Calculating Relative Molar Mass for the ESAT

Introduction to Relative Masses

To perform quantitative chemistry in the laboratory or on an industrial scale, chemists must count atoms, ions, and molecules by weighing them. This is because chemical particles are far too small to count individually. The basis for these calculations is the relative atomic mass ($A_r$).

The relative atomic mass of an element is the weighted mean of the mass numbers of the isotopes of an element, taking into account their abundances. These values are relative to $\frac{1}{12}$ the mass of an atom of carbon-12 ($^{12}_{6}\text{C}$). Because $A_r$ and $M_r$ are ratios of masses compared to a standard, they are dimensionless and do not have any units.

Defining Relative Molar Mass ($M_r$)

Many elements and all compounds exist as groups of atoms bonded together. The relative molar mass (also referred to as the relative molecular mass for covalent compounds or relative formula mass for ionic compounds), represented by the symbol $M_r$, is the total mass of the formula unit.

To calculate the $M_r$ of a substance, you must add together the $A_r$ values of every atom or ion shown in its chemical formula.

Step by Step Calculation Method

1. Identify every element present in the chemical formula.
2. Determine the number of atoms for each element, paying close attention to subscripts and brackets.
3. Look up the $A_r$ value for each element.
4. Multiply the $A_r$ of each element by the number of atoms of that element present.
5. Sum these values together to find the final $M_r$.

Worked Examples of $M_r$ Calculations

The following examples demonstrate how to calculate $M_r$ for different types of chemical species, using the provided $A_r$ values: $H = 1.0$; $C = 12.0$; $N = 14.0$; $O = 16.0$; $S = 32.0$; $Cu = 63.5$; $Ba = 137$.

Example 1: Simple Covalent Molecules

To find the $M_r$ of oxygen gas ($O_2$): $M_r(O_2) = 2 \times A_r(O) = 2 \times 16.0 = 32.0$

Example 2: Larger Organic Molecules

To find the $M_r$ of glucose ($C_6H_{12}O_6$): $M_r = (6 \times A_r(C)) + (12 \times A_r(H)) + (6 \times A_r(O))$ $M_r = (6 \times 12.0) + (12 \times 1.0) + (6 \times 16.0)$ $M_r = 72.0 + 12.0 + 96.0 = 180.0$

Example 3: Ionic Compounds with Brackets

When a formula contains brackets, such as barium nitrate, $Ba(NO_3)_2$, the subscript outside the bracket multiplies everything inside the bracket. $M_r(Ba(NO_3)_2) = A_r(Ba) + 2 \times [A_r(N) + (3 \times A_r(O))]$ $M_r = 137 + 2 \times [14.0 + (3 \times 16.0)]$ $M_r = 137 + 2 \times [14.0 + 48.0]$ $M_r = 137 + 2 \times [62.0] = 261.0$

Example 4: Hydrated Compounds

Hydrated salts like copper(II) sulfate crystals, $CuSO_4 \cdot 5H_2O$, contain water of crystallisation. The dot in the formula indicates that the mass of the water molecules must be added to the mass of the salt. $M_r = A_r(Cu) + A_r(S) + (4 \times A_r(O)) + 5 \times [(2 \times A_r(H)) + A_r(O)]$ $M_r = 63.5 + 32.0 + (4 \times 16.0) + 5 \times [18.0]$ $M_r = 63.5 + 32.0 + 64.0 + 90.0 = 249.5$

Connection to Molar Mass

While $M_r$ itself is a dimensionless ratio, it is numerically equal to the molar mass of a substance. The molar mass is the mass of one mole of that substance and is measured in grams per mole ($g \text{ mol}^{-1}$). For example, since the $M_r$ of water ($H_2O$) is 18.0, its molar mass is $18.0 \text{ g mol}^{-1}$.