A balanced equation tells you how many particles react, and because moles are counts of particles, it also tells you the ratio in moles. It says nothing directly about grams.
This lesson belongs to conservation reasoning across a reaction. If converting grams to moles is still slow, revise converting between mass, moles and particles first.
What does a coefficient actually count?
A coefficient counts particles in the same way a recipe counts eggs. In 2H₂ + O₂ → 2H₂O, two hydrogen molecules and one oxygen molecule make two water molecules.
Because one mole is a fixed number of particles, the same numbers describe moles: 2 mol H₂ react with 1 mol O₂ to give 2 mol H₂O. The masses are different, since each kind of molecule has its own mass.
Worked example: from particles to grams
The question: 4.0 g of hydrogen burns completely in oxygen. What mass of water forms? (Relative atomic masses: H = 1, O = 16.)
- Write the equation: 2H₂ + O₂ → 2H₂O.
- Moles of H₂: 4.0 ÷ 2 = 2.0 mol, because one mole of H₂ is 2 g.
- Use the ratio H₂ : H₂O = 2 : 2, so moles of H₂O = 2.0 mol.
- Mass of H₂O: 2.0 × 18 = 36 g.
Now check conservation. The oxygen used is 1.0 mol, which is 32 g, and 4.0 + 32 = 36 g. The mass before equals the mass after, even though the coefficients 2, 1 and 2 never matched any of the masses.
The mistake that costs marks
The slip is to read “2 H₂ gives 2 H₂O” as “4 g gives 4 g”, or to give the answer as a mass equal to the coefficient. The working looks short and neat, so the error is easy to miss.
| Step | Coefficient-as-mass slip | Correct route |
|---|---|---|
| Reading 2H₂ | 2 g of hydrogen | 2 mol of hydrogen |
| Given 4.0 g H₂ | 4.0 g H₂O formed | 2.0 mol H₂, so 2.0 mol H₂O |
| Final answer | 4.0 g | 36 g |
| Conservation check | Fails, because the oxygen that joins in is ignored | Holds, 4.0 + 32 = 36 |
A quick test catches the slip. If water is made from hydrogen and oxygen, the water must weigh more than the hydrogen alone.
A second ratio, not 1 : 1
Try N₂ + 3H₂ → 2NH₃ with 0.60 mol of hydrogen. The ratio H₂ : NH₃ is 3 : 2, so the moles of ammonia are 0.60 × 2 ÷ 3 = 0.40 mol.
Multiply by 17 g mol⁻¹ and the mass is 6.8 g. The ratio step is the only place the coefficients are used, and it always sits between two mole values. The mole and stoichiometry steps tool lays out the same sequence if you want to test your own numbers.
Check yourself
2.4 g of magnesium reacts completely with excess dilute hydrochloric acid: Mg + 2HCl → MgCl₂ + H₂. Find the volume of hydrogen at room conditions, where one mole of gas occupies 24 dm³. (Mg = 24.)
Answer
Moles of Mg = 2.4 ÷ 24 = 0.10 mol.
The ratio Mg : H₂ is 1 : 1, so moles of H₂ = 0.10 mol.
Volume = 0.10 × 24 = 2.4 dm³.
The coefficient 2 in front of HCl does not change the answer, because the question asks about H₂ and the ratio only compares Mg with H₂.
What to study next
The next step is to decide which reactant runs out first, in identifying the limiting reactant from a complete dataset. Test the whole cluster with the conservation reasoning practice set.
If you want a teacher to trace exactly where your mole reasoning breaks, see online one-to-one Chemistry tuition.