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GCSE / IGCSE Chemistry – Quantitative Chemistry

Moles, reacting masses, concentration, titrations, yield.

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12 questions  ·  31 marks  ·  answers included  ·  GCSE / IGCSE (Edexcel / Cambridge style)
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This worksheet covers:

  • Moles
  • Reacting masses
  • Concentration
  • Titration calculations
  • Percentage yield

A calculator is needed throughout; molar gas volume at rtp is taken as $24\,\text{dm}^3\text{mol}^{-1}$ and $N_A = 6.02\times10^{23}\,\text{mol}^{-1}$.

Questions

Section A – Skills

Calculator allowed
1

Calculate the relative formula mass, $M_r$, of $\text{CaCO}_3$. ($A_r$: Ca 40, C 12, O 16)

Show answer
$M_r = 40 + 12 + 3\times16 = 100$
[1]
2

Calculate the number of moles in 22.0 g of $\text{CO}_2$. ($M_r = 44$)

Show answer
$n = \dfrac{22.0}{44} = 0.500\,\text{mol}$
[2]
3

Calculate the mass of 0.250 mol of $\text{NaCl}$. ($M_r = 58.5$)

Show answer
mass $= 0.250\times58.5 = 14.6\,\text{g}$
[2]
4

Calculate the concentration, in $\text{mol dm}^{-3}$, of a solution made by dissolving 4.00 g of $\text{NaOH}$ ($M_r = 40$) in water and making up to $250\,\text{cm}^3$.

Show answer
$n = \dfrac{4.00}{40} = 0.100\,\text{mol}$; $c = \dfrac{0.100}{0.250} = 0.400\,\text{mol dm}^{-3}$
[3]
5

Calculate the number of molecules in 0.0200 mol of $\text{H}_2\text{O}$. ($N_A = 6.02\times10^{23}\,\text{mol}^{-1}$)

Show answer
molecules $= 0.0200\times6.02\times10^{23} = 1.20\times10^{22}$
[2]

Section B – Exam-style

Calculator allowed
6

$\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$. Calculate the maximum mass of $\text{CaO}$ that can be made from 50.0 g of $\text{CaCO}_3$. ($M_r$: $\text{CaCO}_3 = 100$, $\text{CaO} = 56$)

Show answer
$n(\text{CaCO}_3) = \dfrac{50.0}{100} = 0.500\,\text{mol} = n(\text{CaO})$; mass $= 0.500\times56 = 28.0\,\text{g}$
[3]
7

$\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$. Calculate the volume, in $\text{cm}^3$ at rtp, of hydrogen gas produced from 0.0600 mol of magnesium reacting with excess acid. (1 mol of gas occupies $24\,\text{dm}^3$ at rtp)

Show answer
$n(\text{H}_2) = n(\text{Mg}) = 0.0600\,\text{mol}$; $V = 0.0600\times24 = 1.44\,\text{dm}^3 = 1440\,\text{cm}^3$
[3]
8

$25.0\,\text{cm}^3$ of sodium hydroxide solution was exactly neutralised by $22.5\,\text{cm}^3$ of $0.100\,\text{mol dm}^{-3}$ hydrochloric acid: $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$. Calculate the concentration of the sodium hydroxide solution.

Show answer
$n(\text{HCl}) = 0.100\times\dfrac{22.5}{1000} = 2.25\times10^{-3}\,\text{mol} = n(\text{NaOH})$; $c(\text{NaOH}) = \dfrac{2.25\times10^{-3}}{25.0/1000} = 0.0900\,\text{mol dm}^{-3}$
[3]
9

A reaction has a theoretical yield of 4.50 g of product. A student actually obtains 3.60 g. Calculate the percentage yield.

Show answer
percentage yield $= \dfrac{3.60}{4.50}\times100 = 80.0\text{ percent}$
[2]
10

Explain, in terms of particle collisions, why increasing the concentration of a reactant increases the rate of reaction.

Show answer
Increasing concentration increases the number of reactant particles in a given volume, so particles are closer together and collide more frequently; this increases the frequency of successful collisions (with energy $\geq$ activation energy) per unit time, increasing the rate of reaction.
[2]

Section C – Challenge

Calculator allowed
11

A 2.00 g sample of impure calcium carbonate reacts fully with excess dilute hydrochloric acid. The carbon dioxide released has a mass of 0.660 g. Calculate the percentage purity of the calcium carbonate sample. ($M_r$: $\text{CaCO}_3 = 100$, $\text{CO}_2 = 44$)

Show answer
$n(\text{CO}_2) = \dfrac{0.660}{44} = 0.0150\,\text{mol} = n(\text{CaCO}_3)$ reacted; mass $\text{CaCO}_3 = 0.0150\times100 = 1.50\,\text{g}$; purity $= \dfrac{1.50}{2.00}\times100 = 75.0\text{ percent}$
[4]
12

A $10.0\,\text{cm}^3$ sample of a stock $\text{NaOH}$ solution is diluted to $100\,\text{cm}^3$ with water. $25.0\,\text{cm}^3$ of the diluted solution requires $20.0\,\text{cm}^3$ of $0.0500\,\text{mol dm}^{-3}$ $\text{HCl}$ for exact neutralisation. Calculate the concentration of the original stock $\text{NaOH}$ solution.

Show answer
$n(\text{HCl}) = 0.0500\times\dfrac{20.0}{1000} = 1.00\times10^{-3}\,\text{mol} = n(\text{NaOH})$ in $25.0\,\text{cm}^3$ diluted; $c(\text{diluted}) = \dfrac{1.00\times10^{-3}}{0.0250} = 0.0400\,\text{mol dm}^{-3}$; moles in $100\,\text{cm}^3$ diluted $= 0.0400\times0.100 = 4.00\times10^{-3}\,\text{mol}$, all from $10.0\,\text{cm}^3$ of stock, so $c(\text{stock}) = \dfrac{4.00\times10^{-3}}{0.0100} = 0.400\,\text{mol dm}^{-3}$
[4]

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