Question

Combustion analysis of an unknown compound containing only carbon and hydrogen produced 2.845 g of CO₂ and 1.744 g of H₂O. What is the empirical formula of the compound?

		C4H10
		CH3
		C5H2
		CH2

Answer 1

Option CH3 is correct.

Solution-

The number of moles of CO₂ = 2.845 / 44 = 0.06465 moles

The number of moles of H₂O = 1.744 / 18 = 0.09689 moles

i) Since 2.0 moles “H” is present in 1.0 mole of H₂O

Therefore, 0.09689 moles of H₂O= 2 x 0.09689 (=0.1938 moles) moles of “H” in the compound

ii) Since 1.0 mole of CO₂ = 1.0 mole of “C”

Since 0.06465 mole of CO₂ = 0.06465 moles of “C” in the compound

Thus the ratio of C:H in compound

= 0.06465 : 0.1938

= 1: (0.1938/0.06465)

= 1:3

So the C:H ratio is 1:3

Therefore the empirical formula for compound = C₁H₃ or CH₃ or C₂H₆