Every calculation in relative masses and amounts can be checked in under a minute. Two checks do most of the work: follow the units through each line, and ask whether the size of the answer is sensible. This lesson shows both so you can use them on any question in the module.
What does a units check look like?
Write the unit beside each number and treat the units like algebra.
- mass ÷ molar mass: g ÷ (g/mol) = mol
- amount × molar mass: mol × (g/mol) = g
- amount × ratio: mol × (mol ÷ mol) = mol
If the units you end with are not the units the question asks for, something has been multiplied or divided the wrong way.
Two more sense checks
Size check. The Mr tells you the mass of 1 mol in grams. So 10 g of a substance with Mr 100 is 0.1 mol, and 10 g of one with Mr 10 is 1 mol. The answer should fit that scale.
Mass conservation. In a reaction model, total mass of reactants used equals total mass of products. This checks the ratio step.
Worked example
A student is asked: what mass of CO₂ forms when 10.0 g of CaCO₃ decomposes fully? The equation is CaCO₃ → CaO + CO₂. (Ca = 40, C = 12, O = 16)
Calculation with units:
n(CaCO₃) = 10.0 g ÷ 100 g/mol = 0.100 mol.
Ratio 1 : 1, so n(CO₂) = 0.100 mol.
m(CO₂) = 0.100 mol × 44 g/mol = 4.4 g.
Checks:
- Units: g ÷ (g/mol) = mol, then mol × (g/mol) = g. The final unit is g, as asked.
- Size: 4.4 g is less than 10.0 g, which suits a model where a gas leaves a solid.
- Mass conservation: CaO = 0.100 × 56 = 5.6 g. Then 5.6 + 4.4 = 10.0 g, equal to the start.
A plausible mistake
A student writes: n = 10.0 × 100 = 1000 mol.
Mistaken answer: 1000 mol
The unit of this line is g × (g/mol) = g²/mol, which is not mol. A size check also fails: 10 g of a substance with Mr 100 cannot be 1000 moles.
The correction is to divide: 10.0 ÷ 100 = 0.100 mol. The units check would have flagged the error before the next step.
Check yourself
1. What unit results from 0.20 mol × 98 g/mol? Calculate the value.
Show answer
mol × g/mol leaves g. 0.20 × 98 = 19.6 g.
2. A student says 2.0 g of hydrogen reacts with 16 g of oxygen to form 18 g of water, in 2H₂ + O₂ → 2H₂O. Use moles to check this. (H = 1, O = 16)
Show answer
H₂: 2.0 ÷ 2 = 1.0 mol. O₂: 16 ÷ 32 = 0.50 mol. The ratio 2 : 1 holds, so both are used up. H₂O = 1.0 mol × 18 = 18 g. Mass is conserved: 2.0 + 16 = 18 g. The statement is correct.
3. A student finds the amount in 8.0 g of NaOH as 8.0 × 40 = 320 mol. Explain the error and give the right answer. (Na = 23, O = 16, H = 1)
Show answer
Multiplying g by g/mol gives g²/mol, not mol. The correct step is division: 8.0 ÷ 40 = 0.20 mol.
Where this leads next
Try your new habit on the mixed practice set for this module. The mole and equation-ratio tutor writes units on every line, which is a good pattern to copy, and the equation balance reasoning trainer supports the equation step.
Having a routine for checking your own work is a skill a teacher can build with you, question by question, in online one-to-one Chemistry tuition.