Question

What is the empirical formula of a compound that is 64.3 % c, 7.2 % h, and 28.5 % o by mass?

Answer 1

Asnwer : Empirical formula of a compound is : $C_{3}H_{4}O$

Given information : C = 64.3 % , H = 7.2 % , O = 28.5 %

Step 1 : Convert the given percentage (%) to grams.

Explanation : Let the total mass of the compound be 100 grams.

Mass of C = 64.3 g

$(100g)\times \frac{(64.3percent)}{(100percent)} = 64.3g$

Mass of H = 7.2 g

$(100g)\times \frac{(7.2percent)}{(100percent)} = 7.2g$

Mass of O = 28.5 g

$(100g)\times \frac{(28.5percent)}{(100percent)} = 28.5g$

Step 2 : Convert the grams of each compound to moles.

$Moles = \frac{Grams}{Molar mass}$

Molar mass of C = 12.0g/mol  

Molar mass of H = 1.0 g/mol

Molar mass of O = 16.0g/mol

$Moles of C = \frac{64.3g}{12.0\frac{g}{mol}}$

Moles of C = 5.36 mol

$Moles of H = \frac{7.2g}{1.0\frac{g}{mol}}$

Moles of H = 7.2 mol

$Moles of O = \frac{28.5g}{16.0\frac{g}{mol}}$

Moles of O = 1.78 mol

Step 3 : Find the mole ratio of C , H and O

Mole ratio is calculated by dividing the mole values by the smallest value.

Mole of C = 5.36 mol , Mole of H = 7.2 mol , Mol of O = 1.78 mol

Out of the three mole values , mole value of O that is 1.78 mol is less , so we divide all the mole values by 1.78 mol.

$Mole of C = \frac{5.36mol}{1.78mol} = 3.0$

$Mole of H = \frac{7.2mol}{1.78mol} = 4.0$

$Mole of O = \frac{1.78mol}{1.78mol} = 1.0$

C : H : O = 3 : 4 : 1

So empirical formula of the compound is $C_{3}H_{4}O_{1}$ or $C_{3}H_{4}O$