Answer:
Empirical formula
C3H2Cl
Molecular formula
C6H4Cl2
Explanation:
Firstly, we can get the mass of carbon and hydrogen produced by calculating their number of moles from that of carbon iv oxide and water respectively.
To get the number of moles of carbon iv oxide, we simply make a division. The number of moles is the mass of carbon iv oxide divided by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol. The number of moles is thus 3.02/44 = 0.069
One atom of carbon is present in Carbon iv oxide. This means 0.069 moles of carbon is present. The mass of carbon present equals the atomic mass of carbon multiplied by the number of moles = 0.069 * 12 = 0.83g
The mass of water is 0.412g, the number of moles is 0.412/18 = 0.023
In water, there are two atoms of hydrogen. Hence the number of moles of hydrogen is 2 * 0.023 = 0.046
The mass of hydrogen is the number of moles of hydrogen multiplied by the atomic mass unit of hydrogen = 0.046 * 1 = 0.046g
Now we need to know the mass of chlorine. We simply subtract the masses of hydrogen and carbon.
1.68-0.046-0.83 = 0.804
Now we proceed to calculate the empirical formula. We simply divide the mass of each by their atomic masses and then proceed from there.
H = 0.046/1 = 0.046
C = 0.83/12 = 0.069
Cl = 0.804/35.5 = 0.023
We divide by the smallest number which is that of that of carbon
H = 0.046/0.023= 2
C = 0.069/0.023= 3
Cl = 0.023/0.023 = 1
The empirical formula would thus be C3H2Cl
The molecular formula can be calculated using the molar mass;
(C3H2Cl)n = 147
(36+ 2 + 35.5 ) = 147
73.5n = 147
n = 147/73.5 = 2
The molecular formula is thus C6H4Cl2
