Calculate empirical formulae and molecular formulae, given appropriate data

Stoichiometry – The Mole, Avogadro Constant, Concentration & Formulae

Learning Objectives

• Define the mole, Avogadro constant, relative atomic mass (Ar) and relative molecular mass (Mr).
• Convert between mass, amount of substance and number of particles.
• Express concentrations as g dm⁻³ and mol dm⁻³ and inter‑convert them.
• Determine empirical formulae from experimental data.
• Use the empirical formula and a given molar mass to obtain the molecular formula.
• Carry out simple stoichiometric (mass‑mass) calculations.
• Recall the core electro‑lysis, energetics and rate‑of‑reaction ideas that appear in the Cambridge IGCSE/A‑Level syllabus.

1. The Mole Concept

The mole (mol) is the SI unit for amount of substance. One mole of any substance contains exactly

$$N_{\text A}=6.022\times10^{23}\ \text{particles}$$

where N_A is the Avogadro constant.

2. Relative Atomic and Molecular Masses

• Relative atomic mass (Ar) – the average mass of an element’s atoms relative to 1/12 of the mass of a carbon‑12 atom. It is a dimension‑less number (e.g. Ar(C)=12.01, Ar(O)=16.00).
• Relative molecular mass (Mr) – the sum of the Ar values of all atoms in a molecule or formula unit. It is numerically equal to the molar mass (g mol⁻¹) but carries no units.

Example – calculate Mr for sodium chloride (NaCl):

$$\text{Mr(NaCl)} = \underbrace{22.99}_{\text{Na}} + \underbrace{35.45}_{\text{Cl}} = 58.44$$

The molar mass of NaCl is therefore 58.44 g mol⁻¹.

3. Conversions Between Mass, Moles and Particles

Conversion	Equation	Worked Example
Moles → mass	$$m = n\times M$$	0.025 mol C × 12.01 g mol⁻¹ = 0.300 g
Mass → moles	$$n = \dfrac{m}{M}$$	0.300 g C ÷ 12.01 g mol⁻¹ = 0.025 mol
Moles → particles	$$N = n\times N_{\text A}$$	0.025 mol × 6.022 × 10²³ = 1.51 × 10²² particles
Particles → moles	$$n = \dfrac{N}{N_{\text A}}$$	1.51 × 10²² ÷ 6.022 × 10²³ = 0.025 mol

4. Concentration

Two concentration units are used in IGCSE chemistry:

• Mass concentration – g dm⁻³ (grams of solute per cubic decimetre of solution).
• Molar concentration (molarity) – mol dm⁻³ (moles of solute per cubic decimetre of solution).

Conversion using the molar mass M of the solute:

$$c_{\text{mol}} = \frac{c_{\text{mass}}}{M}\qquad\text{and}\qquad c_{\text{mass}} = c_{\text{mol}}\times M$$

Worked conversion

A solution contains 0.50 g cm⁻³ of NaCl. Convert to g dm⁻³ and mol dm⁻³.

1. 1 cm³ = 0.001 dm³ → 0.50 g cm⁻³ = 0.50 g × 1000 cm³ dm⁻³ = 500 g dm⁻³.
2. Molar mass of NaCl = 58.44 g mol⁻¹.
3. Molar concentration: $$c_{\text{mol}} = \frac{500\ \text{g dm}^{-3}}{58.44\ \text{g mol}^{-1}} = 8.55\ \text{mol dm}^{-3}.$$

5. Determining an Empirical Formula

1. Convert the mass of each element to moles (using atomic masses).
2. Divide all mole values by the smallest mole value obtained.
3. If any ratio is not a whole number, multiply all ratios by the same factor (2, 3, 4 …) until whole numbers are obtained.
4. Write the formula using the whole‑number subscripts.

Worked Example 1 – Simple Empirical Formula

A 1.80 g sample of an unknown compound contains 0.40 g C, 0.0667 g H and 0.533 g O. Find the empirical formula.

Element	Mass (g)	Atomic mass (g mol⁻¹)	Moles	Ratio to smallest
C	0.40	12.01	0.0333	0.0333 ÷ 0.0167 = 2
H	0.0667	1.008	0.0662	0.0662 ÷ 0.0167 = 4
O	0.533	16.00	0.0333	0.0333 ÷ 0.0167 = 2

Dividing by the smallest value (0.0167 mol) gives the whole‑number ratio C₂H₄O₂, which can be reduced to CH₂O. The empirical formula is CH₂O.

Worked Example 2 – Empirical Formula from Combustion Data

Combustion of a hydrocarbon yields 2.20 g CO₂ and 0.90 g H₂O. Determine the empirical formula of the hydrocarbon.

1. Convert products to moles:
   • CO₂: $n = 2.20\ \text{g} ÷ 44.01\ \text{g mol}^{-1} = 0.0500\ \text{mol}$ → 0.0500 mol C.
   • H₂O: $n = 0.90\ \text{g} ÷ 18.02\ \text{g mol}^{-1} = 0.0500\ \text{mol}$ → $2\times0.0500 = 0.100\ \text{mol H}$.
2. Assume the hydrocarbon contains only C and H. The mole ratio C : H = 0.0500 : 0.100 = 1 : 2.
3. Empirical formula = CH₂.

6. Determining a Molecular Formula

1. Calculate the empirical‑formula mass ($M_{\text{emp}}$) by adding the atomic masses of the atoms in the empirical formula.
2. Find the integer factor $n$ such that $n\times M_{\text{emp}} \approx M_{\text{molar}}$ (the given molar mass).
3. Multiply each subscript in the empirical formula by $n$ to obtain the molecular formula.

Worked Example – Molecular Formula

The compound from Example 1 has a molar mass of 180 g mol⁻¹. Determine its molecular formula.

Empirical‑formula mass:

$$M_{\text{emp}} = 12.01 + 2(1.008) + 16.00 = 30.03\ \text{g mol}^{-1}$$

Factor $n$:

$$n = \frac{180}{30.03} = 5.99 \approx 6$$

Multiplying each subscript by 6 gives the molecular formula C₆H₁₂O₆.

7. Simple Stoichiometric (Mass–Mass) Calculations

Core syllabus requirement: calculate the mass of one reactant when the mass of the other is known, using the mole concept.

Example – Magnesium reacts with excess oxygen to form magnesium oxide:

$$\mathbf{2\,Mg\ (s) + O_2\ (g) \rightarrow 2\,MgO\ (s)}$$

Given 3.00 g of Mg, find the mass of MgO produced.

1. Convert Mg to moles: $n_{\text{Mg}} = 3.00\ \text{g} ÷ 24.31\ \text{g mol}^{-1} = 0.123\ \text{mol}$.
2. From the balanced equation, 2 mol Mg → 2 mol MgO, so $n_{\text{MgO}} = 0.123\ \text{mol}$.
3. Convert to mass: $m_{\text{MgO}} = 0.123\ \text{mol} × 40.30\ \text{g mol}^{-1} = 4.96\ \text{g}$.

8. Electrolysis (Core Syllabus)

9. Chemical Energetics (Core)

Reactions are classified by the sign of the enthalpy change (ΔH).

• Exothermic – ΔH < 0; heat is released to the surroundings (e.g. combustion).
• Endothermic – ΔH > 0; heat is absorbed from the surroundings (e.g. dissolution of NH₄NO₃).

In IGCSE exams a simple energy‑profile diagram is often required:

10. Rate of Reaction (Core)

The rate of a chemical reaction depends on four main factors. Remember these when answering qualitative questions.

• Concentration of reactants – higher concentration → higher rate.
• Temperature – higher temperature → more energetic collisions → higher rate.
• Surface area (for solids) – greater surface area → higher rate.
• Catalyst – provides an alternative pathway with lower activation energy.

All of these factors are expressed in terms of molarity or temperature, linking back to the mole concept.

11. Acids, Bases & Salts – Link to pH

In the syllabus, the concentration of hydrogen ions determines acidity:

$$\text{pH} = -\log_{10}[\,\text{H}^+\,]$$

For a strong acid that dissociates completely, the molarity of the acid solution equals the concentration of H⁺. Thus, once the molarity is known (using the methods in Section 4), the pH can be calculated directly.

12. Common Pitfalls & Tips

• When a ratio is close to 0.33, 0.66, 0.125, etc., multiply by 3, 1.5, 8, etc., to obtain whole numbers.
• Keep at least three significant figures in intermediate steps; round only in the final answer.
• Check that the calculated molecular mass is as close as possible to the given molar mass – a difference of ≤ 2 % is acceptable.
• Use the latest atomic masses (e.g. C = 12.01, H = 1.008, O = 16.00 g mol⁻¹).

13. Summary Table

Step	Action	Key Equation
1	Convert masses to moles	$$n = \dfrac{m}{M}$$
2	Find simplest whole‑number ratio	Divide each $n$ by the smallest $n$; multiply if needed
3	Write empirical formula	Use subscripts from the ratio
4	Calculate $M_{\text{emp}}$	$$M_{\text{emp}} = \sum (\text{subscript}\times\text{Ar})$$
5	Determine factor $n$	$$n = \dfrac{M_{\text{molar}}}{M_{\text{emp}}}$$
6	Obtain molecular formula	Multiply each empirical subscript by $n$
7	Convert concentration units (if required)	$$c_{\text{mol}} = \dfrac{c_{\text{mass}}}{M}$$
8	Perform simple mass–mass stoichiometry	Use balanced equations and $m = nM$

14. Practice Questions

1. A compound contains 40.0 % C, 6.7 % H and 53.3 % O by mass. Its molar mass is 180 g mol⁻¹. Determine its empirical and molecular formulae.
2. A 2.00 g sample of a hydrate yields 0.90 g of water on heating. The anhydrous residue (1.10 g) contains 52.0 % C, 13.0 % H and 35.0 % O. Find the formula of the hydrate.
3. Calculate the molarity of a solution that contains 25.0 g of glucose (C₆H₁₂O₆, $M = 180.16\ \text{g mol}^{-1}$) dissolved in 500 mL of water.
4. Magnesium metal reacts with excess oxygen to give magnesium oxide: $$\mathbf{2\,Mg + O_2 \rightarrow 2\,MgO}.$$ If 4.86 g of Mg are used, what mass of MgO is formed?
5. Write the balanced overall equation for the electrolysis of aqueous NaCl and state the gases evolved at each electrode.

15. Answers to Practice Questions

1. Assume 100 g total.

• $n_{\text C}=40.0/12.01=3.33$ mol
• $n_{\text H}=6.7/1.008=6.65$ mol
• $n_{\text O}=53.3/16.00=3.33$ mol
• Divide by 3.33 → C₁H₂O₁ → empirical formula CH₂O.
• $M_{\text{emp}}=12.01+2(1.008)+16.00=30.03\ \text{g mol}^{-1}$.
• $n=180/30.03=6$ → molecular formula C₆H₁₂O₆.

2. Water released: $n_{\text{H₂O}}=0.90/18.02=0.0500\ \text{mol}$.

Anhydrous part (1.10 g):

• C: $0.52\times1.10=0.572\ \text{g}$ → $0.572/12.01=0.0476\ \text{mol}$
• H: $0.13\times1.10=0.143\ \text{g}$ → $0.143/1.008=0.1419\ \text{mol}$
• O: $0.35\times1.10=0.385\ \text{g}$ → $0.385/16.00=0.0241\ \text{mol}$

Divide by the smallest (0.0241): C ≈ 2, H ≈ 6, O ≈ 1 → empirical formula C₂H₆O.

Empirical‑formula mass $M_{\text{emp}}=58.08\ \text{g mol}^{-1}$.

Total mass of one mole of hydrate = $\dfrac{2.00\ \text{g}}{0.0500\ \text{mol}}=40.0\ \text{g mol}^{-1}$.

Let $x$ be the number of water molecules: $58.08 + x(18.02) = 40.0$ → $x = 2$.

Hydrate formula: C₂H₆O·2H₂O (often written C₂H₁₀O₃).

3. Moles of glucose $=25.0\ \text{g} ÷ 180.16\ \text{g mol}^{-1}=0.1387\ \text{mol}$.

Volume $=0.500\ \text{L}$ → $c = 0.1387 ÷ 0.500 = 0.277\ \text{mol dm}^{-3}$.

4. $n_{\text{Mg}}=4.86\ \text{g} ÷ 24.31\ \text{g mol}^{-1}=0.200\ \text{mol}$.

From the equation, $n_{\text{MgO}} = 0.200\ \text{mol}$.

Mass of MgO $=0.200\ \text{mol} × 40.30\ \text{g mol}^{-1}=8.06\ \text{g}$.

5. Overall reaction for aqueous NaCl:

$$\mathbf{2\,NaCl_{(aq)} + 2\,H_2O_{(l)} \rightarrow H_2_{(g)} + Cl_2_{(g)} + 2\,NaOH_{(aq)}}$$

• Cathode (reduction): $2\,H_2O + 2e^- \rightarrow H_2(g) + 2\,OH^-$ – hydrogen gas observed.
• Anode (oxidation): $2\,Cl^- \rightarrow Cl_2(g) + 2e^-$ – chlorine gas observed (yellow‑green colour).