Question

A compound has a mass percentage of 53.46% C, 6.98% H, and 39.56% O. What is the empirical formula for this compound

Answer 1

Answer:

The empirical formula is: C₂H₃O

Explanation:

The empirical formula, also known as the “minimum formula”, is the simplest expression to represent a chemical compound and indicates the elements that are present and the minimum integer ratio between its atoms.

The percentage composition is the percentage by mass of each of the elements present in a compound.

Having 100 g of the compound as a base, it is possible to express the percentages in grams. That is, assuming you have 100 g of the compound, you have 53.46 g of  C , 6.98 g of  H , and 39.56 g of  O.

Taking into account the molecular mass of each substance, the number of relative atoms of each chemical element is calculated:

C: $53.46 g *\frac{1 mol}{12.01 g } = 4.45 moles$

H:$6.98 g *\frac{1 mol}{1.01 g } = 6.91 moles$

O:$39.56 g *\frac{1 mol}{16g } = 2.47 moles$

Now you divide each value obtained by the least of them:

C: $\frac{4.45 moles}{2.47 moles}= 1.80$

H:$\frac{6.91 moles}{2.47 moles}= 2.8$

O:$\frac{2.47 moles}{2.47 moles}=1$

Decimals approach the nearest integer, then:

C: 2

H: 3

O: 1

Therefore the empirical formula is: C₂H₃O