Moles and Molar Mass Conversions for the ESAT

Introduction to the Mole

In chemical reactions, substances are processed in quantities involving trillions of particles. Because atoms, ions, and molecules are too small to count individually, chemists use the mole (unit: $mol$) to measure the amount of a substance. This allows us to relate the microscopic scale of atoms to the macroscopic scale of the laboratory.

Avogadro's constant ($N_A$) is the fundamental conversion factor. It defines the number of particles in one mole of a substance. To four significant figures, its value is $6.022 \times 10^{23} \text{ mol}^{-1}$. Just as a dozen refers to twelve items, one mole of helium atoms contains $6.022 \times 10^{23}$ atoms, and one mole of water contains $6.022 \times 10^{23}$ molecules.

Worked Example: Counting Particles

Question: A balloon contains $0.200 \text{ mol}$ of helium. How many atoms of helium are contained in the balloon? [$N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$]

Solution: To find the number of particles, multiply the amount in moles by Avogadro's constant.

$\text{Number of atoms} = 0.200 \text{ mol} \times 6.022 \times 10^{23} \text{ mol}^{-1} = 1.204 \times 10^{23} \text{ atoms}$.

Relative Molar Mass ($M_r$)

Many substances exist as molecules or ionic lattices rather than isolated atoms. The relative molar mass (or relative formula mass), $M_r$, is calculated by summing the relative atomic masses ($A_r$) of every atom or ion shown in the chemical formula. Since these are relative values compared to carbon-12, $M_r$ values have no units.

Worked Example: Calculating $M_r$

Calculate the relative molar masses of the following substances using these $A_r$ values: $H = 1.0$; $C = 12.0$; $N = 14.0$; $O = 16.0$; $S = 32.0$; $Cu = 63.5$; $Ba = 137$.

1. Oxygen gas ($O_2$): $2 \times 16.0 = 32.0$.
2. Glucose ($C_6H_{12}O_6$): $(6 \times 12.0) + (12 \times 1.0) + (6 \times 16.0) = 72.0 + 12.0 + 96.0 = 180.0$.
3. Barium nitrate ($Ba(NO_3)_2$): $137 + 2 \times [14.0 + (3 \times 16.0)] = 137 + 2 \times [62.0] = 261.0$.
4. Hydrated copper sulfate ($CuSO_4 \cdot 5H_2O$): $63.5 + 32.0 + (4 \times 16.0) + 5 \times [(2 \times 1.0) + 16.0] = 159.5 + 90.0 = 249.5$.

Conversions Between Mass and Moles

The mass of one mole of a substance in grams is numerically equal to its $A_r$ or $M_r$. This is known as the molar mass (unit: $g \text{ mol}^{-1}$). The relationship between mass, molar mass, and the amount of substance is defined by the formula:

$\text{Amount (mol)} = \frac{\text{mass (g)}}{\text{molar mass (g mol}^{-1}\text{)}}$

Worked Example: Mass to Moles

Question: What is the mass of $0.010 \text{ mol}$ of magnesium? [$A_r \text{ of Mg} = 24.3$]

Solution: Rearrange the formula to find mass:

$\text{Mass} = \text{Amount} \times \text{Molar mass}$

$\text{Mass} = 0.010 \text{ mol} \times 24.3 \text{ g mol}^{-1} = 0.243 \text{ g}$.

Working with Large Units (Kilograms and Tonnes)

In industrial chemistry, masses are often given in kilograms ($kg$) or tonnes. You must convert these to grams before using the mole equation:

• $1 \text{ kg} = 1,000 \text{ g}$ ($10^3 \text{ g}$)
• $1 \text{ tonne} = 1,000,000 \text{ g}$ ($10^6 \text{ g}$)

Worked Example: Unit Conversions

Question A: How many moles of aluminium are in $5.4 \text{ tonnes}$? [$A_r \text{ of Al} = 27.0$]

Solution: First, convert tonnes to grams: $5.4 \times 10^6 \text{ g}$.

$\text{Amount} = \frac{5.4 \times 10^6 \text{ g}}{27.0 \text{ g mol}^{-1}} = 200,000 \text{ mol}$.

Question B: What is the mass, in $kg$, of $100 \text{ mol}$ of sodium hydroxide ($NaOH$)? [$A_r \text{ values: } Na=23.0, O=16.0, H=1.0$]

Solution: First, find the $M_r$: $23.0 + 16.0 + 1.0 = 40.0$.

$\text{Mass (g)} = 100 \text{ mol} \times 40.0 \text{ g mol}^{-1} = 4,000 \text{ g}$.

Convert grams to kilograms: $\frac{4,000}{1,000} = 4 \text{ kg}$.