Cambridge IGCSE Chemistry · Syllabus 0620 · Stoichiometry
Mole
What is Mole?
The mole, symbol mol, is the unit of amount of substance; one mole of any substance contains 6.02 times 10 to the 23 specified particles, which may be atoms, ions, molecules or formula units.
This definition is part of the Stoichiometry chapter in Cambridge IGCSE Chemistry.
Common mistakes with Mole
- 8. Not saying which particle you counted Repair "\(6.02\times10^{23}\) particles" is an incomplete answer. One mole of \(\mathrm{CO_2}\) is \(6.02\times10^{23}\) molecules, and \(1.806\times10^{24}\) atoms. Name the entity every time. Lesson 3.3A Supplement 3.3.2
- "One mole of \(\mathrm{Cl_2}\) contains \(6.02\times10^{23}\) chlorine atoms." Fix It contains \(6.02\times10^{23}\) \(\mathrm{Cl_2}\) molecules, which is \(1.204\times10^{24}\) chlorine atoms.
- "A mole is a very large mass." Fix A mole is an amount of substance, not a mass. One mole of hydrogen molecules has a mass of about \(2\ \mathrm{g}\); one mole of lead atoms has a mass of about \(207\ \mathrm{g}\). Same amount, very different masses.
- Applying the ratio to grams. Fix The ratio belongs to the middle of the route, between two mole values. Cross into moles first, every time.
- Reading \(\mathrm{g\,dm^{-3}}\) as if it were \(\mathrm{mol\,dm^{-3}}\). Fix Check the unit before choosing a route. A \(\mathrm{g\,dm^{-3}}\) value has to pass through the molar mass before it can meet an equation ratio, because ratios only speak in moles.
- Using \(c_1V_1 = c_2V_2\) on a non-1 : 1 reaction. Fix Go through moles. The ratio line is the only place the chemistry enters the calculation, and a method without it cannot be right except by luck.
Examiner tips on Mole
- Why percentages can be used as if they were masses A percentage by mass is the mass present in \(100\ \mathrm{g}\) of the compound. Dividing \(54.55\) by \(12\) is therefore genuinely "the moles of carbon in \(100\ \mathrm{g}\)". Since the method only ever uses the ratio of the amounts, the choice of \(100\ \mathrm{g}\) does not affect the answer — which is why you can start from percentages without converting them first.
- Marking yourself honestly Core paper, out of 25. Mark Q1–Q6 only. A score below about 18 usually means one of two things, and they need different repairs: if the marks went in Q1, Q2, Q3 or Q5 the problem is reading and writing formulae, so return to Lessons 3.1A and 3.1C. If they went in Q4, the problem is the proportion method, and Lesson 3.2B is the whole repair. Supplement paper, out of 57. Mark Q7–Q19 as well, and total out of 82. Sort every lost mark into one of three piles before you decide what to revise: a unit error, a ratio error, or a chemistry error (a wrong formula, a wrong state, an unbalanced equation). The largest pile names the section to return to, and it is almost never the one you expected. Do not average the two papers. They are different papers, and a strong Core score with a weak Supplement score is a clear and useful diagnosis: the foundations are sound and the mole route is not.
- The five-minute version Core route. Write out one balanced equation, put the relative masses under it, multiply each by its coefficient, and write the sentence “x grams of A gives y grams of B”. Then write, underneath, “scale factor = given mass ÷ that substance's own number”. Those two lines are the whole of statement 3.2.3 on one piece of paper. Extended route. Do the above, then draw the mole map, write the four bridge operations onto it with their unit conditions, and write the three percentage fractions in words. That is the part of Topic 3 which, once it is on paper in front of you, makes every other part findable.