Home / IB DP / IB DP Chemistry 2026, 2027 & 2028 / IB DP Chemistry SL & HL Study Notes / IB DP Chemistry – S1.4.3 Molar mass – Study Notes

IB DP Chemistry – S1.4.3 Molar mass – Study Notes

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

IITian Academy excellent Study Notes and effective strategies will help you prepare for your IB DP Chemistry  exam.

IB DP Chemistry Study Notes – All Topics

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:
\( M = \frac{m}{n} = \frac{3.60}{0.050} = 72.0 \, \text{g mol}^{-1} \)

Scroll to Top