Percentage Composition, Empirical Formula, and Molecular Formula

You already know how to count particles using the mole and how to find the mass of one mole of a substance. But what if someone hands you an unknown compound and asks, “What is it made of?” The first step is to figure out which elements are present and how much of each. That is exactly what percentage composition does. And once you have those numbers, you can work backwards to find the compound’s formula.

Fig 1.11: One mole of various substances

What is Percentage Composition?

The mass percentage (also called mass per cent) of an element in a compound tells you what share of the total mass comes from that element. Think of it this way: if you had 100 g of the compound, how many grams would be contributed by that particular element?

The formula is straightforward:

$\text{Mass \% of an element} = \frac{\text{mass of that element in one formula unit of the compound}}{\text{molar mass of the compound}} \times 100$

To find the “mass of that element in one formula unit,” you multiply the number of atoms of that element in the formula by its atomic mass.

Working It Out for Water

Let us see how this works with a familiar compound: water ($H_2O$).

Water contains hydrogen and oxygen. Its molar mass is 18.02 g.

Mass percentage of hydrogen:

Water has 2 hydrogen atoms, each with an atomic mass of 1.008 u. So the total mass of hydrogen in one formula unit is $2 \times 1.008 = 2.016$ g.

$\text{Mass \% of H} = \frac{2.016}{18.02} \times 100 = 11.18\%$

Mass percentage of oxygen:

There is 1 oxygen atom with an atomic mass of 16.00 u.

$\text{Mass \% of O} = \frac{16.00}{18.02} \times 100 = 88.79\%$

Notice something interesting: even though water has twice as many hydrogen atoms as oxygen atoms, oxygen accounts for nearly 89% of the mass. This is because each oxygen atom is roughly 16 times heavier than a hydrogen atom. Percentage composition is about mass contributions, not atom counts.

Also note that $11.18 + 88.79 \approx 100\%$. The percentages of all elements in any compound must add up to 100%. If they do not, you have made an error somewhere. This quick check is always worth doing.

Ethanol: A More Detailed Example

Now let us try a slightly more complex compound: ethanol ($C_2H_5OH$). Its molecular formula tells us it contains carbon, hydrogen, and oxygen.

First, work out the molar mass by adding up the atomic masses of every atom:

$\text{Molar mass of } C_2H_5OH = (2 \times 12.01) + (6 \times 1.008) + 16.00 = 24.02 + 6.048 + 16.00 = 46.068 \text{ g}$

Notice that the formula $C_2H_5OH$ has 2 carbon atoms, 6 hydrogen atoms (five from $H_5$ plus one from the $OH$ group), and 1 oxygen atom.

Now calculate each element’s mass percentage:

Carbon:

$\text{Mass \% of C} = \frac{24.02}{46.068} \times 100 = 52.14\%$

Hydrogen:

$\text{Mass \% of H} = \frac{6.048}{46.068} \times 100 = 13.13\%$

Oxygen:

$\text{Mass \% of O} = \frac{16.00}{46.068} \times 100 = 34.73\%$

Quick check: $52.14 + 13.13 + 34.73 = 100.00\%$. The numbers add up perfectly.

Why Percentage Composition is Useful

Percentage composition data serves two important purposes:

• Identifying unknown compounds. If you analyse a new substance in the lab and find that it contains 52.14% carbon, 13.13% hydrogen, and 34.73% oxygen, you can use these numbers to determine its formula (we will learn how to do this in the next section).
• Checking purity. For a known compound, the mass percentages are fixed and well-established. If you analyse a sample that is supposed to be pure ethanol but the carbon percentage comes out as 48% instead of 52.14%, you know the sample is not pure. Something else is mixed in.

From Percentages to Formulas: Empirical and Molecular Formulas

So far, you have been going from a known formula to its percentage composition. But the real power of this calculation lies in going the other direction: starting with percentage data and finding the formula.

Before we get into the method, two terms need to be clear:

• The empirical formula gives the simplest whole-number ratio of the atoms in a compound. It tells you the relative proportions, but not the actual count. For example, $CH_2O$ says “for every carbon atom, there are two hydrogens and one oxygen.”
• The molecular formula gives the actual number of each type of atom in one molecule. It may be the same as the empirical formula, or it may be a whole-number multiple of it. Glucose has the molecular formula $C_6H_{12}O_6$, but its empirical formula is just $CH_2O$ (the ratio 6:12:6 simplifies to 1:2:1).

The empirical formula can be found from percentage composition data alone. To find the molecular formula, you also need the molar mass of the compound.

The Five-Step Method: Finding Empirical and Molecular Formulas

Here is a systematic procedure that works for any compound, illustrated with a worked example.

The Problem

A compound contains 4.07% hydrogen, 24.27% carbon, and 71.65% chlorine. Its molar mass is 98.96 g. Find the empirical and molecular formulas.

Step 1: Convert mass percentages to grams

The simplest approach is to imagine a 100 g sample. Then each percentage becomes a mass in grams directly:

• Hydrogen: 4.07 g
• Carbon: 24.27 g
• Chlorine: 71.65 g

This works because “4.07% of 100 g” is just 4.07 g. No calculation needed.

Step 2: Convert grams to moles

Divide each mass by the element’s atomic mass. This converts the masses into mole amounts, because the formula depends on the ratio of atoms (moles), not the ratio of masses.

$\text{Moles of H} = \frac{4.07}{1.008} = 4.04$

$\text{Moles of C} = \frac{24.27}{12.01} = 2.021$

$\text{Moles of Cl} = \frac{71.65}{35.453} = 2.021$

Step 3: Find the simplest ratio

Divide every mole value by the smallest mole value. Here, the smallest is 2.021:

$\text{H} : \frac{4.04}{2.021} = 2.0 \qquad \text{C} : \frac{2.021}{2.021} = 1.0 \qquad \text{Cl} : \frac{2.021}{2.021} = 1.0$

The ratio is $\text{H : C : Cl} = 2 : 1 : 1$, which is already in whole numbers.

What if the ratios are not whole numbers? Sometimes you will get values like 1 : 1.5 : 1. In that case, multiply all values by the smallest integer that makes them whole numbers. Here, multiplying by 2 gives 2 : 3 : 2. If you get something like 1 : 1.33 : 1, multiply by 3 to get 3 : 4 : 3.

Step 4: Write the empirical formula

Place each element with its ratio number as a subscript:

$\text{Empirical formula} = CH_2Cl$

Step 5: Find the molecular formula

This is a three-part calculation:

(a) Calculate the empirical formula mass. Add up the atomic masses of all atoms in the empirical formula:

$\text{Empirical formula mass of } CH_2Cl = 12.01 + (2 \times 1.008) + 35.453 = 12.01 + 2.016 + 35.453 = 49.48 \text{ g}$

(b) Find the multiplier $n$. Divide the actual molar mass by the empirical formula mass:

$n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{98.96}{49.48} = 2$

This tells you that the real molecule contains exactly twice the atoms shown in the empirical formula.

(c) Multiply through. Multiply every subscript in the empirical formula by $n$:

$\text{Molecular formula} = (CH_2Cl) \times 2 = C_2H_4Cl_2$

So the compound is $C_2H_4Cl_2$ (dichloroethane).

Quick Summary of the Method

Step	What you do	Example value
1	Treat percentages as grams (100 g sample)	H: 4.07 g, C: 24.27 g, Cl: 71.65 g
2	Divide each by its atomic mass to get moles	H: 4.04, C: 2.021, Cl: 2.021
3	Divide all by the smallest mole value	H: 2, C: 1, Cl: 1
4	Write the empirical formula	$CH_2Cl$
5	Divide molar mass by empirical formula mass to get $n$, then multiply	$n = 2$, so $C_2H_4Cl_2$

The key insight behind this entire method is that the empirical formula is about the mole ratio of the elements, not their mass ratio. That is why Step 2 (converting grams to moles) is essential. Masses alone can be misleading because different atoms have very different weights. Converting to moles puts every element on the same footing: each value now represents an actual count of atoms.