3.1.2.4 - Empirical and Molecular Formula

What each formula means

An empirical formula and a molecular formula do not tell you exactly the same thing, so it is worth separating the ideas before doing any calculations.

The key word is simplest. If a substance has carbon and hydrogen in a 2:6 ratio, that can be simplified to 1:3, so its empirical formula would be $\mathrm{CH_3}$.

So ethene has molecular formula $\mathrm{C_2H_4}$, but its empirical formula is $\mathrm{CH_2}$. By contrast, carbon dioxide has molecular formula $\mathrm{CO_2}$ and empirical formula $\mathrm{CO_2}$ because the ratio is already in its simplest form.

This gives the key relationship between the two ideas: the molecular formula is always a whole-number multiple of the empirical formula. Later in the lesson, we will find that multiple by comparing $M_r$ with the mass of one empirical-formula unit.

Finding empirical formula from masses

To calculate an empirical formula from mass data, always follow the same sequence:

1. Write down the mass of each element.
2. Convert each mass into moles using $n = \frac{m}{A_r}$.
3. Divide every mole value by the smallest mole value.
4. If needed, multiply the ratio so that every number becomes a whole number.

Here, $n$ is the amount in mol, $m$ is the mass in g, and $A_r$ is the relative atomic mass of the element. The most common mistake is to compare masses directly. Formulae come from mole ratios, not mass ratios.

The diagram below summarises the standard sequence for turning composition data into an empirical formula, and following the same order each time helps prevent ratio mistakes.

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Worked example: a compound contains 4.01 g calcium, 1.20 g carbon, and 4.80 g oxygen.

The simplest ratio is therefore Ca : C : O = 1 : 1 : 3, so the empirical formula is $\mathrm{CaCO_3}$.

Notice that it is sensible to keep a few decimal places until the ratio stage. If you round too early, the final ratio can come out wrong.

Finding composition from a metal oxide

A crucible, lid and magnesium can be weighed before heating in air. Heat with the lid slightly raised from time to time to admit oxygen while limiting loss of white oxide powder. Cool before weighing, then heat, cool and weigh again until the mass is constant. The final mass gain is the mass of oxygen combined with the magnesium, provided no product was lost and the reaction was complete.

For example, 0.486 g magnesium forming 0.806 g oxide has gained 0.320 g oxygen. The amounts are $0.486/24.3=0.0200\,\mathrm{mol}$ Mg and $0.320/16.0=0.0200\,\mathrm{mol}$ O, giving MgO. Loss of product lowers the apparent oxygen gain and makes the inferred oxygen-to-magnesium ratio too small. Constant mass is evidence that further heating causes no measurable change; it is not by itself proof of a pure product.

Using percentage composition and awkward ratios

Percentage composition questions use exactly the same chemistry. The only extra step is to imagine a 100 g sample, so each percentage becomes a mass in grams. For example, 43.7% phosphorus means 43.7 g phosphorus in 100 g of the compound.

Suppose a compound is 43.7% phosphorus and 56.3% oxygen by mass:

• P moles $= \frac{43.7}{31.0} = 1.41$
• O moles $= \frac{56.3}{16.0} = 3.52$

Now divide by the smallest value:

• P : O $= 1 : 2.50$

That is not a whole-number ratio yet, so you must scale the whole ratio. Multiplying both numbers by 2 gives 2 : 5, so the empirical formula is $\mathrm{P_2O_5}$.

This is another place where ratios can go wrong. If you see 1 : 1.5, do not round 1.5 to 2. Instead, multiply the entire ratio by 2 to get 2 : 3. The same idea applies to values close to 1.33 or 1.25, which often need multiplying by 3 or 4.

If the question only gives percentages for some of the elements, first work out the missing percentage by subtracting from 100. That missing percentage can then be treated as a mass in a 100 g sample.

From empirical formula to molecular formula

Once you know the empirical formula, finding the molecular formula is a scaling problem. First calculate the mass of one empirical-formula unit. Then compare that value with the given $M_r$.

For a molecular substance:

Worked example: a compound has empirical formula $\mathrm{CH_2O}$ and $M_r = 60.0$.

1. Empirical formula mass $= 12.0 + (2 \times 1.0) + 16.0 = 30.0$
2. Multiplier $= \frac{60.0}{30.0} = 2$
3. Multiply every subscript in $\mathrm{CH_2O}$ by 2
4. Molecular formula $= \mathrm{C_2H_4O_2}$

If the multiplier is 1, then the molecular formula and empirical formula are the same. Also remember that $M_r$ is a relative quantity, so it has no unit.

One final trap to avoid: do not add atoms one by one. If the empirical formula is $\mathrm{CH_2O}$ and the multiplier is 2, the molecular formula is not something like $\mathrm{C_2H_3O_2}$. Every subscript must be multiplied by the same whole number.