Question

An unknown compound contains only C, H, and O. Combustion of 3.90 g of this compound produced 9.18 g of CO2 and 2.51 g of H2O.

What is the empirical formula of the unknown compound?

Answer 1

An unknown compound contains only C, H, and O. Combustion of 3.90 g of this compound produced 9.18 g of CO2 and 2.51 g of H2O.

What is the empirical formula of the unknown compound?

solution :-

We are given with the masses of the CO2 and H2O

so lets first calculate the masses of the C and H using the mass of the CO2 and H2O

mass of C = 9.18 g CO2 * 27.29 % / 100 % = 2.505 g C

mass of H = 2.51 g H2O * 11.2% H / 100 % = 0.2811 g H

now lets find the mass of oxygen in the sample

mass of O = mass of sample - (mass of C + mass of H)

               = 3.90 g - ( 2.505 g + 0.2811 g)

               = 1.114 g O

now lets calculate the moles of the each element

moles = mass / molar mass

moles of C = 2.505 g / 12.01 g per mol = 0.209 mol C

moles of H = 0.2811 g / 1.0079 g per mol = 0.279 mol H

moles of O = 1.114 g / 16 g per mol = 0.0696 mol O

moles of the oxygen are lowest therefore kets find the ratio of the moles by dividing each mole value by the smallest mole number

C= 0.209 /0.0696 = 3

H = 0.279 / 0.0696 = 4

O= 0.0696 / 0.0696 = 1

Therefore the empirical formula = C3H4O