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Refresher Practice: Moles and Stoichiometry

Twelve short problems on the mole and on reaction stoichiometry, in different formats: more or less than a mole, true or false, matching molar masses, balancing equations, a quick-fire round with new numbers every time, finding a mistake, and multi-part problems that end in a workplace decision. If moles are new to you, or it has been a while, start with Problem 1 and open How to do a stoichiometry calculation. Each answer is checked as you go, and the working is shown once each step is correct.

Toolbox: atomic masses and key relationships

Atomic masses

  • H = 1 g mol−1
  • C = 12 g mol−1
  • N = 14 g mol−1
  • O = 16 g mol−1

The mole

  • 1 mol = 6.02 × 1023 particles
  • n = m ÷ M
  • m = n × M
  • N = n × 6.02 × 1023 mol−1

Molar mass

  • Add up the atomic masses in the formula
  • M(H2O) = 2 × 1 + 1 × 16 = 18 g mol−1

Mass units

  • 1 kg = 1,000 g
  • 1 g = 1,000 mg
  • 1 mg = 1,000 µg
How to do a stoichiometry calculation
  1. Write the balanced equation. The coefficients give the ratio of moles that react.
  2. Convert the mass you are given into moles: n = m ÷ M.
  3. Use the coefficients to find the moles of the substance you want, for example 1 mol CH4 gives 2 mol H2O.
  4. Work out the molar mass of the substance you want.
  5. Convert its moles into mass: m = n × M. Check: the total mass of the products must equal the total mass of the reactants.

About the questions

  • The atomic masses are whole numbers, as in the lecture slides: H = 1, C = 12, N = 14 and O = 16 g mol−1. They are collected in the toolbox.
  • Small rounding differences are accepted, so you do not need to match every decimal place.
  • Large and small numbers can be typed as 6.02e23 or 6.02 x 10^23.
  • The quick-fire round gives different questions and numbers every time, so you can come back to it as often as you like.
  • Your answers are saved in this browser, so you can leave and come back later.