Skip to content
Dabour HubLEARN · PRACTICE · EXPLORESign inStudent

5.1.2Additional lessons

Reacting Mass Calculations — IB Chemistry SL Revision Notes & Topical Questions

5 min read 6 parts Lesson 60 of 98

📚 IB Chemistry SL Hub / Topical Revision
🩺
100% Curriculum Specification Match Lesson Doctor Verified
The only platform with 100% curriculum specification match and verified interactive question banks.
0%read

Part 1 of 6The Mole Link

5.1.2 Reacting Mass Calculations

The mole links the mass of a substance to the number of particles in the equation.

Amount
  • The amount in moles equals mass in grams divided by molar mass in g mol⁻¹, written n = m/M.
Molar mass
  • Add the relative atomic masses of all atoms in the formula, using the values in the data booklet.
Mole ratio
  • The coefficients of a balanced equation give the ratio in which amounts of substances react and form.
Mass ratio
  • Reacting masses are not in the ratio of the coefficients, since different substances have different molar masses.

Part 2 of 6Method

1
Write the balanced equation for the reaction and identify the substance whose mass is given and the one whose mass is required.
2
Convert the given mass to moles using n = m/M.
3
Use the coefficients to find the moles of the required substance by ratio.
4
Convert these moles into a mass using m = n × M.
5
Round the answer to the same number of significant figures as the least precise data and include units.

Part 3 of 6Worked Pattern

Worked Pattern
Given
  • Always the starting point, whether it is a reactant or a product.
Ratio
  • The relation that makes the calculation specific to the equation, such as 1 : 1 or 2 : 3.
Required
  • The quantity asked for, found by the last step of converting from moles to mass.
Check
  • A product mass should not exceed the total mass of reactants, which follows from conservation of mass.
StepFormula
Moles from massn = m ÷ M
Moles by ration(B) = n(A) × coefficient B ÷ coefficient A
Mass from molesm = n × M

Part 4 of 6Typical Contexts

Typical Contexts
Thermal decomposition
  • Heating a carbonate gives an oxide and CO₂, and the loss in mass can be predicted from the equation.
Metals and acids
  • The mass of metal that reacts can be used to find the mass of a salt formed.
Industrial scale
  • Chemical manufacturers use these calculations to decide how much raw material to buy.
Experimental analysis
  • Measured masses can be compared with theoretical values to find yield and purity.

Part 5 of 6Worked Example

Question

  • Calculate the mass of calcium oxide formed when 25.0 g of calcium carbonate is heated: CaCO₃(s) → CaO(s) + CO₂(g).
  • (Mr CaCO₃ = 100.09, Mr CaO = 56.08)

  1. Amount of CaCO₃ = 25.0 ÷ 100.09 = 0.2498 mol.
  2. The ratio CaCO₃CaO is 1 : 1, so n(CaO) = 0.2498 mol.
  3. Mass of CaO = 0.2498 × 56.08 = 14.0 g.

Answer 14.0 g of calcium oxide

Part 6 of 6Exam Checklist & Tips

5.1.2 – Key Points for the Exam

  • Molar mass The mass of one mole of a substance, in g mol⁻¹.
  • Mole ratio The ratio of coefficients in a balanced equation.
  • Amount of substance The quantity measured in moles.
  • Theoretical mass The mass predicted from the balanced equation, assuming complete reaction.

Common Mistakes

  • Using masses directly in the ratio of the equation rather than converting to moles first.
  • Using an unbalanced equation to get the mole ratio.
  • Using the wrong molar mass because a subscript or bracket was ignored.
  • Giving the answer to too many significant figures.

Last updated:

\n