Convert the known mass to moles with n = mass / molar mass, apply the mole ratio from the balanced equation, then convert back with mass = moles x molar mass.
This page walks through mass of product or reactant for SPM Chemistry, step by step: the formula you need, the units to watch, and a worked example.
When you use this
This is the workhorse stoichiometry calculation in SPM Chemistry: you are given the mass of one substance and asked for the mass of another. Three fixed steps always solve it, convert to moles, use the ratio, convert back.
The formula and units
- moles: n = mass / molar mass, so n = m / M
- rearranged: m = n x M
- molar mass M is the relative formula mass in grams per mole (g mol−1)
Units. Mass m in grams (g), molar mass M in g mol−1, amount n in moles (mol). Use the relative atomic masses given in the exam (for example H = 1, C = 12, O = 16, Mg = 24, Ca = 40, Fe = 56).
The three steps: (1) write the balanced equation; (2) n(known) = m / M and apply the mole ratio to get n(wanted); (3) m(wanted) = n x M.
Worked example 1 (easy)
Magnesium burns completely: 2Mg + O2 → 2MgO. Find the mass of magnesium oxide formed from 6.0 g of magnesium. (Mg = 24, O = 16)
- n(Mg) = 6.0 / 24 = 0.25 mol.
- Ratio Mg : MgO = 2 : 2 = 1 : 1, so n(MgO) = 0.25 mol.
- M(MgO) = 24 + 16 = 40 g mol−1.
- m(MgO) = 0.25 x 40 = 10 g.
Worked example 2 (medium)
Calcium carbonate decomposes on heating: CaCO3 → CaO + CO2. Find the mass of calcium oxide formed from 25 g of calcium carbonate. (Ca = 40, C = 12, O = 16)
- M(CaCO3) = 40 + 12 + (16 x 3) = 100 g mol−1.
- n(CaCO3) = 25 / 100 = 0.25 mol.
- Ratio CaCO3 : CaO = 1 : 1, so n(CaO) = 0.25 mol.
- M(CaO) = 40 + 16 = 56 g mol−1.
- m(CaO) = 0.25 x 56 = 14 g.
Worked example 3 (SPM level)
Iron(III) oxide is reduced in a blast furnace: Fe2O3 + 3CO → 2Fe + 3CO2. Find the mass of iron produced from 320 g of iron(III) oxide. (Fe = 56, O = 16)
- M(Fe2O3) = (56 x 2) + (16 x 3) = 160 g mol−1.
- n(Fe2O3) = 320 / 160 = 2.0 mol.
- Ratio Fe2O3 : Fe = 1 : 2, so n(Fe) = 2.0 x 2 = 4.0 mol.
- m(Fe) = 4.0 x 56 = 224 g.
Notice the ratio step is where marks are won: skip it and you would wrongly get 112 g.
The method in words
Every mass-to-mass question is the same journey with different names: grams in, moles across, grams out. Start by writing the balanced equation at the top of your working, because the ratio you need lives there and nowhere else. Convert the mass you are given into moles, because the equation only understands moles, never grams. Cross the bridge using the mole ratio. Then turn the moles of the substance you want back into grams. If you keep this left-to-right layout every time, an examiner can follow each step and award the method marks even when the final arithmetic slips. It also makes checking easy: a quick glance tells you whether you multiplied where you should have divided, which is the single most common slip under time pressure in the exam hall.
Common traps
- Forgetting the mole ratio. Converting to moles and straight back, without the ratio, is the most common error.
- Wrong molar mass. Multiply each atom by its subscript; M(Fe2O3) is 160, not 72.
- Rounding relative atomic masses. Use the exact values on the exam sheet (Cu = 64, Cl = 35.5).
- Mixing up which substance is which. Label every n and M with the formula it belongs to.
Our teachers have students write the three steps as a fixed template so no mark is dropped, whatever the reaction. Once the layout is a habit, even an unfamiliar equation is just numbers slotted into the same frame.
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