Reacting Masses, Yield and Limiting Reagents

AS · 9 min

A balanced equation is a recipe written in moles. Once you can turn masses into moles and back, you can predict how much product a reaction should give, work out which reactant runs out first, compare the actual yield with the theoretical one, and even deduce an unknown equation from experimental masses. These calculations appear in every Paper 2 and in Paper 3 data analysis, usually as a multi-step question where each step carries a mark.

The mole ratio is the bridge

The coefficients in a balanced equation give the ratio of amounts in moles, not masses. In

\ce{2Mg(s) + O2(g) -> 2MgO(s)}

two moles of magnesium react with one mole of oxygen molecules to make two moles of magnesium oxide. The masses are in the ratio $48.6 : 32.0 : 80.6$ , which is not $2 : 1 : 2$ .

Method

Reacting-mass calculation

Write the balanced equation.
Convert the known mass to moles: $n = m / M$ .
Use the mole ratio from the equation to find moles of the substance you want.
Convert back to mass: $m = n \times M$ .
Round the final answer sensibly (normally to the same number of significant figures as the data, usually 3).

Mass of product from a thermal decomposition

Calculate the mass of calcium oxide formed when $25.0\ \text{g}$ of calcium carbonate is heated until it has completely decomposed.

Solution

\ce{CaCO3(s) -> CaO(s) + CO2(g)}

$n(\ce{CaCO3}) = \dfrac{25.0}{100.1} = 0.2498\ \text{mol}$

Mole ratio $\ce{CaCO3} : \ce{CaO} = 1 : 1$ , so $n(\ce{CaO}) = 0.2498\ \text{mol}$ .

$m(\ce{CaO}) = 0.2498 \times 56.1 = 14.0\ \text{g}$

Percentage yield

The mass calculated from the equation is the theoretical yield: the maximum mass of product if every particle of the limiting reactant reacts as the equation says. In practice the actual yield is less, because:

the reaction may be reversible and not go to completion;
side reactions make other products;
product is lost during transfer, filtration, distillation or purification;
some reactant may be impure.

Key result

\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100

Both yields must be in the same units (both masses or both amounts in moles).

Percentage yield of an ester

$6.00\ \text{g}$ of ethanoic acid is heated with excess ethanol and a little concentrated sulfuric acid. $6.16\ \text{g}$ of ethyl ethanoate is collected. Calculate the percentage yield.

\ce{CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O}

Solution

$n(\ce{CH3COOH}) = \dfrac{6.00}{60.0} = 0.100\ \text{mol}$

Mole ratio $1 : 1$ , so the theoretical $n(\text{ester}) = 0.100\ \text{mol}$ .

Theoretical mass $= 0.100 \times 88.0 = 8.80\ \text{g}$ .

\text{percentage yield} = \frac{6.16}{8.80} \times 100 = 70.0\%

A large part of the shortfall here is because esterification is reversible and reaches equilibrium.

Limiting and excess reagents

Reactants are often not mixed in the exact ratio of the equation. The reactant that is used up first is the limiting reagent; it decides how much product forms. The other reactant is in excess, and some of it is left over.

Method

Finding the limiting reagent

Calculate the moles of each reactant.
Divide each by its coefficient in the equation. The smallest result identifies the limiting reagent.
Use the moles of the limiting reagent, with the mole ratio, to find the product.
To find the excess left: moles of excess reagent present minus moles that react.

Iron and sulfur

$5.58\ \text{g}$ of iron is heated with $4.00\ \text{g}$ of sulfur: $\ce{Fe(s) + S(s) -> FeS(s)}$ . Identify the limiting reagent and calculate the mass of iron(II) sulfide formed and the mass of the reactant left over.

Solution

$n(\ce{Fe}) = \dfrac{5.58}{55.8} = 0.100\ \text{mol}$ ; $n(\ce{S}) = \dfrac{4.00}{32.1} = 0.1246\ \text{mol}$ .

The ratio is $1 : 1$ , and there is less iron, so iron is limiting.

$n(\ce{FeS}) = 0.100\ \text{mol}$ ; $m = 0.100 \times (55.8 + 32.1) = 0.100 \times 87.9 = 8.79\ \text{g}$ .

Sulfur left $= 0.1246 - 0.100 = 0.0246\ \text{mol}$ , mass $= 0.0246 \times 32.1 = 0.790\ \text{g}$ .

A limiting reagent when the ratio is not 1 : 1

$2.70\ \text{g}$ of aluminium reacts with $7.10\ \text{g}$ of chlorine: $\ce{2Al(s) + 3Cl2(g) -> 2AlCl3(s)}$ . Find the mass of aluminium chloride formed.

Solution

$n(\ce{Al}) = \dfrac{2.70}{27.0} = 0.100\ \text{mol}$ ; $n(\ce{Cl2}) = \dfrac{7.10}{71.0} = 0.100\ \text{mol}$ .

Divide by coefficients: $\ce{Al}$ : $0.100 / 2 = 0.0500$ ; $\ce{Cl2}$ : $0.100 / 3 = 0.0333$ . Chlorine gives the smaller value, so chlorine is limiting (0.100 mol of Al would need 0.150 mol of $\ce{Cl2}$ ).

$n(\ce{AlCl3}) = 0.100 \times \tfrac{2}{3} = 0.0667\ \text{mol}$ ; $m = 0.0667 \times 133.5 = 8.90\ \text{g}$ .

Watch out

Equal numbers of moles do not mean neither is limiting. Always compare against the equation's ratio. In the aluminium example, 0.100 mol of each looks balanced, but the equation needs 1.5 times as much chlorine as aluminium.

Deducing an equation from reacting masses

Experimental masses can reveal the stoichiometry of a reaction you have not seen before. Convert every mass to moles and find the simplest whole-number ratio.

Exam-style: deducing the formula of a chloride

$1.12\ \text{g}$ of iron reacts completely with chlorine to form $3.25\ \text{g}$ of an iron chloride. Deduce the formula of the chloride and write an equation for its formation.

Solution

Mass of chlorine combined $= 3.25 - 1.12 = 2.13\ \text{g}$ .

$n(\ce{Fe}) = \dfrac{1.12}{55.8} = 0.02007\ \text{mol}$ ; $n(\ce{Cl}) = \dfrac{2.13}{35.5} = 0.0600\ \text{mol}$ .

Ratio $\ce{Fe} : \ce{Cl} = 0.02007 : 0.0600 = 1 : 2.99 \approx 1 : 3$ . The chloride is $\ce{FeCl3}$ , iron(III) chloride.

\ce{2Fe(s) + 3Cl2(g) -> 2FeCl3(s)}

Exam-hard: composition of a mixture from mass loss

A $10.00\ \text{g}$ mixture of sodium hydrogencarbonate and sodium chloride is heated to constant mass. The mass decreases by $2.17\ \text{g}$ . Sodium chloride does not decompose. Calculate the percentage by mass of sodium hydrogencarbonate in the mixture.

\ce{2NaHCO3(s) -> Na2CO3(s) + CO2(g) + H2O(g)}

Solution

The mass lost is the carbon dioxide and water that escape. For every 2 mol of $\ce{NaHCO3}$ , 1 mol of $\ce{CO2}$ (44.0 g) and 1 mol of $\ce{H2O}$ (18.0 g) are lost: $62.0\ \text{g}$ per 2 mol of $\ce{NaHCO3}$ .

n(\ce{NaHCO3}) = 2 \times \frac{2.17}{62.0} = 0.0700\ \text{mol}

$m(\ce{NaHCO3}) = 0.0700 \times 84.0 = 5.88\ \text{g}$

\text{percentage} = \frac{5.88}{10.00} \times 100 = 58.8\%

Significant figures in calculations

Cambridge expects answers to reflect the precision of the data.

Give the final answer to the same number of significant figures as the least precise data value, unless the question says otherwise. With data to 3 s.f., give 3 s.f.
Never round an intermediate value to fewer figures than the final answer needs. Carry at least one extra figure, or keep the full value in your calculator.
Do not "lose" significant figures: $14.0\ \text{g}$ is 3 s.f., but $14\ \text{g}$ is only 2.

Exam tip

Show every step: equation, moles of the known, ratio, moles of the unknown, mass. Method marks are awarded for each, so a slip in arithmetic costs one mark, not all of them.
Label each quantity: " $n(\ce{CaCO3}) = \ldots$ ". It helps the examiner follow you and stops you dividing by the wrong molar mass.
In limiting reagent questions, state explicitly which reagent is limiting and why.
A percentage yield over 100% means a mistake (or an impure, wet product in a practical question).

Summary

Equations give mole ratios, not mass ratios.
Method: mass to moles, use the ratio, moles to mass.
Percentage yield $= \dfrac{\text{actual}}{\text{theoretical}} \times 100$ . Yields are below 100% because of incomplete or reversible reactions, side reactions and losses.
The limiting reagent has the smallest value of moles divided by its coefficient; it fixes the amount of product.
Experimental masses can be turned into a mole ratio to deduce a formula or an equation.
Keep extra figures during working; round at the end to the precision of the data.

Practice

Question

Calculate the mass of iron that could be extracted from $1.00\ \text{kg}$ of iron(III) oxide: $\ce{Fe2O3 + 3CO -> 2Fe + 3CO2}$ .
Calculate the mass of oxygen needed to burn $11.6\ \text{g}$ of butane completely: $\ce{2C4H10 + 13O2 -> 8CO2 + 10H2O}$ .
$12.0\ \text{g}$ of magnesium is burnt and $18.0\ \text{g}$ of magnesium oxide is collected. Calculate the percentage yield.
$10.0\ \text{g}$ of hydrogen reacts with $40.0\ \text{g}$ of oxygen to form water. Identify the limiting reagent, the mass of water formed, and the mass of the excess reagent left.
$2.00\ \text{g}$ of a copper oxide is reduced by hydrogen to $1.60\ \text{g}$ of copper. Deduce the formula of the oxide.
Dehydration of $9.20\ \text{g}$ of ethanol gives $3.92\ \text{g}$ of ethene: $\ce{C2H5OH -> C2H4 + H2O}$ . Calculate the percentage yield.
Propan-1-ol is oxidised to propanoic acid with a 75.0% yield. What mass of propan-1-ol is needed to produce $6.00\ \text{g}$ of propanoic acid?
Calculate the mass of magnesium oxide formed when $7.42\ \text{g}$ of magnesium nitrate decomposes: $\ce{2Mg(NO3)2 -> 2MgO + 4NO2 + O2}$ .
A $2.00\ \text{g}$ mixture of calcium carbonate and magnesium carbonate is heated to constant mass, leaving $1.055\ \text{g}$ of a mixture of $\ce{CaO}$ and $\ce{MgO}$ . Calculate the percentage by mass of calcium carbonate in the original mixture.
$1.00\ \text{g}$ of calcium reacts completely with a halogen, $\ce{X2}$ , to form $2.77\ \text{g}$ of $\ce{CaX2}$ . Identify X.

Answers

$n(\ce{Fe2O3}) = 1000 / 159.6 = 6.266\ \text{mol}$ ; $n(\ce{Fe}) = 2 \times 6.266 = 12.53\ \text{mol}$ ; $m = 12.53 \times 55.8 = 699\ \text{g}$ .
$n(\ce{C4H10}) = 11.6 / 58.0 = 0.200\ \text{mol}$ ; $n(\ce{O2}) = 0.200 \times \tfrac{13}{2} = 1.30\ \text{mol}$ ; $m = 1.30 \times 32.0 = 41.6\ \text{g}$ .
Theoretical: $n(\ce{Mg}) = 12.0 / 24.3 = 0.4938\ \text{mol} = n(\ce{MgO})$ ; $m = 0.4938 \times 40.3 = 19.9\ \text{g}$ . Yield $= 18.0 / 19.9 \times 100 = 90.4\%$ .
$\ce{2H2 + O2 -> 2H2O}$ . $n(\ce{H2}) = 10.0 / 2.0 = 5.00$ ; $n(\ce{O2}) = 40.0 / 32.0 = 1.25$ . Divide by coefficients: $2.50$ and $1.25$ , so oxygen is limiting. $n(\ce{H2O}) = 2.50\ \text{mol}$ , $m = 45.0\ \text{g}$ . Hydrogen used $= 2.50\ \text{mol}$ ; left $= 2.50\ \text{mol} = 5.0\ \text{g}$ .
$n(\ce{Cu}) = 1.60 / 63.5 = 0.0252$ ; $n(\ce{O}) = 0.40 / 16.0 = 0.0250$ . Ratio $1 : 1$ : $\ce{CuO}$ .
$n(\ce{C2H5OH}) = 9.20 / 46.0 = 0.200\ \text{mol}$ ; theoretical ethene $= 0.200 \times 28.0 = 5.60\ \text{g}$ ; yield $= 3.92 / 5.60 \times 100 = 70.0\%$ .
$n(\text{acid}) = 6.00 / 74.0 = 0.08108\ \text{mol}$ . With 75.0% yield, $n(\text{alcohol}) = 0.08108 / 0.750 = 0.1081\ \text{mol}$ ; $m = 0.1081 \times 60.0 = 6.49\ \text{g}$ .
$M(\ce{Mg(NO3)2}) = 148.3$ ; $n = 7.42 / 148.3 = 0.05003\ \text{mol} = n(\ce{MgO})$ ; $m = 0.05003 \times 40.3 = 2.02\ \text{g}$ .
Let the mass of $\ce{CaCO3}$ be $x$ g; $\ce{MgCO3}$ is $(2.00 - x)$ g. Mass of $\ce{CaO}$ $= x \times \tfrac{56.1}{100.1} = 0.5604x$ ; mass of $\ce{MgO}$ $= (2.00 - x) \times \tfrac{40.3}{84.3} = 0.4781(2.00 - x)$ . So $0.5604x + 0.9561 - 0.4781x = 1.055$ , giving $0.0824x = 0.0989$ and $x = 1.20\ \text{g}$ . Percentage $= 1.20 / 2.00 \times 100 = 60.0\%$ .
$n(\ce{Ca}) = 1.00 / 40.1 = 0.02494\ \text{mol} = n(\ce{CaX2})$ . Mass of X $= 2.77 - 1.00 = 1.77\ \text{g}$ , which is $2 \times 0.02494 = 0.04988\ \text{mol}$ of X atoms. $A_r(\ce{X}) = 1.77 / 0.04988 = 35.5$ : X is chlorine.

The mole ratio is the bridge

Percentage yield

Limiting and excess reagents

Deducing an equation from reacting masses

Significant figures in calculations

Practice

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