# A compound produced as a by-product in an industrial synthesis of polymers was found to contain C, H, and I. A combustion analysis of 1.70 g of the compound produced 1.32 g of CO2 and 0.631 g of...

A compound produced as a by-product in an industrial synthesis of polymers was found to contain C, H, and I. A combustion analysis of 1.70 g of the compound produced 1.32 g of CO2 and 0.631 g of H20. The mass percentage of I in the compound was determined by converting the I in a 0.850 g sample of the compound into 2.31 g of PbI2. What is the empirical formula of the compound?

jeew-m | Certified Educator

Molar mass in `g/(mol)`

`CO_2 = 44`

`H_2O = 18`

`PbI_2 = 461`

Amount of produced `CO_2 = 1.32/44 = 0.03mol`

Amount of C moles in 1.7g of compound `= 0.03mol`

`2H+1/2O_2 rarr H_2O`

Amount of produced `H_2O = 0.631/18 = 0.035mol`

Amount of H moles in 1.7g of compound `= 0.035x2 = 0.07mol`

Amount of produced `PbI_2` = 2.31/461 = 0.005mol

`2I^(-)+Pb^(2+) rarr PbI_2`

Amount of I in 0.85g of compound `= 0.005xx2 = 0.01mol`

Amount of I in 1.7g of the compound `= (0.01/0.85)xx1.7 = 0.02mol`

Molar ratio

`C:H:I = 0.03:0.07:0.02 = 3:7:2`

So the empirical formulae is `C_3H_7I_2` .