Question

Determine the empirical formulas for compounds with the following percent compositions.

1. 35.97% carbon and 64.03% sulfur

2. 37.48% carbon, 12.58% hydrogen, and 49.93% oxygen

- A compound of carbon and hydrogen contains 85.63% C and has a molar mass of 84.16 g/mol. What is its molecular formula?

Answer 1

1) per 100g we have C mass= 35.97 g , S mass = 64.03g

C moles = mass / atomic mass of C = ( 35.97g) / ( 12g/mol) = 3

S moles = ( 64.03g / 32g/mol) = 2

hence emperical formula is C3S2

2) C moles = ( 37.48g) / ( 12g/mol) = 3.12 , H moles = ( 12.58g/1g/mol) = 12.58 mol

O moles = ( 49.93g /16g/mol) = 3.12

now dividing all by smallest number 3.12 we get C/O = 1 , H/O = 4 , O/O = 1

hence emperical formula is CH4O

3) C mass per 1mol compound = ( 85.63/100) x ( 84.16g) = 72g

H mass = 84.16 - 72g = 12 g

C moles = ( 72g /12g/mol) = 6 , H moles = ( 12g/1g/mol) = 12

Molecular formula is C6H12