Question

Series 100 contains carbon, hydrogen and oxygen, and has a molecular weight of 136 g/mol. In the combustion of a 5.000 g sample, 12.943 g of Co, and 2.655 g H2O were produced.

Answer 1

1)

let in compound number of moles of C, H and O be x, y and z respectively

Number of moles of CO2 = mass of CO2 / molar mass CO2

= 12.943/44

= 0.2942

Number of moles of H2O = mass of H2O / molar mass H2O

= 2.655/18

= 0.1475

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.2942

so, x = 0.2942

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.1475 = 0.295

Molar mass of O = 16 g/mol

mass O = total mass - mass of C and H

= 5.0 - 0.2942*12 - 0.295*1

= 1.175

number of mol of O = mass of O / molar mass of O

= 1.175/16.0

= 7.344*10^-2

so, z = 7.344*10^-2

Divide by smallest to get simplest whole number ratio:

C: 0.2942/7.344*10^-2 = 4

H: 0.295/7.344*10^-2 = 4

O: 7.344*10^-2/7.344*10^-2 = 1

So empirical formula is:C4H4O

Molar mass of C4H4O,

MM = 4*MM(C) + 4*MM(H) + 1*MM(O)

= 4*12.01 + 4*1.008 + 1*16.0

= 68.072 g/mol

Now we have:

Molar mass = 136.0 g/mol

Empirical formula mass = 68.072 g/mol

Multiplying factor = molar mass / empirical formula mass

= 136.0/68.072

= 2

So molecular formula is:C8H8O2

Answer:

empirical formula is:C4H4O

molecular formula is:C8H8O2