Chemistry • Year 11 • Module 2 • Lesson 2

Molar Mass

Apply n = m ÷ MM to real data, spot and fix calculation errors, and connect molar mass to real-world industrial and pharmaceutical contexts.

Apply · Data & Reasoning

1. Calculation series — n, m and MM

Show all working: calculate molar mass first, identify known and unknown, write the formula, substitute, calculate, and check units. 12 marks (3 each)

A_r values (g mol⁻¹): H = 1.008 · C = 12.011 · N = 14.007 · O = 15.999 · Na = 22.990 · S = 32.060 · Cl = 35.450 · Ca = 40.078 · Fe = 55.845

1.1 Calculate the number of moles in 117.0 g of sodium chloride (NaCl).

1.2 What mass of iron (Fe) contains 0.750 mol?

1.3 Calculate the molar mass of an unknown substance if 2.500 mol has a mass of 245.2 g.

1.4 A student dissolves 9.80 g of sulfuric acid (H₂SO₄) in 100 mL of distilled water. Calculate the amount of H₂SO₄ in moles.

Stuck? Revisit Worked Examples 1 and 2 in the lesson. Always write MM before substituting.

2. Interpret experimental data — fertiliser quality control

A fertiliser manufacturer tests five product samples. The table records the mass of each sample and the moles calculated by their technician. 8 marks

Sample	Compound (formula)	Mass (g)	Technician’s n (mol)
A	Urea CO(NH₂)₂	12.01	0.200
B	Ammonium nitrate NH₄NO₃	40.04	0.250
C	Calcium nitrate Ca(NO₃)₂	32.80	0.200
D	Potassium chloride KCl (K = 39.098 g mol⁻¹)	74.55	1.000
E	Ammonium sulfate (NH₄)₂SO₄	66.10	0.250

2.1 For each sample, verify the technician’s n value by calculating MM and applying n = m ÷ MM. Complete the “Correct?” and “Correct n” columns. 6 marks (1 per sample, deduct for missing working)

2.2 For any sample where the technician made an error, identify the most likely mistake (e.g. wrong formula, bracket error, wrong operation). 2 marks

Stuck? Expand brackets before summing. For urea CO(NH₂)₂: there are 2 N and 4 H from the (NH₂)₂ group.

3. Graph interpretation — mass vs moles of glucose

A student weighed out five different samples of glucose (C₆H₁₂O₆) and recorded the mass and number of moles for each. The graph below shows the data. 7 marks

Figure 3. Mass (g) of glucose C₆H₁₂O₆ vs amount (mol). Each point represents a separately weighed sample. Illustrative data.

3.1 Describe the relationship shown in the graph between mass and amount of substance. 2 marks

3.2 Use the graph to calculate the gradient of the line. State what physical quantity the gradient represents and give its unit. 3 marks

3.3 Verify your gradient value by calculating MM(C₆H₁₂O₆) from its formula. Show all working. 2 marks

Stuck? Gradient = rise ÷ run = Δmass ÷ Δmol. For MM of glucose: C = 12.011, H = 1.008, O = 15.999.

4. Predict and justify — a pharmaceutical context

A hospital pharmacist prepares an intravenous saline solution by dissolving sodium chloride (NaCl) in sterile water. Each 1-litre bag must contain exactly 9.00 g of NaCl (normal saline: 0.9% w/v). 5 marks

4.1 Calculate how many moles of NaCl are in each 1-litre bag. 2 marks

4.2 A nurse accidentally uses 18.0 g of NaCl instead of 9.00 g. Without further calculation, predict what happens to the number of moles of NaCl in the bag and explain your reasoning using n = m ÷ MM. 2 marks

4.3 The pharmacist checks by calculating n for the 18.0 g bag. State the value and confirm your prediction. 1 mark

Stuck? Revisit Worked Example 1 in the lesson. MM(NaCl) = Na + Cl = 22.990 + 35.450.

Answers — Do not peek before attempting

Q1 — Calculations

1.1 NaCl: MM = 22.990 + 35.450 = 58.440 g mol⁻¹. n = 117.0 ÷ 58.440 = 2.002 mol ≈ 2.00 mol. Units: g ÷ g mol⁻¹ = mol ✓

1.2 Fe: MM(Fe) = 55.845 g mol⁻¹. m = n × MM = 0.750 × 55.845 = 41.88 g. Units: mol × g mol⁻¹ = g ✓

1.3 Unknown: MM = m ÷ n = 245.2 ÷ 2.500 = 98.08 g mol⁻¹. (This is close to the molar mass of H₂SO₄ = 98.072 g mol⁻¹.)

1.4 H₂SO₄: MM = 2(1.008) + 32.060 + 4(15.999) = 2.016 + 32.060 + 63.996 = 98.072 g mol⁻¹. n = 9.80 ÷ 98.072 = 0.0999 mol ≈ 0.100 mol.

Q2.1 — Fertiliser quality control

Sample A — Urea CO(NH₂)₂: Expand: 1 C + 1 O + 2 N + 4 H = 12.011 + 15.999 + 2(14.007) + 4(1.008) = 12.011 + 15.999 + 28.014 + 4.032 = 60.056 g mol⁻¹. n = 12.01 ÷ 60.056 = 0.2000 mol. Technician’s value: 0.200 mol. Correct: Y.

Sample B — NH₄NO₃: 2 N + 4 H + 3 O = 2(14.007) + 4(1.008) + 3(15.999) = 28.014 + 4.032 + 47.997 = 80.043 g mol⁻¹. n = 40.04 ÷ 80.043 = 0.5003 mol. Technician’s value: 0.250 mol. Correct: N. Correct n = 0.500 mol.

Sample C — Ca(NO₃)₂: Expand: 1 Ca + 2 N + 6 O = 40.078 + 2(14.007) + 6(15.999) = 40.078 + 28.014 + 95.994 = 164.086 g mol⁻¹. n = 32.80 ÷ 164.086 = 0.1999 mol ≈ 0.200 mol. Correct: Y.

Sample D — KCl: MM = 39.098 + 35.450 = 74.548 g mol⁻¹. n = 74.55 ÷ 74.548 = 1.0000 mol. Correct: Y.

Sample E — (NH₄)₂SO₄: Expand: 2 N + 8 H + 1 S + 4 O = 2(14.007) + 8(1.008) + 32.060 + 4(15.999) = 28.014 + 8.064 + 32.060 + 63.996 = 132.134 g mol⁻¹. n = 66.10 ÷ 132.134 = 0.5003 mol ≈ 0.500 mol. Technician’s value: 0.250 mol. Correct: N. Correct n = 0.500 mol.

Q2.2 — Errors identified

Sample B: The technician likely used only one nitrogen atom (14.007) instead of two (NH₄NO₃ contains 2 N), giving a MM of about 50 g mol⁻¹. Alternatively, they may have multiplied instead of divided, giving half the correct answer.

Sample E: The technician likely failed to expand the bracket (NH₄)₂ — treating NH₄ as having only 1 N and 4 H rather than 2 N and 8 H. This halves the contribution from nitrogen and hydrogen, approximately doubling the calculated moles compared to the correct answer.

Q3.1 — Graph description

Mass increases linearly (in a straight line through the origin) as the amount of substance increases. This is a direct proportion: doubling the moles doubles the mass, consistent with m = n × MM where MM is a constant.

Q3.2 — Gradient

Gradient = Δmass ÷ Δmol. Using the endpoints: (450 − 0) g ÷ (2.5 − 0) mol = 450 ÷ 2.5 = 180 g mol⁻¹. The gradient represents the molar mass of glucose, in units of g mol⁻¹. This is because m = n × MM, so the graph of m vs n has slope MM.

Q3.3 — Verification by formula

MM(C₆H₁₂O₆) = 6(12.011) + 12(1.008) + 6(15.999) = 72.066 + 12.096 + 95.994 = 180.156 g mol⁻¹. This agrees with the graphical gradient of 180 g mol⁻¹ ✓

Q4.1 — Moles of NaCl in saline bag

MM(NaCl) = 22.990 + 35.450 = 58.440 g mol⁻¹. n = 9.00 ÷ 58.440 = 0.154 mol (3 s.f.).

Q4.2 — Prediction with 18.0 g

Since n = m ÷ MM and MM is constant, doubling the mass (from 9.00 g to 18.0 g) will exactly double the number of moles of NaCl. The amount becomes 2 × 0.154 = 0.308 mol.

Q4.3 — Confirmation

n = 18.0 ÷ 58.440 = 0.308 mol. This confirms the prediction: exactly twice the amount in the original bag.