Moles and stoichiometry

Chemistry · Senior school chemistry · about 10 min

Chemical equations describe particles reacting in fixed ratios, but laboratory measurements are usually masses. The mole connects these two descriptions. Stoichiometry then uses the balanced equation to predict how much reactant is needed or how much product can form.

The key idea

Convert mass to amount of substance using

n=\frac{m}{M}

Here $n$ is amount in $\mathrm{mol}$ , $m$ is mass in $\mathrm{g}$ , and $M$ is molar mass in $\mathrm{g\,mol^{-1}}$ . Balanced equation coefficients give mole ratios.

Draw a single horizontal reaction schematic. On the left place two distinct diatomic hydrogen molecules (each shown as two small joined circles) and one diatomic oxygen molecule (two larger joined circles). Place a rightward reaction arrow in the centre. On the right place two water molecules, each with one large oxygen circle joined to two smaller hydrogen circles in a bent shape. Keep every molecule separate so atoms can be counted. Under the groups write their chemical formulae and coefficients exactly as provided. — Balanced coefficients relate molecule counts and mole amounts.

Convert between quantities

One mole contains exactly $6.02214076\times10^{23}$ specified entities. If the sample contains $N$ entities, then $n=N/N_A$ , where $N_A$ is the Avogadro constant. For mass calculations, find the molar mass from the chemical formula, then divide the measured mass by it. The mass and molar mass units must agree. For example, using grams with grams per mole gives moles. A subscript in a chemical formula counts atoms inside one molecule; a coefficient in a balanced equation counts the relative numbers of whole molecules or moles.

Follow the balanced ratio

Balance the reaction before calculating anything. For $2\mathrm{H_2}+\mathrm{O_2}\rightarrow2\mathrm{H_2O}$ , two moles of hydrogen react with one mole of oxygen to form two moles of water. A reliable calculation route is mass to moles, mole ratio, then moles to the required final unit. Never use a mass ratio taken directly from equation coefficients. The coefficients represent amounts of substance, not masses. Keep extra digits until the end so a small rounding error does not change the final answer.

Find the limiting reactant

When more than one starting reactant amount is given, determine the maximum product each could make. The reactant giving the smaller maximum product amount runs out first and limits the reaction. You can also divide each available mole amount by its coefficient; the smaller ratio identifies the limiting reactant for the stated reaction. Use that reactant to calculate theoretical yield. If an actual product mass is supplied, percentage yield is $(\text{actual}/\text{theoretical})\times100\%$ . This yield can be below $100\%$ because of incomplete reaction or product loss.

Worked example: Carbon dioxide from methane combustion

Methane reacts completely with oxygen according to $\mathrm{CH_4}+2\mathrm{O_2}\rightarrow\mathrm{CO_2}+2\mathrm{H_2O}$ . A reaction starts with $8.00\,\mathrm{g}$ of methane and $30.0\,\mathrm{g}$ of oxygen. Determine the limiting reactant and the theoretical mass of carbon dioxide. Use $M(\mathrm{CH_4})=16.04\,\mathrm{g\,mol^{-1}}$ , $M(\mathrm{O_2})=32.00\,\mathrm{g\,mol^{-1}}$ and $M(\mathrm{CO_2})=44.01\,\mathrm{g\,mol^{-1}}$ .

Draw two feed containers at left, one containing methane and labelled CH4 with 8.00 g, the other oxygen labelled O2 with 30.0 g. Draw separate arrows from each container into a central simple reaction vessel. One arrow from the vessel leads to a product area labelled CO2 and H2O with no numerical output amounts. Show the methane and oxygen quantities as input data only. Do not show which reactant is limiting or any answer. — Reactant amounts determine the maximum carbon dioxide formed.

Show the solution

Convert both reactant masses to moles:

n(\mathrm{CH_4})=\frac{8.00}{16.04}\approx0.49875\,\mathrm{mol}

n(\mathrm{O_2})=\frac{30.0}{32.00}=0.9375\,\mathrm{mol}

Each mole of methane needs two moles of oxygen. Consuming all the methane would require $2(0.49875)\approx0.99751\,\mathrm{mol}$ of oxygen, more than is available. Oxygen is limiting.
The equation forms one mole of carbon dioxide for every two moles of oxygen:

n(\mathrm{CO_2})=\frac{0.9375}{2}=0.46875\,\mathrm{mol}

Convert the product amount to mass:

m(\mathrm{CO_2})=0.46875(44.01)=20.6296875\,\mathrm{g}

Answer: oxygen is the limiting reactant; the theoretical carbon dioxide mass is $20.6\,\mathrm{g}$ (3 significant figures).

Common mistakes

Using an unbalanced equation

Balance atoms on both sides before taking mole ratios.

Comparing reactant masses directly

Convert each mass to moles, then account for the balanced coefficients.

Using the excess reactant for product yield

Calculate theoretical product from whichever reactant is limiting.

Rounding too early

Retain intermediate precision and round only the final result to suitable significant figures.

Check your understanding

1. How many moles are in $4.00\,\mathrm{g}$ of $\mathrm{O_2}$ if $M=32.00\,\mathrm{g\,mol^{-1}}$ ?

2. For $2\mathrm{H_2}+\mathrm{O_2}\rightarrow2\mathrm{H_2O}$ , how much water can form from $0.30\,\mathrm{mol}$ of oxygen with excess hydrogen?

3. Approximately how many molecules are in $0.50\,\mathrm{mol}$ of a molecular substance?

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