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}](https://tex.z-dn.net/?f=C_%7B2%7D%20H%20Cl_%7B3%7D)
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}](https://tex.z-dn.net/?f=0.872g%20CO%20_%7B2%7D%20%2A%5Cfrac%7B12g%20C%7D%7B44g%20CO_%7B2%7D%20%7D%20%3D%200.238g%20CO_%7B2%7D)
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](https://tex.z-dn.net/?f=0.089g%20H_%7B2%7DO%2A%5Cfrac%7B2g%20H%7D%7B18g%20H_%7B2%7DO%20%7D%20%20%3D%200.010g%20H)
The percentage by mass of carbon and hydrogen are calculated as follows:
%C
= 18.3%
%H
=0.77%
From the precipation data it is possible obtain the amount of chlorine present in the compound:
= 0.43g AgCl
Let's calculate the percentage by mass of chlorine:
%Cl=
= 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](https://tex.z-dn.net/?f=18.3g%20C%2A%5Cfrac%7B1mol%20C%7D%7B12g%20C%7D%20%3D%201.52mol%20C)
![0.77g H*\frac{1mol H}{1g H} = 0.77mol H](https://tex.z-dn.net/?f=0.77g%20H%2A%5Cfrac%7B1mol%20H%7D%7B1g%20H%7D%20%3D%200.77mol%20H)
![80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl](https://tex.z-dn.net/?f=80.37gCl%2A%5Cfrac%7B1molCl%7D%7B35.45g%20Cl%7D%20%3D%202.27mol%20Cl)
Dividing each of the quantities above by the smallest (0.77mol), the subscripts in a tentative formula would be
≈ 2
![H = \frac{0.77}{0.77} = 1](https://tex.z-dn.net/?f=H%20%3D%20%5Cfrac%7B0.77%7D%7B0.77%7D%20%3D%201)
≈3
The empirical formula for the compound is:
![C_{2} H Cl_{3}](https://tex.z-dn.net/?f=C_%7B2%7D%20H%20Cl_%7B3%7D)
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.