Why masses must become moles

Stoichiometry describes the relative amounts of substances that react and form products. The coefficients in a balanced equation give their mole ratio. For example, $\ce{2H2 + O2 -> 2H2O}$ means 2 mol of hydrogen molecules react with 1 mol of oxygen molecules to make 2 mol of water molecules.

This is not a gram ratio. Different substances have different masses per mole, so convert each mass to moles before comparing them.

Use the formula of the substance actually reacting. For $\ce{O2}$, the relative formula mass is $2 \times 16 = 32$, not 16. Keep the small numbers inside formulae fixed: changing $\ce{O2}$ to O would change the substance.

Worked example: magnesium and oxygen

4.8 g of magnesium reacts with 3.2 g of oxygen, forming 8.0 g of magnesium oxide. These are the masses actually consumed and formed.

The mole ratio is 0.20 : 0.10 : 0.20. Dividing every term by 0.10 gives 2 : 1 : 2, so the equation is $\ce{2Mg + O2 -> 2MgO}$. There are two Mg atoms and two O atoms on each side.

Turn fractional ratios into coefficients

Dividing by the smallest mole value makes the smallest term 1. It does not always give whole numbers. If a simple fraction remains, multiply every term by the same small integer. For example, multiply by 2 to remove halves or by 3 to remove thirds, then reduce to the smallest whole-number ratio.

Worked example: iron(III) oxide

11.2 g of iron reacts with 4.8 g of oxygen to form 16.0 g of iron(III) oxide, $\ce{Fe2O3}$.

Divide by 0.10:

$\ce{Fe} : \ce{O2} : \ce{Fe2O3} = 2 : 1.5 : 1$

Multiply all three terms by 2:

$\ce{Fe} : \ce{O2} : \ce{Fe2O3} = 4 : 3 : 2$

Therefore $\ce{4Fe + 3O2 -> 2Fe2O3}$. Check: each side contains four Fe atoms and six O atoms. Rounding 1.5 to 2 would change the ratio and give an equation that does not conserve atoms.

You can also divide the original mole values by another common amount, such as 0.05 mol here, to get 4 : 3 : 2 directly. Any method that preserves the relative amounts and gives the simplest whole-number ratio is valid.

Check which masses the data describe

Use masses actually consumed or formed, not an initial excess reactant mass. If 5.0 g is supplied but 1.0 g remains unreacted, only 4.0 g belongs in the reaction mole ratio.

Product masses are evidence too. Include every supplied reactant and product in the mole ratio, keeping them in a named order. For example, 1.20 g C, 3.20 g O2 and 4.40 g CO2 correspond to 0.100 : 0.100 : 0.100 mol, using molar masses of 12, 32 and 44 g mol⁻¹. This gives $\ce{C + O2 -> CO2}$. The masses also add up: $1.20 + 3.20 = 4.40\,\mathrm{g}$. Conservation of total mass is a useful check, but it does not replace the mole calculation.

Keep enough digits until you have compared the mole values. Experimental data may give a ratio near a simple value: 1 : 1.99 : 1 could support 1 : 2 : 1 within measurement uncertainty. A value such as 1.5 must instead be handled by multiplying the ratio. If the numbers disagree substantially with the proposed equation, recheck formulae, calculations, unreacted material and measurement uncertainty rather than forcing a match.

To finish any calculation, check the coefficients against the atoms on both sides. Change coefficients in front of formulae, never the small numbers within them.