Question

62. An unknown sample with a total mass of 440.85 g contains carbon, hydrogen, and oxygen. Show all work

a. If the sample contains 300.25 g of carbon and 60.6 g of hydrogen, how much oxygen (in g) does the unknown sample contain?

b. What is the empirical formula for the unknown compound?

c. If the formula mass for the unknown is 88.17 g/mol, what is the molecular formula for the unknown compound?

Answer 1

Total mass of sample = 440.85 g

C = 300.25 g

H = 60.6 g

O = ?

a) O = total mass - (mass of C + H) = 440.85 - (300.25 + 60.6) = 80 g

b)

Elements	Mass	Atomic mass	moles	Simplest ratio
C	300.25 g	12.01	300.25/12.01 = 25	25/5 = 5
H	60.6 g	1.00784	60.6/1.00784 = 60.13	60/5 = 12
O	80 g	16	80/16 = 5	5/5 = 1

Empirical formula = C₅H₁₂O

c) Molecular formula = n x Empirical formula

Molecular mass = n x Empirical mass

88.17 (Formula mass given ) = n x (12.01 x 5 + 12 x 1.00784 + 16)

88.17 = n x 88.14

n = 1

Therefore, Molecular formula = C₅H₁₂O