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Module 2 · L2 of 20 ~35 min ⚡ +50 XP in Learn · +25 to complete

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.

Today's hook, Every dose in an IV bag, every gram of fertiliser, calculated using molar mass. It's the workhorse formula of practical chemistry, and it lives on the periodic table you already have.

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Warm up and orient yourself

Recall earlier work, make a prediction, then scan the formula you are about to use.

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

Recall, your gut answer first

+5 XP warm-up

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?

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Formula reference · this lesson

core formula

$$n = \dfrac{m}{MM}$$

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: $MM = m \div n$

What you'll master, and the words for it

What you'll master

Know

Key facts

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

Concepts

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

Skills

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

Key terms

Molar mass (M)

The mass of one mole of a substance, expressed in g mol⁻¹.

Relative atomic mass (Aᵣ)

The weighted average mass of an element's atoms relative to ¹²C = 12; dimensionless but numerically equal to M in g mol⁻¹.

Relative molecular mass (Mᵣ)

The sum of relative atomic masses in a molecule; numerically equal to molar mass in g mol⁻¹.

Formula mass

Same calculation as Mᵣ but used for ionic compounds (which don't have discrete molecules).

n = m ÷ M

The mole–mass equation: amount (mol) = mass (g) ÷ molar mass (g mol⁻¹).

Subscript rule

In a formula, multiply the element's Aᵣ by its subscript, then sum all contributions to get M.

What molar mass actually is

What is molar mass?

core concept · +3 XP at end

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.

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 breakdown of H2SO4 as a comparison table across elements

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

Molar mass (MM) is the mass of one mole of a substance in g mol⁻¹, numerically equal to the relative atomic or molecular mass from the periodic table. For compounds, multiply each element's RAM by its subscript (expanding brackets first) and sum all contributions.

Pause, copy the highlighted definition into your book before moving on.

Did you get this? True or false: the molar mass of O₂ is 2 × 15.999 = 31.998 g mol⁻¹, not 15.999 g mol⁻¹.

Quick check: Which is the correct molar mass calculation for Ca(OH)₂?

The formula that links n, m and MM

The formula: n = m ÷ MM

core concept

We just saw how to calculate molar mass for any element or compound using the periodic table. That raises a question: once you know MM, how do you convert a weighed mass into moles (or back)? This card answers it → with 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.

The m-n-MM triangle links mass, moles and molar mass.

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 ✓

n = m ÷ MM converts mass (g) to moles; rearranged, m = n × MM and MM = m ÷ n. Units confirm the formula: g ÷ (g mol⁻¹) = mol. Always convert kg to g before substituting.

Pause, write the highlighted equation into your book.

Fill the blanks: drag each token into the matching blank.

divide multiply MM mol

To find moles from grams, ___ by ___. To find grams from moles, ___ by MM. n is measured in ___ .

Did you get this? True or false: if a question gives mass in kilograms, you must convert to grams before substituting into n = m ÷ MM.

Two truths, one lie about molar mass. Pick the lie.

The practical: measuring molar mass in the lab

Practical investigation, measuring molar mass experimentally

practical investigation · core concept

We just saw how to calculate molar mass from a formula using the periodic table. That raises a question: can you find the molar mass of an unknown volatile liquid without already knowing its formula? This card answers it → the volatile liquid (vapour density) method, a real HSC practical.

The volatile liquid method determines the molar mass of a liquid that evaporates easily (for example ethanol or propanone) by weighing the vapour that exactly fills a flask of known volume at a measured temperature and pressure.

Aim

To determine the molar mass of a volatile liquid experimentally, then compare it with the value calculated from its molecular formula.

A volatile liquid is vaporised in a foil-capped flask; the condensed vapour's mass gives the molar mass.

Method: (1) Weigh a small conical flask fitted with a loose foil cap pierced with a fine pinhole, this is mass m₁. (2) Add 2 to 3 mL of the volatile liquid and replace the cap. (3) Stand the flask in a beaker of water heated to a steady boil, a water bath rather than a naked flame, since these liquids are flammable, until all the liquid has vaporised and excess vapour has escaped through the pinhole. (4) Record the temperature of the boiling water bath and the atmospheric pressure. (5) Remove the flask, allow it to cool to room temperature so the vapour condenses back to liquid, dry the outside, and reweigh, this is mass m₂. (6) Fill the flask completely with water and measure the volume, this is the volume the vapour occupied inside the flask.

Calculating molar mass from the data: mass of vapour = m₂ − m₁. Convert the recorded volume, temperature and pressure of the flask into moles of vapour using PV = nRT, the ideal gas equation developed fully in Lesson 15. Then MM = mass of vapour ÷ moles of vapour, exactly the same n = m ÷ MM relationship you just learned, just with the moles found experimentally instead of from a formula.

Why a foil cap with a pinhole?

The pinhole lets excess vapour and displaced air escape as the liquid boils, so the flask ends up containing pure vapour at atmospheric pressure, not a pressurised mixture of air and vapour. Sealing the flask completely would be a safety hazard and would also trap air, giving a molar mass that is too high.

The volatile liquid (vapour density) method finds molar mass experimentally: heat a known mass of liquid in a flask of known volume until it fully vaporises through a pinholed foil cap, record temperature and pressure, cool and reweigh to find the mass of vapour, then use PV = nRT to find moles and MM = mass ÷ moles.

Pause, copy the highlighted method into your book before moving on.

Did you get this? True or false: the foil cap in the volatile liquid method must be completely sealed with no pinhole, so no vapour can escape.

Sort the steps+7 XP

Put these steps of the volatile liquid method for determining molar mass into the correct order.

Cool the flask, dry it, and reweigh to find the mass of condensed vapour.
Weigh the empty flask fitted with a pinholed foil cap.
Add 2 to 3 mL of the volatile liquid and replace the cap.
Heat the flask in a boiling water bath until all the liquid has vaporised and excess vapour has escaped through the pinhole.
Record the water bath temperature and the atmospheric pressure.
Fill the flask with water to measure its volume, then use PV = nRT and MM = m ÷ n to calculate the molar mass.

Two worked examples

Worked examples · reveal as you go

Worked example 1 · finding moles from mass +5 XP on full reveal

Calculate the number of moles in 54 g of water (H₂O).

$\mathrm{MM(H_2O)} = 2(1.008) + 15.999 = 18.015\ \mathrm{g}\ \mathrm{mol}^{-1}$

Calculate molar mass

$m = 54\ \mathrm{g}$ | $MM = 18.015\ \mathrm{g}\ \mathrm{mol}^{-1}$ | $n = ?$

Identify known values

$n = m \div MM$

Write the formula

$n = 54 \div 18.015$

Substitute values

$n = 2.998\ \mathrm{mol} \approx 3.0\ \mathrm{mol}$ of H₂O ✓

Calculate · units: g ÷ g mol⁻¹ = mol ✓

Worked example 2 · finding mass from moles +5 XP on full reveal

What mass of sodium chloride (NaCl) contains 0.50 mol?

$\mathrm{MM(NaCl)} = 22.990 + 35.450 = 58.440\ \mathrm{g}\ \mathrm{mol}^{-1}$

Calculate molar mass

$n = 0.50\ \mathrm{mol}$ | $MM = 58.440\ \mathrm{g}\ \mathrm{mol}^{-1}$ | $m = ?$

Identify known values

$m = n \times MM$

Rearrange the formula

$m = 0.50 \times 58.440$

Substitute

$m = 29.22\ \mathrm{g}$ of NaCl ✓

Calculate · units: mol × g mol⁻¹ = g ✓

Sort the steps+7 XP

Put these steps for "What mass of NaCl contains 0.50 mol?" into the correct order.

Calculate: m = 0.50 × 58.440 = 29.22 g of NaCl.
List known and unknown values: n = 0.50 mol, MM = 58.440 g mol⁻¹, m = ?
Calculate molar mass: MM(NaCl) = 22.990 + 35.450 = 58.440 g mol⁻¹.
Rearrange the formula: m = n × MM.

The three traps, then quick-fire drill

Common errors · the 3 traps that cost 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.

Quick-fire practice · 5 reps +2 XP per reveal

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

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

A technician-prepared solution contains 14.7 g of dissolved sulfuric acid (H₂SO₄). Calculate the amount in moles. (S = 32.06 g mol⁻¹)

Calculate the mass of 0.40 mol of glucose (C₆H₁₂O₆). (C = 12.011, H = 1.008, O = 15.999)

A sample of an unknown compound has a mass of 23.4 g and contains 0.30 mol. Calculate the molar mass of the compound.

Revisit your 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.

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Do this next

Practise independently

Attempt at least one calculation or explanation in your own words before checking the model answers.

Multiple choice

+2 XP per correct · +5 bonus if perfect

Pick your answer, then rate your confidence. That tells the system what to drill next.

Short answer

UnderstandLower-order3 marks

Q1. 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.

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ApplyLower-order4 marks

Q2. 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)

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AnalyseMid-order4 marks

Q3. 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)

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EvaluateHigher-order5 marks

Q4. 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 a hypothetical batch-release specification of ≥97.0% purity by mass (illustrative figure, not an actual pharmacopoeia limit). Show full working.

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CreateHigher-order6 marks

Q5. A student has a balance, distilled water, a volumetric flask, a pipette, a burette, and a standard solution that reacts with the unknown soluble salt in a known 1:1 mole ratio. Design a procedure to determine the molar mass of the 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.

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📖 Comprehensive answers (click to reveal)

Multiple choice, drill bank

MC answers and feedback are shown inline as you complete each question. Use the retry button to attempt a fresh set.

Key MC notes:

Molar mass of MgSO₄. MgSO₄: 24.305 + 32.06 + 4(15.999) = 24.305 + 32.06 + 63.996 = 120.361 g mol⁻¹

Rearranging the formula. m = n × MM. Rearranging n = m ÷ MM gives m = n × MM.

Moles from mass, a diatomic gas. MM(Cl₂) = 2 × 35.45 = 70.90 g mol⁻¹. n = 71.0 ÷ 70.90 = 1.00 mol

Molar mass with brackets, Al(OH)₃. Al(OH)₃: 26.982 + 3(15.999) + 3(1.008) = 26.982 + 47.997 + 3.024 = 78.003 g mol⁻¹

Moles of iron from mass. n = 167.2 ÷ 55.845 = 2.994 mol ≈ 2.99 mol

Brackets multiply everything inside them. MM(Ca(H₂PO₄)₂) = 40.078 + 2(2×1.008 + 30.974 + 4×15.999) = 234.05 g mol⁻¹

Identifying an unknown metal from its molar mass. MM(XCl₂) = 55.75 ÷ 0.500 = 111.50 g mol⁻¹. MM(X) = 111.50 − 2(35.453) = 40.59 ≈ calcium (Ca = 40.078)

Short answer model answers

Q1 (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].

Q2 (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, 1 for correct formula, 1 for substitution, 1 for final answer with units.

Q3 (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.

Q4 (5 marks): (a) % purity = 9.76 ÷ 10.00 = 97.6%. m(pure) per tablet = 0.500 g × 0.976 = 0.488 g. n = 0.488 ÷ 180.16 = 2.71 × 10⁻³ mol. (b) Since 97.6% ≥ 97.0%, the batch passes the quality standard.

Q5 (6 marks): (a) Measurements: (i) accurately weigh the dry salt sample (tare the container, then weigh); (ii) dissolve it in distilled water and make up to the mark in the volumetric flask of known volume (e.g. 250.0 mL); (iii) pipette aliquots and titrate against the standard solution to find the average titre. (b) Calculation: n(standard) = c × V(titre); since the reaction is 1:1, n(salt in aliquot) = n(standard); scale up by (V(flask) ÷ V(aliquot)) to get total n(salt); MM = m(salt weighed) ÷ n(salt total). (c) Systematic error: the salt may absorb moisture from air (hygroscopic), so m(salt) appears larger than the true mass, making MM too large. Fix: dry the salt in an oven and cool in a desiccator before weighing.

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