Mole Ratios, Percentage Yield and Atom Economy
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).
Example:
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}
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\%
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
Example:
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\%
Mole Ratios, Percentage Yield and Atom Economy Example Questions
Question 1: In the Solvay Process, sodium carbonate is produced in two stages.
\begin{aligned}&\text{Stage }1:&\text{NaCl}+\text{NH}_3+\text{CO}_2+\text{H}_2\text{O}&\rightarrow\text{NaHCO}_3+\text{NH}_4\text{Cl}\\ &\text{Stage }2:& 2\text{NaHCO}_3&\rightarrow\text{Na}_2\text{CO}_3+\text{H}_2\text{O}+\text{CO}_2\end{aligned}
Calculate the maximum mass of sodium carbonate which, theoretically, could be obtained from 546\text{ kg} of sodium chloride.
[4 marks]
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}
Question 2: 3.89\text{ g} of magnesium reacts with oxygen to form 4.07\text{ g} of magnesium oxide in the following reaction.
2\text{Mg}+\text{O}_2\rightarrow 2\text{MgO}
Calculate the percentage yield of the reaction.
[4 marks]
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\%
Question 3: The following reaction is used to form ethanoic acid with water as a waste product.
\text{C}_2\text{H}_6\text{O}+\text{O}_2\rightarrow\text{C}_2\text{H}_4\text{O}_2+\text{H}_2\text{O}
What is the percentage atom economy for the reaction?
[2 marks]
\text{Percentage atom economy}=\dfrac{60}{78}\times100=76.9\%