Introduction

Now that we understand how to convert between grams and moles, we can apply this to real scenarios. Using the Law of Definite Proportions, we can analyze samples of substances to figure out the identity of the substance, along with learning how to calculate the Empirical Formula.

Law of Definite Proportions

The Law of Definite Proportions states that for any pure sample of a substance, the mass ratio of elements will always be the same. For example, in any sample of pure water, $89%$ of the mass will be oxygen and $11%$ will be hydrogen. This is because we can calculate the percent mass of an element in a compound by dividing the molar mass of the element in the compound by the molar mass of the whole compound. Look at the hydrogen in a water molecule. There are 2 hydrogen atoms in a water molecule, each with a molar mass of about 1 gram per mole. Each water molecule has a molar mass of about 18 grams per mole. Therefore we can do $\frac{2}{18}$ to get $11%$ mass percent hydrogen.

Empirical Formula

The empirical formula of a substance is the simplest ratio of elements in a compound. The molecule glucose has a molecular formula of $C X_{6} H X_{12} O X_{2}$ , but an empirical formula of $CH X_{2} O$ . This is an important distinction to make, as sometimes compounds have the same empirical formula but different molecular ones. Ethene is a compound with a molecular formula of $C X_{2} H X_{4}$ , and an empirical formula of $CH X_{2}$ . Butene is a different compound with a molecular formula of $C X_{4} H X_{8}$ , but also has an empirical formula of $CH X_{2}$ . This means that the mass percentages of carbon and hydrogen are the same in pure samples of ethene and butene, and with just mass percentages we cannot tell them apart.

Calculations using the Empirical Formula

You will often be asked to identify the empirical formula of a compound after being given the mass percentages of each of the elements in a sample. Let’s say you have a compound containing $42.5%$ chlorine and $57.5%$ oxygen and you want to find its empirical formula. The empirical formula is based on mole ratios of elements, so we need to go from the mass percentages to moles. To convert to moles, we need to have a set mass for each element. To do this we can imagine a sample with those mass percentages, which is usually 100 grams. This makes the conversion easy, as we can keep the numbers and change the units from percent mass to grams. Now we just need to convert $42.5 g Cl$ and $57.5 g O$ to moles of both. We divide each of the numbers by the molar mass of the element ( $35.45 gmo l^{- 1}$ for chlorine, $57.5 gmo l^{- 1}$ for oxygen) to get the amount of moles of each. This gives us $1.2 mol Cl$ and $3.6 mol O$ . We can divide by the lowest value ( $1.2$ ) to get whole numbers, and our final empirical formula, $ClO X_{3}$ .

Going back to our example of two compounds with the same empirical formula but different molecular formulas, you may also be asked to find the molecular formula when you have the empirical formula and the molar mass of the compound. If we know the empirical formula is $CH X_{2}$ and the molar mass is $56.1 gmo l^{- 1}$ , what is the molecular formula?

Finding the Molecular Formula with Molar Mass

The first step to solve this is to find the molar mass of just the atoms from the empirical formula, which in this case would be $14.03 gmo l^{- 1}$ . We can then divide the molar mass we were given by the molar mass per $CH X_{2}$ in our compound. This gives us a result of 4, which means there is 4 times our empirical formula in our compound, meaning our compound is $C X_{4} H X_{8}$ .

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