Question

Combustion analysis of 0.6943 g of an unknown compound containing carbon, hydrogen, and oxygen, produced 1.471 g CO2 and 0.226 g H2O. Determine the molecular formula, given the molar mass is 166 g/ mol.

Answer 1

Answer – We are given, mass of compound = 0.6943 g

Mass of CO₂ = 1.471 g , mas of H₂O = 0.226 g , molecular mass = 166 g/mol

First we need to calculate moles of CO₂ and H₂O from the given mass

Moles of CO₂ = 1.471 g / 44.0 g.mol^-1 = 0.0334 mole

Moles of H₂O = 0.226 g / 18.015 g.mol^-1 = 0.0125 moles

Moles of C from the moles of CO₂

1 moles of CO₂ = 1 moles of C

So, 0.0344 moles of CO₂ = ?

= 0.0334 moles of C

Moles of H from moles of H₂O

1 moles of H₂O = 2 moles of H

So, 0.0125 moles of H₂O = ?

= 0.02521 moles of H

Mass of C = 0.0334 moles * 12.011 g/mol

                  =0.402 g of C

Mass of H = 0.0251 moles * 1.0079 g/mol

                  = 0.0253 g of H

Total mass of compound = mass of C + mass of H + mass of O

0.6943 g = 0.402 g + 0.0253 g + mass of O

Mass of O = 0.6943 g – 0.402 g – 0.0253 g

                  = 0.267 g of O

Moles of O = 0.267 g / 15.998 g.mol^-1

                   = 0.0167 moles of O

So moles of O both are smallest, so we need to divide each mole by this mole

So , C = 0.0334 /0.0167 = 2

      H = 0.0251 / 0.0167 = 1.5

      O = 0.0167 /0.0167 = 1

Now we need to multiply each by 2 for whole number of H

C = 2*2 = 4

H = 1.5 *2 = 3

O = 1*2 = 2

Empirical formula is C₄H₃O₂

Molecular formula = n * empirical formula

n = molecular formula mass / empirical formula mass

= 166 / 83

= 2

So molecular formula = 2 * C₄H₃O₂

= C₈H₆O₄