Question

18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment company analysed a cleaning solvent, J, and found it to contain the elements carbon, hydrogen, and chlorine only. The chemical composition of J was determined using different analytical chemistry techniques. Combustion reaction: Combustion of 1.30 g of J gave 0.872 g CO2 and 0.089 g H2O. Precipitation reaction with AgNO3(aq): 0.535 g of J gave 1.75 g AgCl precipitate. (a) Determine the percentage by mass of carbon and hydrogen in J, using the combustion data. (3) (b) Determine the percentage by mass of chlorine in J, using the precipitation data. (1) (c) The molar mass was determined to be 131.38 g mol–1. Deduce the molecular formula of J.

Answer 1

Answer:

a) Percentage by mass of carbon: 18.3%

   Percentage by mass of hydrogen: 0.77%

b)  Percentage by mass of chlorine: 80.37%

c) Molecular formula: $C_{2} H Cl_{3}$

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

$0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}$

The same process is used to calculate the amount of hydrogen:

$0.089g H_{2}O*\frac{2g H}{18g H_{2}O } = 0.010g H$

The percentage by mass of carbon and hydrogen are calculated as follows:

%C$\frac{0.238g}{1.3g} *100%$= 18.3%

%H$\frac{0.010g}{1.3g} *100%$=0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

$1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}$= 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl=$\frac{0.43g}{0.535g} * 100%$= 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

$18.3g C*\frac{1mol C}{12g C} = 1.52mol C$

$0.77g H*\frac{1mol H}{1g H} = 0.77mol H$

$80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl$

Dividing each of the quantities above by the smallest (0.77mol), the  subscripts in a tentative formula would be

$C=\frac{1.52}{0.77} = 1.97$ ≈ 2

$H = \frac{0.77}{0.77} = 1$

$Cl =\frac{2.27}{0.77}=2.94$≈3

The empirical formula for the compound is:

$C_{2} H Cl_{3}$

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.