Respuesta :
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: [tex]C_{2} H Cl_{3}[/tex]
Explanation:
Firstly, the mass of carbon must be determined by using a conversion factor:
[tex]0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}[/tex]
The same process is used to calculate the amount of hydrogen:
[tex]0.089g H_{2}O*\frac{2g H}{18g H_{2}O } = 0.010g H[/tex]
The percentage by mass of carbon and hydrogen are calculated as follows:
%C[tex] \frac{0.238g}{1.3g} *100% [/tex]= 18.3%
%H[tex]\frac{0.010g}{1.3g} *100%[/tex]=0.77%
From the precipation data it is possible obtain the amount of chlorine present in the compound:
[tex]1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}[/tex]= 0.43g AgCl
Let's calculate the percentage by mass of chlorine:
%Cl=[tex] \frac{0.43g}{0.535g} * 100% [/tex]= 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:
[tex]18.3g C*\frac{1mol C}{12g C} = 1.52mol C[/tex]
[tex]0.77g H*\frac{1mol H}{1g H} = 0.77mol H[/tex]
[tex]80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl[/tex]
Dividing each of the quantities above by the smallest (0.77mol), the subscripts in a tentative formula would be
[tex]C=\frac{1.52}{0.77} = 1.97[/tex] ≈ 2
[tex]H = \frac{0.77}{0.77} = 1[/tex]
[tex]Cl =\frac{2.27}{0.77}=2.94[/tex]≈3
The empirical formula for the compound is:
[tex]C_{2} H Cl_{3}[/tex]
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.
