Chemistry • Year 11 • Module 2 • Lesson 2

Molar Mass

Lock in the vocabulary, formula, and calculation method for molar mass before tackling harder questions.

Build · Vocab & Recall

1. Term–definition match

The definitions below are shuffled. In the right-hand column write the matching term from this list: molar mass, relative atomic mass (Ar), relative molecular mass (Mr), formula mass, mole, subscript rule, n = m ÷ MM, g mol−1. 8 marks (1 each)

#DefinitionMatching term
1.1The mass of one mole of a substance, expressed in grams per mole.
1.2The weighted average mass of an element’s atoms relative to carbon-12; dimensionless but numerically equal to molar mass in g mol−1.
1.3The sum of relative atomic masses of all atoms in a molecule; used for covalent compounds.
1.4The same calculation as Mr but applied to ionic compounds, which do not contain discrete molecules.
1.5The SI unit of amount of substance; defined as exactly 6.022 × 1023 particles.
1.6The rule that the element’s Ar must be multiplied by the number shown as the subscript in a chemical formula before summing contributions.
1.7The equation linking amount, mass and molar mass: amount (mol) = mass (g) ÷ molar mass (g mol−1).
1.8The unit of molar mass: grams per mole.
Stuck? Revisit the Key Terms panel in the lesson.

2. True or false — with correction

Circle T or F for each statement. If the statement is false, write the corrected version on the line below it. 12 marks (1 T/F + 1 correction each)

2.1 The molar mass of oxygen gas (O2) is 16.00 g mol−1.    T  /  F

2.2 Relative atomic mass is dimensionless, but numerically equals molar mass in g mol−1.    T  /  F

2.3 To find the number of moles in a sample, you multiply the mass by the molar mass.    T  /  F

2.4 For Ca(OH)2, there are 2 oxygen atoms and 2 hydrogen atoms per formula unit.    T  /  F

2.5 The molar mass of water (H2O) is 18.015 g mol−1.    T  /  F

2.6 “Formula mass” and “relative molecular mass” are different terms for the same calculation applied to covalent molecules.    T  /  F

Stuck? Revisit the Key Terms panel, the Misconceptions box, and the Common Mistakes box in the lesson.

3. Fill-in-the-blank paragraph

Use the word bank to complete the passage. Each word or phrase is used once. 8 marks (1 per blank)

Word bank:

divide  ·  g mol−1  ·  grams  ·  molar mass  ·  mole  ·  multiply  ·  periodic table  ·  subscript

The ___________ of a substance is the mass of one ___________ of that substance. Its units are ___________. To read molar mass from the ___________, locate the element’s relative atomic mass — the larger number in each element’s box. For a compound, ___________ each element’s relative atomic mass by its ___________ in the chemical formula, then add the contributions of all elements together to give the total molar mass in ___________. To find the amount in moles, ___________ the mass of the sample (in grams) by the molar mass.

Stuck? Revisit the “What Is Molar Mass?” card and the formula panel in the lesson.

4. Calculate molar mass

For each substance, show the complete MM calculation step-by-step (list each element × Ar, then sum). Use the periodic table values provided. 10 marks (2 per substance)

Ar values (g mol−1): H = 1.008  ·  C = 12.011  ·  N = 14.007  ·  O = 15.999  ·  Na = 22.990  ·  Mg = 24.305  ·  S = 32.060  ·  Cl = 35.450  ·  Ca = 40.078  ·  Fe = 55.845

4.1 NaCl (sodium chloride)

4.2 CO2 (carbon dioxide)

4.3 NH3 (ammonia)

4.4 MgSO4 (magnesium sulfate)

4.5 Ca(OH)2 (calcium hydroxide) — expand the brackets first

Stuck? Revisit the H2SO4 breakdown SVG and the brackets example (Ca(OH)2) in the lesson.

5. Short recall questions

Answer each question in 1–2 sentences using precise terms from the lesson. 8 marks (2 each)

5.1 What does the “larger number” in each element box on the periodic table represent, and what unit does molar mass carry?

5.2 Why must you multiply the Ar of each element by its subscript before summing to find the molar mass of a compound?

5.3 Write the three rearrangements of n = m ÷ MM (one to find n, one to find m, one to find MM).

5.4 What unit cancellation confirms that n = m ÷ MM gives the correct unit for amount of substance?

Stuck? Revisit the formula panel, the formula triangle SVG, and the units check callout in the lesson.
Answers — Do not peek before attempting

Q1 — Term–definition match

1.1 molar mass • 1.2 relative atomic mass (Ar) • 1.3 relative molecular mass (Mr) • 1.4 formula mass • 1.5 mole • 1.6 subscript rule • 1.7 n = m ÷ MM • 1.8 g mol−1.

Q2 — True / false with correction

2.1 False. The molar mass of O2 is 2 × 15.999 = 31.998 g mol−1, not 16.00. The subscript 2 in O2 means there are two oxygen atoms per molecule.

2.2 True. Ar is dimensionless; molar mass carries the unit g mol−1 but has the same numerical value.

2.3 False. To find moles, you divide mass by molar mass: n = m ÷ MM. Multiplying mass by molar mass has no physical meaning and gives the wrong unit.

2.4 True. The subscript 2 outside the bracket applies to both O and H: Ca(OH)2 = 1 Ca + 2 O + 2 H.

2.5 True. MM(H2O) = 2(1.008) + 15.999 = 18.015 g mol−1.

2.6 False. “Formula mass” is used for ionic compounds (which have no discrete molecules). “Relative molecular mass” is used for covalent compounds. They involve the same calculation but are applied in different contexts.

Q3 — Cloze paragraph

In order: molar mass / mole / g mol−1 / periodic table / multiply / subscript / grams / divide.

Q4 — Molar mass calculations

4.1 NaCl: 1 × Na + 1 × Cl = 22.990 + 35.450 = 58.440 g mol−1.

4.2 CO2: 1 × C + 2 × O = 12.011 + 2(15.999) = 12.011 + 31.998 = 44.009 g mol−1.

4.3 NH3: 1 × N + 3 × H = 14.007 + 3(1.008) = 14.007 + 3.024 = 17.031 g mol−1.

4.4 MgSO4: 1 × Mg + 1 × S + 4 × O = 24.305 + 32.060 + 4(15.999) = 24.305 + 32.060 + 63.996 = 120.361 g mol−1.

4.5 Ca(OH)2: Expand brackets: 1 Ca + 2 O + 2 H.
= 40.078 + 2(15.999) + 2(1.008) = 40.078 + 31.998 + 2.016 = 74.092 g mol−1.

Q5.1 — Periodic table reading

The larger number in each element box is the relative atomic mass (Ar). Molar mass carries the unit g mol−1 (grams per mole).

Q5.2 — Why multiply by subscript

The subscript tells you how many atoms of that element appear in one formula unit of the compound. Each atom contributes its own Ar to the total mass, so you must add Ar once for each atom present. Multiplying Ar by the subscript is an efficient way to do this before summing all contributions.

Q5.3 — Three rearrangements

Find n: n = m ÷ MM  •  Find m: m = n × MM  •  Find MM: MM = m ÷ n.

Q5.4 — Unit cancellation

n = m ÷ MM gives: g ÷ (g mol−1) = g × mol g−1 = mol ✓. The grams cancel, leaving the correct unit of amount of substance.

HSCScience • Chemistry • Year 11 • Module 2 • Lesson 2 • Worksheet 1 (Build)