Question

0.14 g of an unknown organic compound was burned in the excess of oxygen and produced 0.44 g of CO₂ and 0.18 g of H₂O. Determine the empirical formula.

a)     CH            

b)    CHO         

c)     CH₂                            

d)    CH₂O       

e) C₂HO₂

Answer 1

Let; We know; The reaction as;

Unknown Organic compound + O₂ CO₂ + H₂O

We have given;

Mass of CO₂ =0.44g

Calculating the moles of CO_2;

0.44g CO₂ x( 1 mole of CO₂/44 g)= 0.01 moles of CO₂

1 mole of CO₂ = 1 mole of C

0.01 mole of CO₂ =0.01 moles of C

Mass of Carbon(C) = 0.01 moles of C x( 12g/1mol)=0.12g of C

Also;

Mass of H₂O = 0.18g

Calculating the moles of H₂O _;

0.18g H₂O x( 1 mole of H₂O /18g)= 0.01 moles of H₂O

1 mole of H₂O = 2 moles of H

0.01 mole of H₂O =0.02 moles of H

Mass of Carbon(C) = 0.02 moles of H x( 1g/1mol)=0.02g of H

Now; Calculating the mass of Oxygen;

Mass of oxygen = Mass of the compound - ( mass of C+ Mass of O)

= 0.14 -(0.12+0.2)

Mass of oxygen = 0.14-0.14

Mass of oxygen = 0

So, The compound contains only Carbon(C) and Hydrogen(H)

Taking mole ratio for C and H;

0.02/0.01 = 2 moles of H

0.01/0.01 = 1 mole of C

So, The empirical formula will be Option(C) CH₂