32.1 g SiH₄)/(\textasciitilde{}\textasciitilde{}1 mol SiH₄\textasciitilde{}\textasciitilde{}) = {\bf{20.9 SiH₄}} \newline % Row Count 6 (+ 3)
\textasciitilde{}\textasciitilde{}50 g H₂O\textasciitilde{}\textasciitilde{} x (1 mole H₂O)/(\textasciitilde{}\textasciitilde{}18.0 g H₂O\textasciitilde{}\textasciitilde{}) x (\textasciitilde{}\textasciitilde{}1 mol SiH₄\textasciitilde{}\textasciitilde{})/(4 mol H₂O) x (32.1 g SiH₄)/(\textasciitilde{}\textasciitilde{}1 mol SiH₄\textasciitilde{}\textasciitilde{}) = {\bf{22.3 g SiH₄}} \newline % Row Count 9 (+ 3)
50g Mg₂Si = 20.9 SiH₄ {\emph{(Limiting reactant)}} \newline % Row Count 10 (+ 1)
50 g H₂O = 22.3 g SiH₄ {\emph{(Excess reactant)}}% Row Count 11 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{-Divide the number of initial molar mass of compound by the final/given molar mass and multiply with the molar mass of required compound to convert \newline -Determine how much of 50 g of H₂O is in excess by 50 g H₂O x (1 mol Mg₂Si)/(4 mol H₂O) = 12.5 g Mg₂Si \newline 50 g Mg₂Si - 12.5 g Mg₂Si {\emph{excess in 50 g H₂O}} = {\bf{37.5 g excess Mg₂Si}}}  \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}{Theoretical and Percent Yield}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Percent Yield - ratio of actual yield to the theoretical yield expressed as a percentage \newline % Row Count 2 (+ 2)
{\emph{Percent Yield = (Actual Yield)/(Theoretical Yield) x 100\%}} \newline % Row Count 4 (+ 2)
Theoretical Yield - maximum/expected amount of product produced from the given amount of reactant \newline % Row Count 6 (+ 2)
Actual Yield - actual amount of product produced from the given amount of reactant (determined experimentally)% Row Count 9 (+ 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}{Calculating Theoretical and Percent Yield}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{\{\{ac\}\}In calculating theoretical yield, always use the limiting reactant.{\emph{from the previous example}} \newline % Row Count 3 (+ 3)
Mg₂Si + 4H₂O 2Mg(OH)₂ + SiH₄ \newline % Row Count 4 (+ 1)
If 19.87 g SiH₄ is formed, what is the percent yield of the reaction? \newline % Row Count 6 (+ 2)
50g Mg₂Si = 20.9 SiH₄ (Limiting reactant) = Theoretical yield \newline % Row Count 8 (+ 2)
{\emph{Percent Yield = (Actual Yield)/(Theoretical Yield) x 100\%}} \newline % Row Count 10 (+ 2)
= 19.87 g SiH₄/20.9 SiH₄ x 100\% = {\bf{94.89\%}} \newline % Row Count 11 (+ 1)
Percent error = 100\% - 98.89\% = {\bf{5.11\%}}% 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}{Module 5 - Gases I}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Pressure - amount of force exerted per unit area \newline % Row Count 1 (+ 1)
Standard atmosphere (atm) - widely used unit for pressure; {\bf{1 atm = 760mmHg}} \newline % Row Count 3 (+ 2)
Torr (or mmHg) - milliliter of mercury equal to 1 atmosphere; named after Italian scientist Evangelista Torricelli (invented barometer) \newline % Row Count 6 (+ 3)
Pounds per square inch (psi) - amount of pressure in pounds that gas exerts in a container per square inch of unit area \newline % Row Count 9 (+ 3)
kilopascal (kPa) - equal to 1000 Pa, modern unit for pressure/default \newline % Row Count 11 (+ 2)
Conversion Factor: \newline % Row Count 12 (+ 1)
{\bf{1 atm = 760mmHg = 760 Torr = 101.3 kPa = 14.7 psi = 1k Pa = 1000 Pa}}% Row Count 14 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{x{2.052 cm} x{2.812 cm} x{2.736 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{3}{x{8.4cm}}{\bf\textcolor{white}{Gas Laws}}  \tn
% Row 0
\SetRowColor{LightBackground}
Boyle's Law & Volume is inversely proportional to its pressure at a {\bf{constant temparature}}; ↑V ⟺ ↓P & P₁V₁ = P₂V₂ \tn 
% Row Count 7 (+ 7)
% Row 1
\SetRowColor{white}
Charles' Law & Volume is directly proportional to its absolute temperature and {\bf{constant pressure}}; ↑V ⟺ ↑T & V₁/T₁ = V₂/T₂ \tn 
% Row Count 15 (+ 8)
% Row 2
\SetRowColor{LightBackground}
\seqsplit{Avogadro's} Law & Volume is directly proportional to the number of moles contained in the volume at {\bf{constant temperature and pressure}}; ↑V ⟺ ↑n & V₁/n₁ = V₂/n₂ \tn 
% Row Count 25 (+ 10)
% Row 3
\SetRowColor{white}
\seqsplit{Gay-Lussac's} Law/Ideal Gas Law & Sums up and combines Boyle's, Charles', and Avogadro's Laws & PV = nRT (where R or universal gas constant = {\bf{0.0821 atm.L/mol.K}}) \tn 
% Row Count 30 (+ 5)
\hhline{>{\arrayrulecolor{DarkBackground}}---}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Gas Mixtures}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Different gases can be present in a container and can be represented as n₁ (gas 1), n₂ (gas 2), or n₃ (gas 3), etc.; the total number of moles as {\bf{nₜₒₜₐₗ}} \newline % Row Count 4 (+ 4)
The pressure exerted by the mixture can be interpreted as {\bf{Pₘᵢₓₜᵤᵣₑ = (nₜₒₜₐₗRT)/V}} \newline % Row Count 7 (+ 3)
It can be simplified as {\bf{P₁= (n₁RT)/V}}; {\bf{P₂ = (n₂RT)/V}}; {\bf{P₃ = (n₃RT)/V}} \newline % Row Count 9 (+ 2)
Pressures P₁, P₂, and P₃ are partial pressure of each gas \newline % Row Count 11 (+ 2)
Dalton's Law of Partial Pressure - pressure exerted by the mixture is the sum of the pressures exerted by each component \newline % Row Count 14 (+ 3)
Get the partial pressure of gas 1 by {\bf{P₁ = Pₘᵢₓₜᵤᵣₑ X₁}} (wherein X₁ is the mole fraction of gas 1)% Row Count 17 (+ 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}{Module 6 - Gases II}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Stoichiometric ratio - dictates the ratio of components to start the reaction \newline % Row Count 2 (+ 2)
Standart Temperature and Pressure (STP) = 0°C (273 K) and pressure of 1 atm \newline % Row Count 4 (+ 2)
Amount of gaseous products are determined using {\bf{n = (PVₛₜₚ)/(RT)}} wherein Vₛₜₚ is the volume of gases involved measured in STP in liters (L) \newline % Row Count 8 (+ 4)
{\bf{n = \seqsplit{(PVₛₜₚ)/(0.0821)(273} K) = Vₛₜₚ/22.4}} \newline % Row Count 10 (+ 2)
{\bf{n  = Vₛₜₚ/22.4}} \newline % Row Count 11 (+ 1)
Gases are also measured in {\emph{Standard Ambient Temperature and Pressure (SATP)}} which is more accurate than STP which is at 25°C (298 K) and 1 atm. \newline % Row Count 14 (+ 3)
{\bf{n = \seqsplit{(PVₛₐₜₚ)/(0.0821)(298} K) = Vₛₐₜₚ/24.5}} \newline % Row Count 16 (+ 2)
{\bf{n = Vₛₐₜₚ/24.5}}% Row Count 17 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{x{1.656 cm} p{0.72 cm} x{1.656 cm} x{3.168 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{4}{x{8.4cm}}{\bf\textcolor{white}{Temperature Conversion}}  \tn
% Row 0
\SetRowColor{LightBackground}
Celsius & → & Kelvin & C + 273.15 \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
Celsius & → & \seqsplit{Farenheight} & C (9/5) + 32 \tn 
% Row Count 3 (+ 2)
% Row 2
\SetRowColor{LightBackground}
\seqsplit{Farenheight} & → & Kelvin & (F - 32)(5/9) + 273.15 \tn 
% Row Count 5 (+ 2)
% Row 3
\SetRowColor{white}
\seqsplit{Farenheight} & → & Celsius & (F - 32)(5/9) \tn 
% Row Count 7 (+ 2)
% Row 4
\SetRowColor{LightBackground}
Kelvin & → & Celsius & K - 273.15 \tn 
% Row Count 8 (+ 1)
% Row 5
\SetRowColor{white}
Kelvin & → & \seqsplit{Farenheight} & (K - 273.15)(9/5) + 32 \tn 
% Row Count 10 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}----}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Kinetic Molecular Theory}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{1. Gases are very small molecules separated by expansive space between them \newline % Row Count 2 (+ 2)
2. Force of attraction between particles is {\bf{negligible}} \newline % Row Count 4 (+ 2)
3. The molcules are in {\bf{constant motion}} and move randomly in all directions \newline % Row Count 6 (+ 2)
4. Sometimes particles {\bf{collide}} with each other or with the walls of container \newline % Row Count 8 (+ 2)
5. The collisions are {\bf{perfectly elastic}}, hence, there is no change in momentum \newline % Row Count 10 (+ 2)
6. The average kinetic energy is determined only by the absolute temparature of the gas \newline % Row Count 12 (+ 2)
To determine the kinetic energy of gas particles, the root-mean-square velocity is used: \newline % Row Count 14 (+ 2)
{\bf{vᵣₘₛ = √(3RT/M)}}  \newline % Row Count 15 (+ 1)
where R = ideal gas constant, T = absolute temperature in K, M = molar mass in g/mol \newline % Row Count 17 (+ 2)
To compare the velocities of gases with different molar masses at the same absolute temperature: \newline % Row Count 19 (+ 2)
{\bf{vᵣₘₛ₁/vᵣₘₛ₂ = √(M₂)/√(M₁)}} \newline % Row Count 21 (+ 2)
where M₁ or M₂ = molar mass of gas 1 or 2 \newline % Row Count 22 (+ 1)
This expression is also known as {\bf{Graham's Law of Diffusion}} which states that the diffusion rate (rate at which the gas moves), is {\bf{inversely proportional}} to the square root of its molar mass% Row Count 26 (+ 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}{YEY! you finished q1, I am so proud of you :)}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{} \tn 
% Row Count 0 (+ 0)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Calculating Empirical Formula with Molar mass}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Given} \tn 
% Row Count 1 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{-Divide the given percent composition to the mass number of each element \newline -Divide each quotient by the smallest number among them \newline -Multiply the quotients by the smallest number that will make them whole}  \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}{Calculating Empirical Formula with Molar mass}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Given} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{28.03\% Mg} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{21.6\% Si} \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{1.16\% H} \tn 
% Row Count 4 (+ 1)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{49.21\% O} \tn 
% Row Count 5 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{-Divide the given percent composition to the mass number of each element \newline -Divide each quotient by the smallest number among them \newline -Multiply the quotients by the smallest number that will make them whole}  \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}{Calculating Empirical Formula with Molar mass}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Given} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{28.03\%} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{1.6\%} \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{1.16\%} \tn 
% Row Count 4 (+ 1)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{49.21} \tn 
% Row Count 5 (+ 1)
% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Answer = Mg₃Si₃H₂O₈} \tn 
% Row Count 6 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{-Divide the given percent composition to the mass number of each element \newline -Divide each quotient by the smallest number among them \newline -Multiply the quotients by the smallest number that will make them whole}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}


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

\end{document}