Calculating the mass of product
A theoretical yieldThe maximum possible mass of a product that can be made in a chemical reaction. is the maximum possible massThe amount of matter an object contains. Mass is measured in kilograms (kg) or grams (g). of a productA substance formed in a chemical reaction. that can be made in a chemical reaction.
It can be calculated using:
- the balanced chemical equationA chemical equation written using the symbols and formulae of the reactants and products, so that the number of units of each element present is the same on both sides of the arrow.
- the mass and relative formula massThe sum of the relative atomic masses of the atoms in a chemical formula. of the limiting reactantThe reacting substance that is completely used up in a chemical reaction and which determines how much product is made.
- the relative formula mass of the product
Worked example 1
Question
Carbon reacts with oxygen to produce carbon dioxide:
C(s) + O2(g) → CO2(g)
Calculate the maximum mass of carbon dioxide that can be made from 6.0 g of carbon and an excess of oxygen. (Relative atomic masses: C = 12.0, O = 16.0)
relative formula mass, Mr, of CO2 = 12.0 + (2 Ă— 16.0) = 44.0
Looking at the balanced equation:
- sum of Ar for C = 12.0
- sum of Mr for CO2 = 44.0
\(theoretical\ yield\ =\ \frac{mass\ of\ limiting\ reactant}{sum\ of\ \mathit{A}_{r}\ or\ \mathit{M}_{r}\ of\ limiting\ reactant}\ \)
\( \times\ sum\ of\ \mathit{A}_{r}\ or \mathit{M}_{r}\ of\ product\)
\(theoretical\ yield\ =\ \frac{6.0}{12.0}\ \times\ 44.0\)
= 22.0 g
Worked example 2
Question
Nitrogen reacts with hydrogen to produce ammonia:
N2(g) + 3H2(g) → 2NH3(g)
Calculate the maximum mass of ammonia that can be made from an excess of nitrogen and 12.0 g of hydrogen. (Relative atomic masses: H = 1.0, N = 14.0)
relative formula mass, Mr, of H2 = (2 Ă— 1.0) = 2.0
relative formula mass, Mr, of NH3 = 14.0 + (3 Ă— 1.0) = 17.0
- sum of Mr for H2 = (3 Ă— 2.0) = 6.0
- sum of Mr for NH3 = (2 Ă— 17.0) = 34.0
\(theoretical\ yield\ =\ \frac{mass\ of\ limiting\ reactant}{sum\ of\ \mathit{A}_{r}\ or\ \mathit{M}_{r}\ of\ limiting\ reactant}\ \times\)
\(\ sum\ of\ \mathit{A}_{r}\ or \mathit{M}_{r}\ of\ product\)
\(theoretical\ yield\ =\ \frac{12.0}{6.0}\ \times\ 34.0\)
= 68.0 g
Question
Lithium hydroxide is used to absorb exhaled carbon dioxide in spacecraft:
2LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l)
Calculate the maximum mass of water that can be made from an excess of carbon dioxide and 95.6 g of lithium hydroxide. (Relative atomic masses: H = 1.0, Li = 6.9, O = 16.0)
relative formula mass, Mr, of LiOH = 6.9 + 16.0 + 1.0 = 23.9
relative formula mass, Mr, of H2O = (2 Ă— 1.0) + 16.0 = 18.0
Looking at the balanced equation:
- sum of Mr for LiOH = (2 Ă— 23.9) = 47.8
- sum of Mr for H2O = 18.0
\(theoretical\ yield\ =\ \frac{mass\ of\ limiting\ reactant}{sum\ of\ \mathit{A}_{r}\ or\ \mathit{M}_{r}\ of\ limiting\ reactant}\ \times\)
\(\ sum\ of\ \mathit{A}_{r}\ or \mathit{M}_{r}\ of\ product\)
\(theoretical\ yield\ =\ \frac{95.6}{47.8}\ \times\ 18.0\)
= 36.0 g