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Mole Ratios, Percentage Yield and Atom Economy

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Mole Ratios, Percentage Yield and Atom Economy Revision

Mole Ratios, Percentage Yield, and Atom Economy

The yield and atom economy of chemical reactions are both important quantities in both lab based experimental chemistry and industrial scale commercial chemistry.  These two numbers will tell us about the efficiency and the efficacy of a given chemical process. To calculate them we rely on balanced chemical equations and mole ratios.

Mole Ratios

Mole ratios will tell us how much a of a given substance we can produce from a given amount of product in a balanced chemical reaction (or conversely how much of a given reacted is needed to produce a given amount of product). 


Calculate the total volume of gas produced in \text{dm}^3 at 400\text{ K} and 100\text{ kPa} when 0.524\text{ g} of magnesium nitrate is heated.

2\text{Mg}\left(\text{NO}_3\right)_{2\left(\text{s}\right)}\rightarrow2\text{MgO}_{\left(\text{s}\right)}+4\text{NO}_{2\left(\text{g}\right)} +\text{O}_{2\left(\text{g}\right)}


\text{Moles of Mg}\left(\text{NO}_3\right)_{2\left(\text{s}\right)}= \dfrac{0.524}{148.3}=3.5334\times10^{-3}


Ratio of Moles of \text{Mg}\left(\text{NO}_3\right)_{2\left(\text{s}\right)} to moles of gas produced =2:5 \left(4\text{NO}_2+\text{O}_2\right)


\text{Moles of gas}=3.5334\times10^{-3}\times\dfrac{5}{2}=8.8335\times10^{-3}


\text{Volume of gas}=\dfrac{\text{nRT}}{\text{p}}=\dfrac{8.8335\times10^{-3}\times8.31\times400}{100000}=2.9363\times10^{-4}\text{m}^3


2.9363\times10^{-4}\times1000=0.294\text{ dm}^{3}


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Percentage Yield

For a variety of reasons, chemical reactions never have a 100% yield. We use percentage yield to compare the amount of product we made and the maximum possible amount that could have been made. This will tell us how effective a given chemical process is at producing a desired product. The percentage yield of a reaction will change according to a number of factors. For example the percentage yield of a reaction involving gasses can be increased by carrying it out in a sealed container. Percentage yield can be calculated using the following equation:

\text{Percentage yield}=\dfrac{\text{Actual yield}}{\text{Theoretical yield}}\times100


Example: Calculate the percentage yield of sodium sulphate when 7.20\text{ g} of sodium reacts with sulphuric acid to form 12.8\text{ g} of sodium sulphate.

2\text{Na}+\text{H}_2\text{SO}_4\rightarrow \text{Na}_2\text{SO}_4+\text{H}_2


\text{Moles of Na}=\dfrac{7.2}{23}=0.31304


Ratio of \text{Na} to \text{Na}_2\text{SO}_4 = 2:1


\text{Moles of Na}_2\text{SO}_4=\dfrac{0.31304}{2}=0.15652


\text{Mass of Na}_2\text{SO}_4=0.15652\times142=22.23


\text{Percentage yield}=\dfrac{12.8}{22.23}\times100=57.6\%


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Percentage Atom Economy

Chemical reactions do not always produce only the desired products; there will often be undesired products that we consider waste. We use percentage atom economy to calculate the efficiency of a reaction at producing the desired product. Unlike percentage yield, the atom economy of a chemical reaction is fixed. It is not possible to increase the percentage of atoms used to create product unless by changing the reaction itself. We calculate the atom economy using the following equation:

\text{Percentage atom economy}= \dfrac{\text{Mass of desired products}}{\text{Mass of all reactants}}\times100



Aspirin \left(\text{M}_r \textcolor{#00bfa8}{180}\right) is made in the following reaction:

\begin{aligned}\text{Salicylic Acid}+\text{Ethanoyl Chloride}&\rightarrow\text{Aspirin}+\text{Hydrogen Chloride}\\\text{C}_7\text{H}_6\text{O}_3+\text{C}_2\text{H}_6\text{OCl}&\rightarrow\text{C}_9\text{H}_8\text{O}_4+\text{HCl}\end{aligned}

Calculate the percentage atom economy of the reaction.


\text{Percentage atom economy}=\dfrac{\textcolor{#00bfa8}{180}}{213.5}\times100=83.1\%


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Mole Ratios, Percentage Yield and Atom Economy Example Questions

\text{Moles of NaCl}=\dfrac{546000}{58.5}=9333


Ratio of \text{NaCl} to \text{NaHCO}_3=1:1


\text{Moles of NaHCO}_3=9333


Ratio of \text{NaHCO}_3 to \text{Na}_2\text{CO}_3=2:1


\text{Moles of Na}_2\text{CO}_3 = 9333\div2=4667


\text{Mass of Na}_2\text{CO}_3=4667\times106=495\text{ kg}
\text{Moles of Mg}=\dfrac{3.89}{24.3}=0.160082


Ratio of \text{Mg} to \text{MgO}= 1:1


\text{Moles of MgO}= 0.160082


\text{Mass of MgO}=0.160082\times40.3=6.451


\text{Percentage yield}=\dfrac{4.07}{6.451}\times100=63.1\%
\text{Percentage atom economy}=\dfrac{\text{Mass of desired products}}{\text{Mass of all reactants}}\times100


\text{Percentage atom economy}=\dfrac{60}{78}\times100=76.9\%


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