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How does the mole let us count particles and predict the masses, gas volumes and concentrations in a reaction?

The Avogadro constant and the mole, molar mass, the ideal gas equation, empirical and molecular formulae, concentration and titration calculations, percentage yield and atom economy.

An Eduqas A-Level Chemistry C1.3 answer on the Avogadro constant, molar mass, the ideal gas equation, empirical and molecular formulae, concentrations, titrations, percentage yield and atom economy.

Generated by Claude Opus 4.812 min answer

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  1. What this topic is asking
  2. The mole and molar mass
  3. Gas volumes and the ideal gas equation
  4. Empirical and molecular formulae
  5. Concentration and titrations
  6. Percentage yield and atom economy
  7. Examples in context
  8. Try this

What this topic is asking

Eduqas topic C1.3 wants you to use the mole and the Avogadro constant, work with molar mass and the ideal gas equation, find empirical and molecular formulae, carry out concentration and titration calculations, and judge reactions by percentage yield and atom economy. This is the calculation engine reused throughout the qualification.

The mole and molar mass

The core relationships are:

Gas volumes and the ideal gas equation

At a stated temperature and pressure, equal volumes of gases contain equal numbers of moles. Eduqas uses the ideal gas equation:

Unit conversions matter: 1 cm3=1×10−6 m31\ \text{cm}^3 = 1 \times 10^{-6}\ \text{m}^3, 1 dm3=1×10−3 m31\ \text{dm}^3 = 1 \times 10^{-3}\ \text{m}^3, 1 kPa=1000 Pa1\ \text{kPa} = 1000\ \text{Pa} and T(K)=θ(∘C)+273T(\text{K}) = \theta(^{\circ}\text{C}) + 273.

Empirical and molecular formulae

The empirical formula is the simplest whole-number ratio of atoms; the molecular formula is the actual number. Divide each element's percentage (or mass) by its ArA_r, divide by the smallest result, then scale the molecular formula using the molecular mass.

Concentration and titrations

Concentration links moles and volume by n=c×Vn = c \times V (with VV in dm3\text{dm}^3). A titration uses a known concentration to find an unknown one, working through the balanced mole ratio.

Percentage yield and atom economy

Examples in context

Example 1. Airbag chemistry. Sodium azide decomposes to nitrogen gas: 2NaN3→2Na+3N22\text{NaN}_3 \rightarrow 2\text{Na} + 3\text{N}_2. Engineers use pV=nRTpV = nRT to find the mass of azide needed to inflate a bag to a set volume at a given temperature and pressure within milliseconds.

Example 2. Atom economy in industry. Making ethene oxide directly from ethene and oxygen has a far higher atom economy than older multi-step routes that produce large amounts of by-product. Comparing atom economies guides chemists toward cleaner, cheaper processes.

Try this

Q1. Calculate the number of moles in 11.0 g11.0\ \text{g} of carbon dioxide. [1 mark]

  • Cue. Mr=44.0M_r = 44.0, so n=11.044.0=0.250 moln = \frac{11.0}{44.0} = 0.250\ \text{mol}.

Q2. Calculate the atom economy for producing hydrogen by Zn+H2SO4→ZnSO4+H2\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2. [2 marks]

  • Cue. Mr(H2)Mr(ZnSO4)+Mr(H2)×100=2.0161.5+2.0×100=1.22%\frac{M_r(\text{H}_2)}{M_r(\text{ZnSO}_4) + M_r(\text{H}_2)} \times 100 = \frac{2.0}{161.5 + 2.0} \times 100 = 1.22\%.

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20195 marksA 25.0 cm325.0\ \text{cm}^3 sample of sodium carbonate solution required 20.00 cm320.00\ \text{cm}^3 of 0.150 mol dm−30.150\ \text{mol dm}^{-3} hydrochloric acid for complete neutralisation. The equation is Na2CO3+2HCl→2NaCl+H2O+CO2\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2. Calculate the concentration of the sodium carbonate solution in mol dm−3\text{mol dm}^{-3}.
Show worked answer →

Moles of HCl =c×V=0.150×20.001000=3.00×10−3 mol= c \times V = 0.150 \times \dfrac{20.00}{1000} = 3.00 \times 10^{-3}\ \text{mol} (1).

The 2:12:1 ratio gives moles of Na2CO3=3.00×10−32=1.50×10−3 mol\text{Na}_2\text{CO}_3 = \dfrac{3.00 \times 10^{-3}}{2} = 1.50 \times 10^{-3}\ \text{mol} (2).

Concentration =nV=1.50×10−325.0/1000=0.0600 mol dm−3= \dfrac{n}{V} = \dfrac{1.50 \times 10^{-3}}{25.0/1000} = 0.0600\ \text{mol dm}^{-3} (2).

Markers reward the moles of acid, the use of the 2:12:1 mole ratio and the final concentration to 3 s.f.

Eduqas 20214 marksA compound contains 40.0%40.0\% carbon, 6.7%6.7\% hydrogen and 53.3%53.3\% oxygen by mass. (a) Determine its empirical formula. (b) Its relative molecular mass is 180180. Determine its molecular formula.
Show worked answer →

(a) Divide each percentage by ArA_r: C=40.012.0=3.33\text{C} = \dfrac{40.0}{12.0} = 3.33, H=6.71.0=6.7\text{H} = \dfrac{6.7}{1.0} = 6.7, O=53.316.0=3.33\text{O} = \dfrac{53.3}{16.0} = 3.33. Dividing by the smallest (3.333.33) gives 1:2:11 : 2 : 1, so the empirical formula is CH2O\text{CH}_2\text{O} (2).

(b) Empirical mass =12+2+16=30= 12 + 2 + 16 = 30. 18030=6\dfrac{180}{30} = 6, so the molecular formula is C6H12O6\text{C}_6\text{H}_{12}\text{O}_6 (2).

Markers reward the mole ratio, the empirical formula, the empirical mass and the molecular formula.

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