The amount of a substance, in moles, equals its mass in grams divided by its molar mass in g/mol. The number is the same as the Mr you found in the previous lesson. This lesson works in both directions, from mass to moles and back, and keeps the units honest.
Why are units part of the method?
A mole links a count of particles to a mass you can weigh. The molar mass, written in g/mol, is the bridge: it says how many grams one mole weighs.
If the mass is in grams and the molar mass is in g/mol, then g ÷ (g/mol) leaves mol. If the mass is in kg, the units do not cancel properly, so you convert first.
Steps for mass to amount
- Write the mass with its unit, and convert to g if needed.
- Find the Mr and write it as g/mol.
- Divide: n = m ÷ Mr.
- Write the unit mol beside the answer.
For the reverse, use m = n × Mr, which leaves g.
Worked example
How many moles are in 2.0 kg of sodium hydroxide, NaOH? Use Na = 23, O = 16, H = 1.
Step 1, convert: 2.0 kg × 1000 = 2000 g.
Step 2, Mr: 23 + 16 + 1 = 40, so the molar mass is 40 g/mol.
Step 3, divide: 2000 g ÷ 40 g/mol = 50 mol.
Check by going back: 50 mol × 40 g/mol = 2000 g, which is 2.0 kg. The answer returns to the start.
A plausible mistake
A student writes: n = 2.0 ÷ 40 = 0.050 mol.
Mistaken answer: 0.050 mol
The 2.0 was in kg but was divided by a molar mass in g/mol. The units do not match, so the answer is 1000 times too small.
The correction is to write the unit next to every number and convert to grams first. Then 2000 ÷ 40 = 50 mol.
Check yourself
1. How many moles are in 9.0 g of water, H₂O? (H = 1, O = 16)
Show answer
Mr = 2 + 16 = 18. n = 9.0 ÷ 18 = 0.50 mol.
2. What mass of sodium chloride, NaCl, is 0.30 mol? (Na = 23, Cl = 35.5)
Show answer
Mr = 23 + 35.5 = 58.5. m = 0.30 × 58.5 = 17.55 g, which is 17.6 g to three significant figures.
3. A sample of 250 mg of copper(II) sulfate, CuSO₄, is weighed. How many moles is it? (Cu = 64, S = 32, O = 16)
Show answer
Convert: 250 mg ÷ 1000 = 0.250 g. Mr = 64 + 32 + 64 = 160. n = 0.250 ÷ 160 = 0.0015625 mol, which is 1.56 × 10⁻³ mol to three significant figures.
Where this leads next
The next step is using an equation ratio to relate reacting amounts, which needs the amount you just found. The mole and equation-ratio tutor shows the same conversion with units on every line, and the equation balance reasoning trainer checks the equation it depends on.
When students keep losing marks on units across topics, a teacher who sees your written working can point to the habit. That is part of what we do in online one-to-one Chemistry tuition.