Question

25. When 3.62 g of a compound containing carbon, hydrogen, and oxygen were burned completely in air, 5.19 g of Co, and 2.83 g of H2O were produced. What is the empirical formula of the compound?

Answer 1

given

3.62 g of compound

5.19 g CO2

2.83 g H2O

Now

Moles = mass/molar mass

Moles CO2=5.19 g/44.01g/mol

= 0.118 moles

Moles of C = 0.118 moles*1

= 0.118 moles

mass of C = Moles of C* molar mass of C

= 0.118 moles*12.01g/mol

= 1.42 g

Moles H2O =2.83 g /18.02g/mol

= 0.157 moles

Moles of H = 0.157 moles*2

= 0.314 moles

mass of H = Moles of H* molar mass of H

= 0.314 moles*1.01g/mol

= 0.32 g

So

Mass of O = Mass of compound - Mass of H - Mass of C

= 3.62 g- 0.32 g-1.42 g

= 1.88 g

Now

Moles of C = 1.42 g/12.01g/mol

= 0.118 moles

Moles of H = 0.32 g/1.01g/mol

= 0.32 moles

Moles of O= 1.88 g/16 g/mol

= 0.118 moles

So

Divide moles by lowest number(i.e. 0.118)

C ==>0.118/0.118 =1

H ==>0.32/0.118 =2.7

O ==>0.118/0.118 =1

So

C:H:O = 1:2.7:1

Multiply by 3

C:H:O = 3:8:3

Empirical formula is C₃H₈O₃