Chemistry 1.44 - 1.45 - Empirical and molecular formulae

From composition to an empirical formula

A mass ratio cannot be copied straight into a formula. Different atoms have different masses: $16 \,\mathrm{g}$ of oxygen atoms and $16 \,\mathrm{g}$ of iron atoms do not contain the same number of atoms. Dividing each element's mass by its $A_r$ removes this mass-per-atom difference and gives numbers proportional to the amounts of atoms present.

Use the same route whether the composition is given as reacting masses or as percentages:

1. Write the mass or percentage for each element.
2. Divide each value by that element's $A_r$.
3. Divide every result by the smallest result.
4. If a simple fraction remains, multiply every ratio number by the same small integer.
5. Use the resulting simplest whole numbers as subscripts; do not write a subscript $1$.

When percentages are given, imagine a $100 \,\mathrm{g}$ sample. For example, 43.7% phosphorus then corresponds to $43.7 \,\mathrm{g}$ of phosphorus in that imagined sample. The scale of the imagined sample does not affect the final ratio.

Worked example: from reacting masses

A compound contains $11.2 \,\mathrm{g}$ of iron combined with $4.8 \,\mathrm{g}$ of oxygen. Use $A_r(\ce{Fe}) = 56$ and $A_r(\ce{O}) = 16$.

The simplest whole-number atom ratio is $\ce{Fe}:\ce{O} = 2:3$, so the empirical formula is $\ce{Fe2O3}$.

Sense-check: using $\ce{FeO1.5}$ would not be a valid empirical formula because formula subscripts must be whole numbers. Rounding 1.5 to 2 would also change the ratio; multiplying both values by 2 preserves it.

Worked example: from percentage composition

A phosphorus oxide is $43.7\%$ phosphorus and $56.3\%$ oxygen by mass. Use $A_r(\ce{P}) = 31$ and $A_r(\ce{O}) = 16$.

The empirical formula is $\ce{P2O5}$.

The percentages were rounded measurements, so the intermediate ratio is approximately 2.50 rather than perfectly exact. Keep enough calculator digits until the ratio step. A value close to a whole number because of measurement rounding may be rounded appropriately, but a clear value such as 1.5, 2.5 or 1.33 signals that the entire ratio should be multiplied by 2 or 3.

Reverse check: $M_r(\ce{P2O5}) = (2 \times 31) + (5 \times 16) = 142$. Phosphorus contributes 62, and $(62/142) \times 100 = 43.66\ldots\%$, which agrees with 43.7% after rounding.

For oxygen that entered from oxygen gas, divide the oxygen mass by $A_r(\ce{O}) = 16$, not by $M_r(\ce{O2}) = 32$. The empirical formula counts oxygen atoms in the compound, even though oxygen arrived as molecules.