Relative Mass and the Mole
Calculate relative masses, distinguish them from molar mass and convert mass and amount with units and magnitude checks.
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The core idea
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Learning objectives
- define relative atomic mass, Ar
- define relative molecular mass, Mr, and calculate relative molecular mass (and relative formula mass) as the sum of relative atomic masses
- define the term mole in terms of the Avogadro constant
1. Outcome and prerequisites
By the end, you should be able to calculate relative molecular or formula mass from supplied atomic masses, distinguish unitless relative mass from molar mass in g mol⁻¹, and convert mass in grams to amount in moles and back.
Use
where n is amount in mol, m is mass in g and M is molar mass in g mol⁻¹.
For CO₂, Mᵣ = 44.0 has no unit. Its molar mass is 44.0 g mol⁻¹. The numerical value is the same, but the meaning and unit are not.
2. Modelled example
Modelled example 1
Count atoms before converting mass to amount
Problem
View solution step by step
Calculate the relative formula mass
Method
Count every atom shown by the formula.Reason
There are three oxygen atoms; missing one changes the molar mass and every later conversion.Working
Mᵣ = 40.0 + 12.0 + 3(16.0) = 100.0 This relative mass has no unit.Attach the molar-mass unit
Method
Express the mass of one mole in grams per mole.Reason
Mass-to-amount conversion needs mass per mole.Working
M = 100.0 g mol⁻¹Convert mass to amount
Method
Divide the sample mass by the mass of one mole.Reason
25.0 g is one quarter of 100.0 g, so the answer should be about one quarter of a mole.Working
n = (25.0 g)/(100.0 g mol⁻¹) = 0.250 mol
3. Guided practice
Guided practice 2
Convert in both directions
Problem
Show formula counting and units
Hints
Hint 1: expand brackets
Hint 2: requested mass
View solution step by step
Expand the brackets
Method
Count two nitrogen, eight hydrogen, one sulfur and four oxygen atoms.Reason
The subscript outside a bracket multiplies every atom inside it.Working
Mᵣ = 2(14.0) + 8(1.0) + 32.0 + 4(16.0) = 132.0.Use m = nM
Method
Multiply the requested amount by the molar mass.Reason
The unknown is mass, so rearranging n = m/M gives m = nM.Working
m = (0.150 mol)(132.0 g mol⁻¹) = 19.8 g
4. Plausible error contrast
A learner calculates the amount in 11.0 g of CO₂ as n = 11.0 × 44.0 = 484 mol. The first invalid step is reversing n = m/M. Correctly,
A sample lighter than one molar mass contains less than one mole. The reported 484 mol would require over 21 kg of carbon dioxide, so it cannot describe an 11.0 g sample.
5. Changed-context transfer
Mind stretcher 1: Interpret a less familiar formulaExtension
Using Mg = 24.3, N = 14.0 and O = 16.0, calculate the amount in 7.41 g of Mg(NO₃)₂. Do not round until the final answer.
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Mᵣ = 24.3 + 2(14.0) + 6(16.0) = 148.3, so M = 148.3 g mol⁻¹. Then n = 7.41/148.3 = 0.04997 mol, which rounds to 0.0500 mol. The mass is about one twentieth of the molar mass, matching the magnitude.
6. Independent evidence
Using Al = 27.0, S = 32.0 and O = 16.0, calculate the relative formula mass and molar mass of Al₂(SO₄)₃. Then calculate (a) the amount in 17.1 g and (b) the mass of 0.0750 mol. Label relative mass without a unit, label molar mass in g mol⁻¹, and use a magnitude check for both conversions.
Continue to Stoichiometric Reacting Quantities when both directions are secure.