Question

an unknown compound contains only c h and o. Combustion of 5.60g of this compound is produced, 13.7 of CO2 and 5.60g of H2O. What is the empirical formula of the unknown compound ?

Answer 1

All the C of compound will form only CO2 and H form only H2O. So first calculate the grams of C and H in the compound as follows:

13.7 g CO2 * 12.01 g/ mol C/ 44.01g/ mol CO2= 3.74 g C

5.60 g H2O *2 *2.016 g/ mol / 18.02 g/ mol H2O= 0.62 g H

Now amount of O = total amount – (amount C+ amount H)

= 5.60 g –(3.74 g+0.62 g)

= 1.24 g O

Now calculate the number of moles of C, H and O:

C: 3.74 g C/12.01 g/ mol=0.311 molC

H: 0.62 g H/1.008 g/mol = 0.62 mol H

O: 1.24 g O/16.00 g/mol= 0.08 mol O

Now determine the ratio of C, H and O in the compound:

C: 0.311 mol / 0.08 mol = 4.0

H: 0.62mol / 0.08 mol = 16

O: 0.08 mol / 0.08 mol = 1

The empirical formula of the unknown compound is C4H16O