Cambridge O Level Chemistry · Syllabus 5070 · Stoichiometry
Concentration
What is Concentration?
Concentration states how much solute is dissolved in a given volume of solution; it is measured either in moles of solute per cubic decimetre of solution or in grams of solute per cubic decimetre of solution, and the two are linked by the molar mass of the solute.
This definition is part of the Stoichiometry chapter in Cambridge O Level Chemistry.
Concentration 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.
Common mistakes with Concentration
- Using the volume of water added rather than the volume of solution. Fix Concentration is per \(\mathrm{dm^3}\) of finished solution. "Made up to \(250\ \mathrm{cm^3}\)" gives you the volume you need; "dissolved in \(250\ \mathrm{cm^3}\) of water" is a different statement, and a question will say which it means.
- Dividing by the titre in step 5. Fix Divide by the volume of the solution whose concentration you are finding. Label both volumes with their solution's name as you write them down.
Examiner tips on Concentration
- Working the selector on a question you have not seen "\(50.0\ \mathrm{cm^3}\) of \(0.200\ \mathrm{mol\,dm^{-3}}\) hydrochloric acid is added to excess magnesium. Calculate the volume of hydrogen produced at r.t.p." I have: a concentration and a volume, of hydrochloric acid. I want: a gas volume in \(\mathrm{dm^3}\), of hydrogen. Different substances, so a ratio is needed. Route: \(n = cV\) (converting \(50.0\ \mathrm{cm^3}\) to \(0.0500\ \mathrm{dm^3}\)) → ratio from \(\mathrm{Mg} + 2\mathrm{HCl} \rightarrow \mathrm{MgCl_2} + \mathrm{H_2}\), which is \(2\ \mathrm{HCl} : 1\ \mathrm{H_2}\) → \(V = n \times 24\). Answer: \(n(\mathrm{HCl}) = 0.200 \times 0.0500 = 0.0100\ \mathrm{mol}\); \(n(\mathrm{H_2}) = 0.00500\ \mathrm{mol}\); \(V = 0.00500 \times 24 = 0.120\ \mathrm{dm^3}\), that is \(120\ \mathrm{cm^3}\) at r.t.p. The word "excess" told you the magnesium never needed checking.
Questions students ask about Concentration
When do you use 24 and when do you use 24 000 in a gas volume calculation?
The molar gas volume at room temperature and pressure (r.t.p.) is \(24\ \mathrm{dm^3}\), which is the same as \(24\,000\ \mathrm{cm^3}\). Read the unit on the volume you were given, then choose: 24 goes with \(\mathrm{dm^3}\) and 24 000 goes with \(\mathrm{cm^3}\). Mixing them is a factor-of-1000 error that looks plausible on a calculator. The same rule applies to concentration: \(n = cV\) needs \(V\) in \(\mathrm{dm^3}\), so divide a \(\mathrm{cm^3}\) volume by 1000 on its own line before you substitute. Always state at r.t.p. in the answer.