Question

a 7.284 gram sample of an organic compound containing Containing C,H,O is analyzed by combustion analysis and 9.242 grams CO2 and 2.523 grams H2O are produced.

I need the molecular and empirical formula

Answer 1

% of carbon in sample = mass of CO2 x12 x100 /[ mass of sample x 44}

= 9.242 x12x100/(7.284 x44)

= 34.60%

% of H in sample = mass of H2O x2x100 /[ mass of sample x 18]

= 2.523 x100x2 /(7.284x18)

= 3.848

% of oxygen = 100-(%C +% H)

= 100-(34.6 + 3.848)

= 61.548

Element carbon hydrogen oxygen

% mass 34.6 3.848 61.548

atomic mass 12 1 16

moles of atoms 2.88 3.848 3.84

Ratio 1 1.33 1.33

or 3 4 4 by mucltiplying with 3 to make whole number ratio

Thus the empirical formula is C3H4O4.