Question

A 150.0 gram sample of a white powder contains 61.05 g of carbon, 7.65 g of hydrogen, and 81.30 g of oxygen. What is the empirical formula of this compound?

Answer 1

C2H3O2 (acetate)

Step-by-step explanation:

61.05 g C x 1 mol C / 12.01 g C = 5.083 mol C

7.65 g H x 1 mol H / 1.01 g H = 7.57 mol H

81.30 g O x 1 mol O / 16.00 g O = 5.081 mol O

Divide by lowest number

5.083 / 5.081 = approx. 1 mol C

7.57 / 5.081 = approx. 1.5 mol H

5.081 / 5.081 = 1 mol O

so ratio of moles in this compound are 1 mol C : 1.5 mol H : 1 mol O

Multiply by 2 to get whole number mol ratios

--> empirical formula is C2H3O2 (acetate)