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\author{fongrsy}
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  /Title (mole-calculations.pdf)
  /Creator (Cheatography)
  /Author (fongrsy)
  /Subject (Mole Calculations Cheat Sheet)
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\noindent
\begin{multicols}{3}
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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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\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Mole Calculations Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{fongrsy} via \textcolor{DarkBackground}{\uline{cheatography.com/65383/cs/16384/}}}
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\noindent
\begin{multicols}{3}
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}fongrsy \\
  \uline{cheatography.com/fongrsy} \\
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 18th July, 2018.\\
   Updated 21st July, 2018.\\
   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{document}
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\begin{multicols*}{3}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Conventions and Rules}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{* {\bf{Molar mass}} is the mass of one mole of atoms in a substance. \newline % Row Count 2 (+ 2)
* {\bf{1 mole of any gas is 24dm\textasciicircum{}3\textasciicircum{} at r.t.p.}} \newline % Row Count 3 (+ 1)
* {\bf{Avogadro's Constant}} is the number of particles in one mole of any substance. The number is {\bf{{\emph{6 x 10\textasciicircum{}23\textasciicircum{}}}}}. \newline % Row Count 6 (+ 3)
* {\bf{{\emph{Relative Atomic Mass}}}} is the mass of one atom of an element compared to 1/12 of the mass of one carbon-12 atom. \newline % Row Count 9 (+ 3)
* {\bf{{\emph{Relative Molecular Mass}}}} is the mass of one molecule of an element compared to 1/12 of the mass of one carbon-12 atom. \newline % Row Count 12 (+ 3)
* {\bf{{\emph{Relative Formula Mass}}}} is the mass of one formula unit of an ionic compound. It is the sum of the Ar of all the ions in the formula unit. \newline % Row Count 15 (+ 3)
* {\bf{{\emph{1000cm\textasciicircum{}3\textasciicircum{} = 1 dm\textasciicircum{}3\textasciicircum{}}}}}% Row Count 16 (+ 1)
} \tn 
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\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{1.46464 cm} x{1.00694 cm} x{2.10542 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{3}{x{5.377cm}}{\bf\textcolor{white}{Formula Triangles}}  \tn
% Row 0
\SetRowColor{LightBackground}
Top & Bottom 1 & Bottom 2 \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
No. of Particles & No. of Moles & Avogadro's No. \tn 
% Row Count 3 (+ 2)
% Row 2
\SetRowColor{LightBackground}
Mass in Grams & No. of Moles & Molar Mass in g/mol \tn 
% Row Count 5 (+ 2)
% Row 3
\SetRowColor{white}
Volume of Gas in dm\textasciicircum{}3\textasciicircum{} & No. of Moles & 24dm\textasciicircum{}3\textasciicircum{} \tn 
% Row Count 7 (+ 2)
% Row 4
\SetRowColor{LightBackground}
Mass in g & Volume in dm\textasciicircum{}3\textasciicircum{} & Mass Concentration in g/dm\textasciicircum{}3\textasciicircum{} \tn 
% Row Count 9 (+ 2)
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No. of Moles & Volume in dm\textasciicircum{}3\textasciicircum{} & Molar Concentration in mol/dm\textasciicircum{}3\textasciicircum{} \tn 
% Row Count 11 (+ 2)
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\mymulticolumn{3}{x{5.377cm}}{* To find top, multiply bottom values. To find bottom, take top divided by other bottom value. \newline  \newline 1. No. of Particles // Moles // Avogadro's No. \newline 2. Mass // Moles // Molar Mass \newline 3. Volume // Moles // 24dm\textasciicircum{}3\textasciicircum{} \newline 4. Mass // Volume // Mass Concentration \newline 5. Moles // Volume // Molar Concentration}  \tn 
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Concentration}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{* The amount of solute dissolved in a unit volume of the solution. \newline % Row Count 2 (+ 2)
* Usually in g/dm\textasciicircum{}3\textasciicircum{} or mol/dm\textasciicircum{}3\textasciicircum{}. \newline % Row Count 3 (+ 1)
* Mol/dm\textasciicircum{}3\textasciicircum{} is also called Molarity. \newline % Row Count 4 (+ 1)
* 1M = 1 mol/dm\textasciicircum{}3\textasciicircum{}% Row Count 5 (+ 1)
} \tn 
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\begin{tabularx}{5.377cm}{x{1.92234 cm} x{1.32733 cm} x{1.32733 cm} }
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\mymulticolumn{3}{x{5.377cm}}{\bf\textcolor{white}{Molecular Formula \& Empirical Formula}}  \tn
% Row 0
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Value & Element A & Element B \tn 
% Row Count 1 (+ 1)
% Row 1
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Mass in Grams & x & x \tn 
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Molar Mass & x & x \tn 
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Moles & x & x \tn 
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Mole Ratio & x & x \tn 
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\mymulticolumn{3}{x{5.377cm}}{1. Given \% Composition, find masses of both substances in 100g. If 70\% is A and 30\% is B, then there is 70g of A and 30g of B in 100g of AB. \newline 2. Find molar mass using periodic table. \newline 3. Find no. of moles by multiplying mass and molar mass. \newline 4. Divide all sides by the smallest number and round off to the nearest whole number to get mole ratio. \newline 5. Molecular Formula is always a multiple of the Empirical Formula.}  \tn 
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{\% Purity and \% Yield}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{\% Purity = Mass of Pure Substance / Total Mass x 100\% \newline % Row Count 2 (+ 2)
\% Yield = Actual Mass / Theoretical Mass x 100\%% Row Count 3 (+ 1)
} \tn 
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\begin{tabularx}{5.377cm}{x{1.29487 cm} x{1.16956 cm} p{0.4177 cm} x{1.29487 cm} }
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\mymulticolumn{4}{x{5.377cm}}{\bf\textcolor{white}{Limiting Reagents \& Reactants in Excess}}  \tn
% Row 0
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x & No. of Available Moles & vs & No. of Moles Needed \tn 
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Limiting Reagents & No. of Available Moles & \textless{} & No. of Moles Needed \tn 
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Reactants in Excess & No. of Available Moles & \textgreater{} & No. of Moles Needed \tn 
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{How to Find Limiting Reagents}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{A + 2B \{\{fa-arrow-right\}\} 2AB \newline % Row Count 1 (+ 1)
1. Find no. of available moles for A. \newline % Row Count 2 (+ 1)
2. Find no. of available moles for B. \newline % Row Count 3 (+ 1)
3. Find no. of moles needed for A/B. \newline % Row Count 4 (+ 1)
E.g. {\emph{Moles needed for A = Available moles for B / 2}} \newline % Row Count 6 (+ 2)
(Refer to mole ratio) \newline % Row Count 7 (+ 1)
If moles available is \textless{} moles needed, then that reactant is the limiting reagent.% Row Count 9 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{The concept of limiting reagents is the available moles for reaction vs the needed moles for reaction.}  \tn 
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\begin{tabularx}{5.377cm}{p{0.62655 cm} x{1.21133 cm} x{1.2531 cm} x{1.08602 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{4}{x{5.377cm}}{\bf\textcolor{white}{Mole Calculations}}  \tn
% Row 0
\SetRowColor{LightBackground}
Given &  &  & \{\{ar\}\}	Find \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
Mass of A & \{\{ac\}\}	 Moles of A & \{\{ac\}\}	 Moles of B & \{\{ar\}\} Mass of B \tn 
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% Row 2
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Vol. of A & \{\{ac\}\}	 Moles of A & \{\{ac\}\}	 Moles of B & \{\{ar\}\} Vol. of B \tn 
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Mass of A & \{\{ac\}\}	 Moles of A & \{\{ac\}\}	 Moles of B & \{\{ar\}\} Vol. of B \tn 
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% Row 4
\SetRowColor{LightBackground}
Vol. of A & \{\{ac\}\}	 Moles of A & \{\{ac\}\}	 Moles of B & \{\{ar\}\} Mass of B \tn 
% Row Count 10 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}----}
\SetRowColor{LightBackground}
\mymulticolumn{4}{x{5.377cm}}{Refer to mole ratio when converting Moles of A to Moles of B. \newline 1 mole of any gas is 24dm\textasciicircum{}3\textasciicircum{} at R.T.P.}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}----}
\end{tabularx}
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\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Acids \& Bases}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Metal + Acid \{\{fa-arrow-right\}\} Salt + Hydrogen Gas \newline % Row Count 2 (+ 2)
Metal Carbonate + Acid \{\{fa-arrow-right\}\} Salt + Water + Carbon Dioxide \newline % Row Count 4 (+ 2)
Metal Oxide + Acid \{\{fa-arrow-right\}\} Salt + Water \newline % Row Count 6 (+ 2)
Metal Hydroxide + Acid \{\{fa-arrow-right\}\} Salt + Water \newline % Row Count 8 (+ 2)
Base + Acid \{\{fa-arrow-right\}\} Salt + Water (Neutralisation) \newline % Row Count 10 (+ 2)
Alkali + Acid \{\{fa-arrow-right\}\} Salt + Water \newline % Row Count 11 (+ 1)
Alkali + Ammonium Salt \{\{fa-arrow-right\}\} Salt + Water + Ammonia Gas \newline % Row Count 13 (+ 2)
Alkali + Salt \{\{fa-arrow-right\}\} Metal Hydroxide + Salt% Row Count 15 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}


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

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