Editing Empirical formula (section)

==Calculation example==
A chemical analysis of a sample of [[methyl acetate]] provides the following elemental data: 48.64% [[carbon]] (C), 8.16% [[hydrogen]] (H), and 43.20% [[oxygen]] (O). For the purposes of determining empirical formulas, it's assumed that we have 100 grams of the compound. If this is the case, the percentages will be equal to the mass of each element in grams.

:Step 1: Change each [[percentage]] to an expression of the mass of each element in grams. That is, 48.64% C becomes 48.64 g C, 8.16% H becomes 8.16 g H, and 43.20% O becomes 43.20 g O.

:Step 2: Convert the amount of each element in grams to its amount in moles
:<math>\left(\frac{48.64 \mbox{ g C}}{1}\right)\left(\frac{1 \mbox{ mol }}{12.01 \mbox{ g C}}\right) = 4.049\ \text{mol}</math>
:<math>\left(\frac{8.16 \mbox{ g H}}{1}\right)\left(\frac{1 \mbox{ mol }}{1.007 \mbox{ g H}}\right) = 8.095\ \text{mol}</math>
:<math>\left(\frac{43.20 \mbox{ g O}}{1}\right)\left(\frac{1 \mbox{ mol }}{16.00 \mbox{ g O}}\right) = 2.7\ \text{mol}</math>

:Step 3: Divide each of the resulting values by the smallest of these values (2.7)
:<math>\frac{4.049 \mbox{ mol }}{2.7 \mbox{ mol }} = 1.5</math>
:<math>\frac{8.095 \mbox{ mol }}{2.7 \mbox{ mol }} = 3</math>
:<math>\frac{2.7 \mbox{ mol }}{2.7 \mbox{ mol }} = 1</math>

:Step 4: If necessary, multiply these numbers by integers in order to get whole numbers; if an operation is done to one of the numbers, it must be done to all of them.
:<math>1.5 \times 2 = 3</math>
:<math>3 \times 2 = 6</math>
:<math>1 \times 2 = 2</math>

Thus, the empirical formula of methyl acetate is {{Format molecular formula|C3H6O2}}. This formula also happens to be methyl acetate's molecular formula.