Introduction to Empirical Formula

Though it’s not quite as useful as a molecular formula, an empirical formula still tells you a lot about the substance, and it’s possible to calculate it with much less information. Naturally, the AP® Chemistry exam is almost guaranteed to ask you to calculate an empirical formula for at least one substance, if not more. You’ll learn how to make these calculations in this section of the AP® Chemistry Crash Course. Let’s get started!

Step 1. Figure Out How Much of Each Element is in the Compound.

Depending on the question, this step can take different forms. For example, let’s say that you have a compound composed of nitrogen and oxygen, where nitrogen makes up 30.4% of the compound by mass. We need to have mass of the two elements in grams in order to calculate an empirical formula. Since we don’t know the mass of the sample, we can just simply assume a weight of 100 \text{ g} to make the calculations easier; this means that the compound consists of 30.4 \text{ g } N and (100-30.4) = 69.6 \text{ g } O.

In another case, we might be given the mass of one element in the compound. For example, let’s say that we have another compound consisting of only nitrogen and oxygen. The total mass of the sample is 65 \text{ g}, and the mass of the nitrogen is 19.8 \text{ g}. Of course, the mass of the oxygen is then (65-19.8) = 45.2 \text{ g}.

Step 2. Convert Those Masses into Moles.

Because the empirical formula is based around the ratio of one element’s molecules to another element’s molecules, grams won’t help us much in calculating the empirical formula. Instead, we’ll convert the masses into moles, which correspond directly to the actual number of molecules of each element.

\text{Molar mass of } N = 14.01 \text{ g/mol}

\text{Molar mass of } O = 16.00 \text{ g/mol}

Compound 1 contains 30.4 \text{ g} of nitrogen, which means (30.4 \div 14.01) = 2.17 \text{ moles} of nitrogen, and 69.6 \text{ g} of oxygen, which means (69.6 \div 16.00) = 4.35 \text{ moles} of oxygen.

Compound 2 contains 19.8 \text{ g} of nitrogen, which means (19.8 \div 14.01) = 1.41 \text{ moles} of nitrogen, and 45.2 \text{ g} of oxygen, which means (45.2 \div 16.00) = 2.83 \text{ moles} of oxygen.

Step 3. Find the Largest Common Divisor.

Next, we have to use these numbers to calculate an empirical formula. You’ve never seen a formula that’s N_{2.17}O_{4.35}, right? The coefficients of the elements are always integers. So, we take the numbers we’ve been given and try to find the largest common divisor, or in other words, we’re trying to make them the smallest integers possible while still keeping the ratio between them the same.

The quickest way to do this is to take the smaller of the numbers and divide all the numbers with it. In the case of Compound 1, we have 2.17 \text{ mol } N and 4.35 \text{ mol } O.

4.35 \div 2.17 = 2.005

This number is very close to 2, so we can assume that the ratio of nitrogen to oxygen in the sample was 1{:}2. This makes the empirical formula NO_2.

Doing the same with Compound 2, we have 1.41 \text{ mol } N and 2.83 \text{ mol }O.

2.83 \div 1.41 = 2.007

Again, this is very close to 2. So we’ve discovered that these two compounds are, in fact, the same!

Or are they? To know for sure, we’d have to figure out their molecular formulas.

More Complex Formulas

NO_2 is a pretty simple formula. However, you may be thrown a far more complicated one. For example, let’s say you find out that your sample has 3.39 \text{ mol } C, 5.65 \text{ mol } H, and 2.26 \text{ mol } N (this compound does not exist, by the way).

We start by dividing the two larger numbers by the smallest, 2.26. We get 1.50 \text{ mol }C, 2.50 \text{ mol } H, and 1.00 \text{ mol } N (by definition). Now, all of the coefficients must be integers, so what should we do? We multiply them all by whatever integer factor is necessary.

1.50 needs to be multiplied by 2 to make 3, so we multiply all the numbers by 2 to get 3.00 \text{ mol } C, 5.00 \text{ mol } H, and 2.00 \text{ mol } N.

So, we have calculated an empirical formula of  C_3H_5N_2.

More Examples

The atomic weight of element X is 100 \text{ g/mol}. 50.0 \text{ g } of X combines with 32.0 \text{ g} of oxygen. Calculate an empirical formula for the resulting compound.

As you can see, in this problem, we don’t need to figure out the masses of the two elements in the compound, since they’re listed in the problem. So, we can skip directly to Step 2.

We have 50\text{ g} of X, and X has a molar mass of 100 \text{ g/mol}. So we have (50 \div 100) = 0.5 \text{ mol of } X.

We have 32.0 \text{ g} of oxygen, and oxygen has a molar mass of 16 \text{ g/mol}. So we have (32.0 \div 16) = 2 \text{ mol of } O.

Now, we need both of these numbers to be integers. What should we do? It turns out that the method we already outlined in Step 3 still works here. We just divide both numbers by the smaller one (0.5). So we then have 1 \text{ mol} of X and 4 \text{ mol} of oxygen, yielding an empirical formula of XO_4.

Let’s do just one more example. This one is a complicated example, which is more like the problems you’ll encounter on the free-response section.

A 1.34 \text{ g} sample of a compound containing carbon and hydrogen was burned in an environment with excess oxygen, forming 6.78 \text{ g } CO_2 and 4.16 \text{ g } H_2O (water). Calculate an empirical formula of the compound.

Steps 1 and 2 are quite hard in this case. The best thing to do is to first find the number of moles of each of the two compounds (CO_2 and H_2O).

The molar mass of CO_2 is (12+16 \times 2) = 44 \text{ g/mol}.

So the reaction produced (6.78 \div 44) = 0.154 \text{ moles of } CO_2.

The molar mass of H_2O is (1 \times 2+16) = 18 \text{ g/mol}

So the reaction produced (4.16 \div 18) = 0.231 \text{ moles of } H_2O

Now, we need to calculate the moles of the elements involved (carbon and hydrogen). Oxygen is not important, because the original compound contained no oxygen.

The reaction produced 0.154 \text{ moles} of CO_2, which corresponds to 0.154 \text{ moles} of carbon.

The reaction produced 0.231 \text{ moles} of H_2O, which corresponds to 0.462 \text{ moles} of hydrogen because there are two hydrogens for every molecule of H_2O.

Now we proceed to Step 3. We divide both numbers by the smaller one, giving us 1 \text{ mole} of carbon and 3\text{ moles} of hydrogen. So, we calculate an empirical formula of CH_3!

That’s all you need to know about how to calculate an empirical formula for the AP® Chemistry exam. Good luck and happy studying!

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