Question

In: Chemistry

1. a) Combustion analysis of an unknown organic compound showed the presence of carbon, hydrogen and...

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.

Solutions

Expert Solution

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


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