Question

A compound containing only C, H and O was subjected to combustion analysis. A sample of 9.550×10^-2 g produced 1.825×10^-1 g of CO₂ and 1.120×10^-1 g of H₂O. Determine the empirical formula of the compound and enter the appropriate subscript after each element.

If the molar mass of the compound is 184.276 g/mol, determine the molecular formula of the compound and enter the appropriate subscript after each element.

Answer 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

= 0.1825/44

= 0.004148

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

= 0.112/18

= 0.006222

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.004148

so, x = 0.004148

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.006222 = 0.012444

mass O = total mass - mass of C and H

= 0.0955 - 0.004148*12 - 0.012444*1

= 0.033283

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

= 0.033283/16

= 0.00208

so, z = 0.00208

Divide by smallest to get simplest whole number ratio:

C: 0.004148/0.00208 = 2

H: 0.012444/0.00208 = 6

O: 0.00208/0.00208 = 1

So empirical formula is:C2H6O

Molar mass of C2H6O,

MM = 2*MM(C) + 6*MM(H) + 1*MM(O)

= 2*12.01 + 6*1.008 + 1*16.0

= 46.068 g/mol

Now we have:

Molar mass = 184.276 g/mol

Empirical formula mass = 46.068 g/mol

Multiplying factor = molar mass / empirical formula mass

= 184.276/46.068

= 4

Hence the molecular formula is : C8H24O4

empirical formula:C₂H₆O

molecular formula : C₈H₂₄O₄