Year 11 Chemistry Module 2 ⏱ ~35 min Lesson 2 of 20

Molar Mass

Every tablet your pharmacist counts, every dose in an IV bag, every gram of fertiliser spread across a paddock — all calculated using molar mass. It's the single most-used formula in practical chemistry, and it lives on the periodic table you already have.

Digital balance
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Think First

You know that atoms have different masses — a carbon atom is heavier than a hydrogen atom. How do you think chemists figure out how many grams of a substance they need to weigh out in order to have exactly one mole of it?

Type your initial response below — you will revisit this at the end of the lesson.

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📐

Formula Reference — This Lesson

$n = \dfrac{m}{MM}$
n = amount of substance (mol) m = mass (g) MM = molar mass (g mol⁻¹)
Find n: $n = m \div MM$   |   Find m: $m = n \times MM$   |   Find MM: $MM = m \div n$

Know

  • What molar mass is and its units (g mol⁻¹)
  • Where to read molar mass from the periodic table
  • The formula n = m ÷ MM

Understand

  • Why molar mass equals relative atomic/molecular mass in g mol⁻¹
  • How to calculate MM for compounds from their formula
  • When to add atomic masses vs when to multiply

Can Do

  • Calculate MM for elements and compounds
  • Find n, m or MM using n = m ÷ MM
  • Set up and check units in every calculation

📚 Core Content

Key Terms
MoleThe SI unit for amount of substance; contains exactly 6.022 × 10²³ particles.
Avogadro's Number6.022 × 10²³ — the number of particles in one mole of a substance.
Molar MassThe mass of one mole of a substance, measured in g/mol.
Limiting ReagentThe reactant that is completely consumed first, limiting the amount of product formed.
Empirical FormulaThe simplest whole-number ratio of atoms in a compound.
Molecular FormulaThe actual number of atoms of each element in a molecule of a compound.

Misconceptions to Fix

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Wrong: A catalyst increases the yield of products in an equilibrium reaction.

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Right: A catalyst speeds up both forward and reverse reactions equally; it does not change equilibrium yield.

Salt crystals

What Is Molar Mass?

In Lesson 1, you learned that one mole contains 6.022 × 10²³ particles. But how much does a mole actually weigh? That's where molar mass comes in.

The molar mass (MM) of a substance is the mass of one mole of that substance, measured in grams per mole (g mol⁻¹). The elegant fact is that molar mass numerically equals the relative atomic mass (or relative molecular mass) you read straight off the periodic table — just with units of g mol⁻¹ attached.

1 H Hydrogen 1.008
6 C Carbon 12.011
8 O Oxygen 15.999
11 Na Sodium 22.990
17 Cl Chlorine 35.450
Reading the periodic table: The larger number in each element's box is the relative atomic mass (RAM). This is the molar mass value you use. For example, oxygen has a RAM of 15.999, so the molar mass of O is 15.999 g mol⁻¹, and of O₂ it is 2 × 15.999 = 31.998 g mol⁻¹.

Calculating Molar Mass for Compounds

For a compound, add up the molar masses of every atom in the formula — multiplied by how many of each appear. The diagram below shows the method for sulfuric acid, H₂SO₄.

MOLAR MASS OF H₂SO₄ H 2 atoms 2 × 1.008 = 2.016 g/mol + S 1 atom 1 × 32.060 = 32.060 g/mol + O 4 atoms 4 × 15.999 = 63.996 g/mol = TOTAL 98.072 g mol⁻¹ MM(H₂SO₄) = 2.016 + 32.060 + 63.996 = 98.072 g mol⁻¹ Sum the contribution of every element: atoms × RAM from the periodic table
Interactive: Formula Builder
Example — Water (H₂O): 2 × H + 1 × O = 2(1.008) + 1(15.999) = 2.016 + 15.999 = 18.015 g mol⁻¹
Example — Sodium chloride (NaCl): 1 × Na + 1 × Cl = 22.990 + 35.450 = 58.440 g mol⁻¹
Example — Glucose (C₆H₁₂O₆): 6(12.011) + 12(1.008) + 6(15.999) = 72.066 + 12.096 + 95.994 = 180.156 g mol⁻¹
Brackets in formulas: For Ca(OH)₂, expand the brackets first. The subscript outside applies to everything inside: Ca(OH)₂ = Ca + 2O + 2H = 40.078 + 2(15.999) + 2(1.008) = 74.092 g mol⁻¹
Balance

The Formula: n = m ÷ MM

Once you know the molar mass, you can convert between mass (grams, something you can weigh) and moles (amount of substance, something you can use in calculations). This formula is the workhorse of quantitative chemistry.

m mass (g) n moles (mol) MM molar mass (g mol⁻¹) n = m ÷ MM | m = n × MM Cover the quantity you want to find

The units do the work for you: if you divide grams by g mol⁻¹, you get mol. If you multiply mol by g mol⁻¹, you get grams. Always write your units and watch them cancel.

Units check: n = m ÷ MM gives: g ÷ g mol⁻¹ = g × mol g⁻¹ = mol ✓
m = n × MM gives: mol × g mol⁻¹ = g ✓

🧮 Worked Examples

Worked Example 1 — Finding moles from mass

Stepwise
Calculate the number of moles in 54 g of water (H₂O).
  1. 1
    Calculate molar mass
    H₂O: 2 × H + 1 × O
    MM = 2(1.008) + 15.999 = 18.015 g mol⁻¹
  2. 2
    Identify known values
    m = 54 g  |  MM = 18.015 g mol⁻¹  |  n = ?
  3. 3
    Write the formula
    n = m ÷ MM
  4. 4
    Substitute values
    n = 54 ÷ 18.015
  5. 5
    Calculate and check units
    n = 2.998 mol ≈ 3.0 mol
    Units: g ÷ g mol⁻¹ = mol ✓
✓ Answer n = 3.0 mol of H₂O

Worked Example 2 — Finding mass from moles

Stepwise
What mass of sodium chloride (NaCl) contains 0.50 mol?
  1. 1
    Calculate molar mass
    NaCl: 1 × Na + 1 × Cl
    MM = 22.990 + 35.450 = 58.440 g mol⁻¹
  2. 2
    Identify known values
    n = 0.50 mol  |  MM = 58.440 g mol⁻¹  |  m = ?
  3. 3
    Rearrange the formula
    m = n × MM
  4. 4
    Substitute values
    m = 0.50 × 58.440
  5. 5
    Calculate and check units
    m = 29.22 g
    Units: mol × g mol⁻¹ = g ✓
✓ Answer m = 29.22 g of NaCl
⚠️

Common Mistakes — Don't Lose Easy Marks

Using atomic mass instead of molar mass for a compound
For O₂, the molar mass is 2 × 16.00 = 32.00 g mol⁻¹, not 16.00 g mol⁻¹. The subscript in the formula tells you how many atoms are in one molecule — you must multiply by that number when calculating MM.
✓ Fix: Always write out the MM calculation step explicitly — e.g. "MM(O₂) = 2 × 15.999 = 31.998 g mol⁻¹" — before substituting into n = m ÷ MM.
Forgetting to expand brackets in compound formulas
For Ca(OH)₂, a student might count only 1 oxygen and 1 hydrogen — missing the ×2 from the subscript. The correct expansion is: Ca + 2(O + H) = Ca + 2O + 2H, giving MM = 40.078 + 2(15.999) + 2(1.008) = 74.092 g mol⁻¹, not the incorrect 57.085 g mol⁻¹.
✓ Fix: When you see brackets with a subscript, always distribute the subscript across every atom inside before summing.
Dividing when you should multiply (and vice versa)
Getting n and m confused in the formula causes the calculation to go backwards. If you have grams and want moles, divide by MM. If you have moles and want grams, multiply by MM.
✓ Fix: Use the triangle. Cover up what you want, and the remaining two values show whether to multiply or divide.

📓 Copy Into Your Books

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📖 Key Definitions

  • Molar mass (MM) — mass of one mole of a substance, in g mol⁻¹
  • Relative atomic mass (RAM) — from the periodic table, used to calculate MM
  • m — mass in grams
  • n — amount in moles

📐 Formula & Rearrangements

  • n = m ÷ MM (find moles)
  • m = n × MM (find mass)
  • MM = m ÷ n (find molar mass)

🔢 Calculating MM

  • Read RAM from the periodic table for each element
  • Multiply RAM by the subscript for that element
  • Add results for all elements in the formula
  • Expand brackets before summing

✅ Units Check

  • n = m ÷ MM → g ÷ g mol⁻¹ = mol ✓
  • m = n × MM → mol × g mol⁻¹ = g ✓
  • MM always in g mol⁻¹
  • m always in grams (convert if given in kg)

📝 How are you completing this lesson?

🧪 Activities

📊 Activity 1 — Practice Drill

Applying n = m ÷ MM

Three problems, increasing difficulty. Attempt each before revealing the answer. Show all working — including the MM calculation.

  1. 1 Calculate the number of moles in 88 g of carbon dioxide (CO₂).

    MM(CO₂) = 12.011 + 2(15.999) = 12.011 + 31.998 = 44.009 g mol⁻¹ n = m ÷ MM = 88 ÷ 44.009 = 2.0 mol

  2. 2 What mass of calcium carbonate (CaCO₃) is in 0.25 mol?

    MM(CaCO₃) = 40.078 + 12.011 + 3(15.999) = 40.078 + 12.011 + 47.997 = 100.086 g mol⁻¹ m = n × MM = 0.25 × 100.086 = 25.02 g

  3. 3 A student dissolves 14.7 g of sulfuric acid (H₂SO₄) in water. Calculate the amount in moles. (S = 32.06 g mol⁻¹)

    MM(H₂SO₄) = 2(1.008) + 32.06 + 4(15.999) = 2.016 + 32.06 + 63.996 = 98.072 g mol⁻¹ n = m ÷ MM = 14.7 ÷ 98.072 = 0.1499 mol ≈ 0.150 mol

Type your working below before revealing answers above:

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🔍 Activity 2 — Error Spotting

Find the Mistake

The worked solution below contains two deliberate errors. Your job is to find them, explain why each is wrong, and write the correct working. This is a high-value exam skill.

⚠️ Student's Working — contains errors
Problem: Calculate the number of moles in 49.0 g of calcium hydroxide, Ca(OH)₂.
Step 1: MM(Ca(OH)₂) = 40.078 + 15.999 + 1.008 = 57.085 g mol⁻¹
Step 2: n = m ÷ MM
Step 3: n = 49.0 ÷ 57.085
Step 4: n = 0.858 mol
Your task:
A) Identify both errors. For each, explain: (i) what the student did wrong, and (ii) what they should have done instead.
B) Write the complete correct working and give the correct answer.

Type your response below:

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Revisit Your Initial Thinking

Earlier you were asked: How do you think chemists figure out how many grams of a substance they need to weigh out in order to have exactly one mole of it?

The answer lies in molar mass — the mass in grams of one mole of a substance, which is numerically equal to the relative atomic (or formula) mass from the periodic table. Because one mole of carbon-12 was defined to be exactly 12 g, all other atomic masses scale consistently, so the periodic table directly gives you the grams per mole for any element or compound.

Now revisit your initial response. What did you get right? What has changed in your thinking?

Look back at your initial response in your book. Annotate it with what you now understand differently.

Annotate your initial response in your book
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MC

Multiple Choice

5 random questions from a replayable lesson bank — feedback shown immediately

✍️ Short Answer

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Extended Questions

UnderstandBand 3

6. Define molar mass and explain why its numerical value equals the relative atomic mass found on the periodic table. In your answer, refer to the definition of the mole. 3 MARKS

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ApplyBand 3

7. A chemistry technician needs to prepare 2.40 mol of glucose (C₆H₁₂O₆) for a fermentation experiment. Calculate the mass of glucose required. Show all working, including the molar mass calculation. (C = 12.011, H = 1.008, O = 15.999) 4 MARKS

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AnalyseBand 4

8. A student weighed out 13.3 g of anhydrous sodium carbonate (Na₂CO₃) and claimed it contained "about one-eighth of a mole." Is the student's claim correct? Show all working to justify your answer. (Na = 22.990, C = 12.011, O = 15.999) 4 MARKS

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EvaluateBand 5

9. A pharmaceutical company manufactures aspirin (C₉H₈O₄, MM = 180.16 g mol⁻¹). A quality-control test on a batch finds that 10.00 g of the product contains only 9.76 g of pure aspirin. (a) Calculate the number of moles of pure aspirin per tablet if the tablet mass is 500 mg. (b) Evaluate whether the batch passes the quality standard that requires ≥97.0% purity by mass. Show full working. 5 MARKS

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CreateBand 6

10. Design a procedure a student could use — with only a balance, distilled water, and a volumetric flask — to determine the molar mass of an unknown soluble salt. In your answer: (a) describe the measurements the student must take, (b) explain the calculation steps that lead from those measurements to the molar mass, and (c) identify one source of systematic error and explain how it would affect the result. 6 MARKS

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✅ Comprehensive Answers

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🔍 Activity 2 — Error Spotting Model Answer

Error 1 (Step 1): The student failed to expand the brackets in Ca(OH)₂. The subscript 2 applies to both O and H inside the brackets, so the correct formula contains 1 Ca, 2 O, and 2 H atoms — not 1 Ca, 1 O, and 1 H. The correct MM calculation is:

MM = 40.078 + 2(15.999) + 2(1.008) = 40.078 + 31.998 + 2.016 = 74.092 g mol⁻¹

Error 2 (Steps 3–4): The student used the wrong molar mass (57.085 instead of 74.092), which propagated through to give the wrong answer. With the correct MM:

n = 49.0 ÷ 74.092 = 0.661 mol

The student's answer of 0.858 mol is incorrect on both counts.

❓ Multiple Choice

1. C — MgSO₄: 24.305 + 32.06 + 4(15.999) = 24.305 + 32.06 + 63.996 = 120.361 g mol⁻¹

2. B — m = n × MM. Rearranging n = m ÷ MM gives m = n × MM.

3. A — MM(Cl₂) = 2 × 35.45 = 70.90 g mol⁻¹. n = 71.0 ÷ 70.90 = 1.00 mol

4. D — Al(OH)₃: 26.982 + 3(15.999) + 3(1.008) = 26.982 + 47.997 + 3.024 = 78.003 g mol⁻¹

5. C — n = 167.2 ÷ 55.845 = 2.994 mol ≈ 2.99 mol

6. B — MM(Ca(H₂PO₄)₂) = 40.078 + 2(2×1.008 + 30.974 + 4×15.999) = 40.078 + 2(2.016 + 30.974 + 63.996) = 40.078 + 2(96.986) = 40.078 + 193.972 = 234.05 g mol⁻¹. The first student treated H₂PO₄ as a single unit without multiplying by 2.

7. C — MM(XCl₂) = m ÷ n = 55.75 ÷ 0.500 = 111.50 g mol⁻¹. MM(X) = 111.50 − 2(35.453) = 111.50 − 70.906 = 40.59 ≈ 40.34 g mol⁻¹ → calcium (Ca = 40.078 g mol⁻¹). Note: exact value 40.59 rounds to calcium; answer C is correct.

📝 Short Answer Model Answers

Q6 (3 marks): Molar mass is the mass in grams of one mole of a substance, expressed in g mol⁻¹ [1]. One mole is defined as 6.022 × 10²³ particles, and was originally defined so that one mole of carbon-12 has a mass of exactly 12 g — equal to its relative atomic mass [1]. Since all other atomic masses are defined relative to carbon-12, the molar mass of any element in g mol⁻¹ is numerically equal to its relative atomic mass from the periodic table [1].

Q7 (4 marks):

MM(C₆H₁₂O₆) = 6(12.011) + 12(1.008) + 6(15.999) = 72.066 + 12.096 + 95.994 = 180.156 g mol⁻¹ m = n × MM = 2.40 × 180.156 = 432.37 g ≈ 432 g

Award 1 mark for correct MM calculation, 1 mark for correct formula, 1 mark for correct substitution, 1 mark for correct final answer with units.

Q8 (4 marks):

MM(Na₂CO₃) = 2(22.990) + 12.011 + 3(15.999) = 45.980 + 12.011 + 47.997 = 105.988 g mol⁻¹ n = m ÷ MM = 13.3 ÷ 105.988 = 0.1255 mol

One-eighth of a mole = 0.125 mol. The student's calculated value (0.1255 mol) rounds to 0.125 mol, so the claim is essentially correct [1 mark for accepting with justification, or 1 mark for identifying that 0.1255 ≈ 0.125 mol]. Full marks require the working shown above.

📝 Additional Short Answer Model Answers

Q9 (5 marks):

(a) Pure aspirin per tablet: % purity = 9.76 ÷ 10.00 = 97.6%. m(pure) per tablet = 0.500 g × 0.976 = 0.488 g

n = m ÷ MM = 0.488 ÷ 180.16 = 2.71 × 10⁻³ mol

(b) % purity = (9.76 ÷ 10.00) × 100 = 97.6%. Since 97.6% ≥ 97.0%, the batch passes the quality standard. [Award 1 for purity calculation, 1 for n(aspirin), 1 for comparison with standard, 1 for conclusion, 1 for units and working.]

Q10 (6 marks):

(a) Measurements: (i) Mass of the empty volumetric flask (tare). (ii) Mass of the flask + a weighed sample of the unknown salt. (iii) Volume of the volumetric flask (e.g. 250.0 mL, marked on the flask). The student dissolves the salt in distilled water, makes up to the mark, and uses the formula c = n ÷ V along with a known reaction to determine n — OR more simply: weigh a known mass, dissolve, calculate n from a titration or from assumed formula.

Simpler accepted approach: Weigh accurately, dissolve in known volume. Then use a titration with a standard solution to find n(salt). MM = m ÷ n.

(b) Calculation: n(salt) from titration; MM = m(salt weighed) ÷ n(salt).

(c) Systematic error: The salt may absorb moisture from air (hygroscopic), so the mass weighed includes water. This makes m(salt) appear larger than the true mass, so the calculated MM is too large. Fix: dry the salt in an oven before weighing.

🏎️
Speed Race

Speed Race — Molar Mass

Answer questions on molar mass calculations, Avogadro's number and formula mass before your opponents cross the line. Fast answers = faster car. Pool: lessons 1–2.

Mark lesson as complete

Tick when you've finished all activities and checked your answers.

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