Cambridge O Level Chemistry · Syllabus 5070 · 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 O Level Chemistry.
Mole in context
Stoichiometry is the part of chemistry that uses the balanced equation to calculate how much of each substance reacts or is produced. In Cambridge O Level Chemistry 5070 almost every stoichiometry question follows one route: work out what you have been given, convert it to moles, apply the ratio of coefficients from the balanced equation, convert back to the quantity asked for, then check the unit and the size of the answer. Mass, number of particles, gas volume at r.t.p. and solution concentration are four different doors into moles, and moles are the only quantity a balanced equation can compare.
The mole (mol) is the unit of amount of substance. One mole contains \(6.02\times10^{23}\) particles — the Avogadro constant — and you must always name which particles: atoms, ions, molecules or formula units. Amount of substance is calculated as mass divided by molar mass, \(n = m/M\), where molar mass \(M\) is measured in \(\mathrm{g\,mol^{-1}}\). Relative atomic mass \(A_r\) and relative molecular mass \(M_r\) share the same numbers as molar mass but carry no unit, because each is a comparison with one-twelfth of the mass of a carbon-12 atom. One mole of any gas occupies \(24\,\mathrm{dm^3}\) at room temperature and pressure.
Common mistakes with Mole
- 2. Reading coefficients as a mass ratio Repair \(2\mathrm{H_2} + \mathrm{O_2} \rightarrow 2\mathrm{H_2O}\) does not mean 2 g of hydrogen with 1 g of oxygen. Coefficients count moles. Convert to moles before you touch the ratio. Lesson 3.3C
- 5. 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
- 8. Applying a 1:1 shortcut to a titration that is not 1:1 Repair The relationship written as \(c_1V_1 = c_2V_2\) hides the equation ratio, and is only correct when that ratio happens to be \(1:1\). Go through moles instead — it costs one extra line and works every time. Lesson 3.3G
- "\(A_r\) of chlorine is 35.5, so a chlorine atom weighs 35.5 g." Fix \(35.5\) is a count of reference shares, not a mass. It is one mole of chlorine atoms that has a mass of \(35.5\ \mathrm{g}\).
- "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
- Five of these ten are unit errors, not chemistry errors Traps 2, 3, 4, 6 and 7 all come down to reading a unit — or the absence of one — correctly. Trap 2 belongs here because a coefficient counts moles, so reading it as grams is reading the wrong unit off the equation. That is why every worked example in this chapter carries its units through the substitution rather than adding them at the end. If you build the habit of writing \(\dfrac{4.8\ \mathrm{g}}{24\ \mathrm{g\,mol^{-1}}}\) instead of \(\dfrac{4.8}{24}\), half of this list stops being able to reach you.
- 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.
- The five-minute version If you have only a few minutes before an exam, do this and nothing else: 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.
Questions students ask about Mole
What is a mole, and why do you have to say which particle you are counting?
The mole (mol) is the unit of amount of substance. One mole contains \(6.02\times10^{23}\) particles, and that number is the Avogadro constant. The particles must be named because the count changes with the entity chosen: one mole of \(\mathrm{CO_2}\) is \(6.02\times10^{23}\) molecules but \(1.806\times10^{24}\) atoms. A mole is an amount, not a mass: one mole of hydrogen molecules has a mass of about 2 g, while one mole of lead atoms has a mass of about 207 g.
What is the difference between relative molecular mass and molar mass?
They share a number and nothing else. Relative molecular mass, \(M_r\), is the sum of the relative atomic masses in a formula; it is a comparison with one-twelfth of the mass of a carbon-12 atom, so it is a pure number with no unit. Molar mass, \(M\), is the mass of one mole and carries the unit \(\mathrm{g\,mol^{-1}}\). \(M_r(\mathrm{CO_2}) = 44\) but \(M(\mathrm{CO_2}) = 44\ \mathrm{g\,mol^{-1}}\), and only the molar mass can go into \(n = m/M\). Writing \(M_r = 44\ \mathrm{g}\) loses the mark.
How do you find the limiting reactant?
Convert the amount of each reactant to moles, then divide each amount by its coefficient in the balanced equation. The reactant with the smaller quotient is the limiting reactant: it is completely used up first, so it alone decides the maximum amount of product that can form. Calculate the product from the limiting reactant, never from the one in excess, and finish by checking that the masses balance, including the mass of the excess reactant left over.