**Percent Composition**

The percent composition of an element in a compound is calculated using the formula.

**Example 1**

What is the percent composition of FeSO_{4}?

The total percent composition adds up to 100% (36.7% + 21.1 % + 42.2% = 100%)

**Determination of Empirical Formulas**

The empirical formula is used to show the relative ratios of various atoms of a compound.

The masses of each element in a compound can be measured using experiments with that the empirical formula is derived as shown in the example below.

Find the empirical formula of a compound that contains 38.7% C, 16.2% H, and 45.1% N.

**Step 1:**

Assume 100 g of the compound then the number of moles of each element can be calculated as follows.

**Step 2: **Determining mole ratio

**C**_{3.22}H_{16.2}N_{3.22}

**Step 3 : **Divide all the moles by the smallest one found

**Step 4: **The empirical formula now can be written as follows

C_{1}H_{5}N_{1} = CH_{5}N

**Derivation of Molecular Formulas**

Ethylene glycol has 38.7% C, 9.75% H, and Oxygen. Its molecular weight is 62.07g. Find the molecular formula of ethylene glycol.

**Step 1**

The amount of oxygen in the compound can be found using the following method since the total percent composition is equal to 100%

O = 100% - (38.7% + 9.75%) = 51.55%

**Step 2**

Assume 100 g of the compound then the number of moles of each element can be calculated as follows.

**Step 3**

Divide all the moles by the smallest one found

**Then the empirical formula becomes CH**_{3}O

**Step 4**

Calculating the ratio of molecular weight to empirical forumal weight

Molecular weight = 62.07g

Empirical formula weight = (1*12.0g)+(3*1.0g)+(1*16.0g) = 31.0g

**Step 5**

Multiply the mols of empirical formula by the ratio calculated in **step 4**

Then the molecular formula becomes C_{2}H_{6}O_{2}