Answer to Question #90108 in General Chemistry for Joseph

Question

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 Jwas determined using different analytical chemistry techniques.

Combustion reaction:
Combustion of 1.30 g of J gave 0.872 g CO2and 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. (b) Determine the percentage by mass of chlorine in J, using the precipitation data.
(c) The molar mass was determined to be 131.38 g mol–1. Deduce the molecular formula of J.

Answer 1

a) J reacts with oxygen to form CO₂ and H₂O.

Calculating the amount of CO₂ formed.

$n(CO_2)=\frac{m(CO_2)}{M(CO_2)}$

Amount of carbon in J is equal to n(CO₂):

$n(C)=n(CO_2)$

Mass of carbon in J can be calculated:

$m(C)=n(C)\cdot M(C) = n(CO_2)\cdot M(C)=\frac{m(CO_2)}{M(CO_2)} \cdot M(C) =\frac{0.872 g}{44\frac{g}{mol}}\cdot 12\frac{g}{mol}=0.2378g$

Percentage by mass of carbon in J is:

$\omega (C) = \frac{m(C)}{m(J)}=\frac{0.2378g}{1.30g}=0.1829 \ or \ 18.29 \%$

Calculating similar way percentage by mass of hydrogen in J taking into account that amount of hydrogen in J is twice the amount of water: $n(H)=2n(H_2O)$

$\omega (H) = \frac{m(H)}{m(J)}=\frac{n(H)\cdot M(H) }{m(J)}=\frac{2n(H_2O)\cdot M(H) }{m(J)}=\frac{2m(H_2O)\cdot M(H) }{M(H_2O) \cdot m(J)}=\frac{2\cdot 0.089g \cdot 1\frac{g}{mol} }{18\frac{g}{mol} \cdot 1.30g}=0.0076 \ or \ 0.76 \%$

b) Knowing the mass of AgCl precipitate we are able to calculate amount of AgCl that equals to amount of Cl in J:

$n(Cl)=n(AgCl)=\frac{m(AgCl)}{M(AgCl)}= \frac{1.75g}{143.5\frac{g}{mol}}=0.0122 mol$

Thus mass of chlorine in J is $m(Cl)=n(Cl) \cdot M(Cl) = 0.4331 g$

Percentage by mass of chlorine in J is:

$\omega (Cl) = \frac{m(Cl)}{m(J)}=\frac{0.4331g}{0.535g}=0.8095 \ or \ 80.95 \%$

c) Empirical formula of J is C_xH_yCl_z. We can find ratios of x:y:z using percentages by mass of the elements:

$x:y:z=\frac{\omega (C)}{M(C)}:\frac{\omega (H)}{M(H)}:\frac{\omega (Cl)}{M(Cl)}=\frac{0.1829}{12}:\frac{0.0076}{1}:\frac{0.8095}{35.5}=0.0152:0.0076:0.0228=2:1:3$

Thus empirical formula of J is C₂HCl₃

Checking molar mass of J: M(C₂HCl₃)=131.5 g/mol that corresponds to given value. Thus molecular formula of J is C₂HCl₃. Name of compound is 1,1,2-trichloroethylene.