Question

An unknown compound contains only C, H, and O. Combustion of 2.90 g of this compound produced 5.79 g of CO2 and 2.37 g of H2O.What is the empirical formula of the unknown compound?

Answer 1

Atomic mass of H=1 g/mole
Atomic mass of C=12 g/mole
Atomic mass of O=16 g/mole
Molar mass of CO2=44 g/mole
Molar mass of H2O=18 g/mole
Mass of C in 5.79g of CO2 = 5.79/44 g/mole*12 g/mole =1,5792 g
Mass of H in 2.37g of H2O=2.37/18 g/mole*2 g/mole =0,2634 g
These two amounts of C and H have been obtained from 2.90 g compound containing C ,H and O.
So the remaining (2.90-1,5792-0,2634)g=1,0574g is the mass of Oxygen present in 2.9 g compound.
Then we must dividing these masses of elements with respective atomic masses of elements and taking their ratio. Then we get the ratio of number of atoms of C, H and O in the molecule of the compound as below
C:H:O=(1,5792/12): (0,2634/1): (1,0574/16)= 0,1316:0,2634:0,0660875=(0,1316/0,0660875): (0,2634/0,0660875): (0,0660875/0,0660875)=2:4:1