IB DP Chemistry - S1.4.3 Molar mass - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – S1.4.3 Molar mass – Study Notes – New Syllabus
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Structure 1.4.3 — Molar Mass and the n = m/M Relationship
Structure 1.4.3 — Molar Mass and the n = m/M Relationship
Molar Mass
The molar mass (denoted \( M \)) is the mass of one mole of a substance. It is expressed in units of grams per mole (g mol–1).
The molar mass is numerically equal to the relative molecular or formula mass (Mr) but expressed in grams.
For example:
- Oxygen, \( O_2 \): \( M = 32.00 \, \text{g mol}^{-1} \)
- Water, \( H_2O \): \( M = 18.02 \, \text{g mol}^{-1} \)
- Sodium chloride, \( NaCl \): \( M = 58.44 \, \text{g mol}^{-1} \)
Key Formula : Mass, Moles and Molar Mass
The relationship between the mass of a substance, its amount in moles, and molar mass is:
\( n = \frac{m}{M} \)
- \( n \) = number of moles (mol)
- \( m \) = mass of substance (g)
- \( M \) = molar mass (g mol–1)
This formula allows you to move easily between the mass, the amount in moles, and molar mass of any substance.
Link to Particles (via Avogadro’s constant):
You can also link mass and particles using:
Number of particles = \( n \times N_A \)
Where \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \) (Avogadro constant)
Example
Calculate the number of moles in 9.00 g of water (\( H_2O \)).
▶️ Answer/Explanation
Molar mass of water:
\( M = 2(1.01) + 16.00 = 18.02 \, \text{g mol}^{-1} \)
Using the formula:
\( n = \frac{m}{M} = \frac{9.00}{18.02} = 0.500 \, \text{mol} \)
Example
What is the mass of 0.25 mol of carbon dioxide (\( CO_2 \))?
▶️ Answer/Explanation
Molar mass of \( CO_2 \):
\( M = 12.01 + 2(16.00) = 44.01 \, \text{g mol}^{-1} \)
Using the formula:
\( m = n \times M = 0.25 \times 44.01 = 11.00 \, \text{g} \)
Example
A student weighs 8.5 g of sodium chloride. Calculate the number of sodium chloride formula units present.
▶️ Answer/Explanation
Molar mass of \( NaCl = 22.99 + 35.45 = 58.44 \, \text{g mol}^{-1} \)
\( n = \frac{m}{M} = \frac{8.5}{58.44} = 0.1454 \, \text{mol} \)
Number of formula units = \( n \times N_A = 0.1454 \times 6.022 \times 10^{23} \)
= \( 8.76 \times 10^{22} \, \text{units} \)
Example
Determine the molar mass of a compound if 0.050 mol of it has a mass of 3.60 g.
▶️ Answer/Explanation
Using the formula:
\( M = \frac{m}{n} = \frac{3.60}{0.050} = 72.0 \, \text{g mol}^{-1} \)