Question

1. a) Combustion analysis of an unknown organic compound showed the presence of carbon, hydrogen and bromine. When a 0.0392 g sample of the compound is combusted, 0.06593 g of carbon dioxide and 0.01124 g of water are formed. Determine the empirical formula of compound.

b) Determine the simplest formula for tin chloride if a 0.398 g sample of tin chloride reacts with AgNO3 to produce tin nitrate and 0.602 g of AgCl.

Answer 1

Q1.

m = 0.0392 g sample

m of CO2 = 0.06593 g of CO2

mol of CO2 = mass/MW = 0.06593 /44 = 0.0014984 mol of CO2

1 mol of CO2 = 1 mol of C

0.0014984 mol of CO2 = 0.0014984 mol of C

mass of C = 0.0014984 *12 = 0.0179808 g

for H2O:

m of H2O = 0.01124 g of H2O

mol of H2O = mass/MW = 0.01124 /18 = 0.000624 mol of H2O

mol of H = 2*0.000624 = 0.001248 mol of H; mass of H = 0.001248 g

Now...

mass of Br =0.0392 - ( 0.0179808 +0.001248 ) = 0.0199712 g of BR

mol of Br = mass/NW = 0.0199712/79.904 = 0.00024993 mol

Ratios:

C:Br = 0.0014984 /0.00024993 = 6

H:Br = 0.001248 /0.00024993 = 5

Br:br = 1

CHBr --> C6H5Br

b)

for

m = 0.398 g of SnClx

mol of Agcl = mass/MW = 0.602/143.32 = 0.00420 mol of AgCl

mol of Cl- = 0.00420

mass of Cl presnet = mol*MW = 0.00420*35.5 = 0.1491 g of Cl

mass of Sn = 0.398-0.1491 = 0.2489 g of Sn

mol of Sn = mass/MW = 0.2489/118.71 = 0.0020

ratio = 0.00420/0.0020 = 2 mol of Cl per 1 mol of Sn

SnCl2