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

m = 0.0392 g of sample

m = 0.06593 g CO2; mol = mass/MW = 0.06593 /44 = 0.001498

m = 0.01124 g of H2O ; mol = mass/MW = 0.01124 /18 = 0.0006244

mol of C (1 mol of C per mol of CO2) = 0.001498 ;

mol of H (1/2 mol of H per mol of H2O)= 0.0006244/2 =0.0003122

mass of Br --> mass of sample - mass of C mass of H

mass of Br --> 0.0392 - (0.001498*12 + 0.0006244*1 ) = 0.020599 g

mol of Br = mass/MW = (0.020599)/(79.9040 ) = 0.000257

Ratios:

C:Br = 0.001498 /0.000257 = 5.8

H:Br = 0.0003122 / 0.000257 = 1.21

C:H = 0.001498 /0.0003122 = 4.8

ratio of empirical formula:

C5.8 H1.21 Br 1

C6HBr

Question 2.

SnClx + AgNO3 --> Sn(NO3)x + AgCl

mol of SnClx = mass/MW = 0.398 / MW

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

mol of Ag+ = 0.00420; mol of Cl- = 0.00420

mol of nitrate:

NO3- = Ag+ = 0.00420

mass of Cl = mol*Mw = 0.00420*35.5 = 0.1491 g of Cl

mass of sample = 0.398 g

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

mol of Sn = mass/MW = 0.2489/118.71 = 0.002096 mol of Sn

ratio Sn:Cl

0.00420/0.002096

2:1

then, x = 2

SnCl2