Benzene reacts with hot concentrated nitric acid dissolved in sulfuric acid to give chiefly nitrobenzene, C6H5NO2. A by-product is often obtained, which consists of 42.86% C, 2.40% H, and 16.67% N (by mass). The boiling point of a solution of 5.5 g of the by-product in 45 g of benzene was 1.84°C higher than that of benzene.
6 answers
I don't see a question here.
What is the molar mass of the by-product?
Calculate the molecular formula of the by-product.
Calculate the empirical formula of the by-product.
Calculate the molecular formula of the by-product.
Calculate the empirical formula of the by-product.
So would the molar mass be 168.2?
Empirical be c3h2n1o2?
Empirical be c3h2n1o2?
Yes, empirical formula is C3H2N10.
Empirical formula mass is 178. (3*12) + (2*1) + (10*14) = 178
The approximate molar mass is given by the boiling point elevation.
delta T = Kb*m
1.84 = 2.53*m
m = about 0.7 but you need to go through this and all other calculations. I've estimated here and there.
m = mols/kg solvent
0.7 = mols/0.045 and mols = about 0.03
Then mols = g/molar mass or molar mass = g/mol = 5.5/0.0327 = about 168 for the molar mass approximately.
168/178 = about 0.94 and round that to 1 so the molar mass is 178 which is the same as the empirical mass..
Empirical formula mass is 178. (3*12) + (2*1) + (10*14) = 178
The approximate molar mass is given by the boiling point elevation.
delta T = Kb*m
1.84 = 2.53*m
m = about 0.7 but you need to go through this and all other calculations. I've estimated here and there.
m = mols/kg solvent
0.7 = mols/0.045 and mols = about 0.03
Then mols = g/molar mass or molar mass = g/mol = 5.5/0.0327 = about 168 for the molar mass approximately.
168/178 = about 0.94 and round that to 1 so the molar mass is 178 which is the same as the empirical mass..
HOW DO I FIGURE OUT THE MOLECULAR FORMULA FROM THE DATA?
Since the empirical formula mass is about the same as the molecular formula mass then the molecular formula is the same as the empirical formula.