Question

Determine the empirical formula for the following compound % compositions: 1.) 52.56% Fe, 2.83%H, 44.91% O 2.) 30.60% C, 3.85%H, 45.16% CI, 20.38%O 3.) 54.53%C, 9.15%H, 36.32% 4.) 71.98%C, 6.71%H21.31%O

Answer 1

Solution:

Let total mass of compound is 100 g.

Then,

Fe = 52.56 % = 52.56

H =2.83 % = 2.83

O = 44.91 % = 44.91

Number of moles are calculated as,

Number of mol of Fe = mass/molar mass

= 52.56 g / 56 g mol-1 = 0.94 mol

Number of moles of H = mass/atomic mass

= 2.83 g / 1.00 g mol-1 = 2.83 mol

Number of moles of O = mass /atomic mass

= 44.91 g / 16 g mol-1 = 2.81 mol

Dividing number of moles by lowest number of moles to obtained ratio.

Thus,

Fe = 0.94 / 0.94 = 1

H = 2.83 / 0.94 = 3

O = 2.81 / 0.94 = 3

Hence, empirical formula = FeH3O3

Note : Calculate othes with same procedure.