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)* ((64.3percent))/((100percent)) = 64.3g`

Mass of H = 7.2 g

`(100g)* ((7.2percent))/((100percent)) = 7.2g`

Mass of O = 28.5 g

`(100g)* ((28.5percent))/((100percent)) = 28.5g`

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

`Moles = (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 = (64.3g)/(12.0(g)/(mol))`

Moles of C = 5.36 mol

`Moles of H = (7.2g)/(1.0(g)/(mol))`

Moles of H = 7.2 mol

`Moles of O = (28.5g)/(16.0(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 = (5.36mol)/(1.78mol) = 3.0`

`Mole of H = (7.2mol)/(1.78mol) = 4.0`

`Mole of O = (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`