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.
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.
So would the molar mass be 168.2?
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..
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.
To solve this problem, we need to determine the molecular formula of the by-product and calculate its molar mass. Let's break down the process step by step:
1. Calculate the percent composition of each element in the by-product:
- C (Carbon): 42.86%
- H (Hydrogen): 2.40%
- N (Nitrogen): 16.67%
2. Assume we have 100g of the by-product. This means we have:
- C: 42.86g of carbon
- H: 2.40g of hydrogen
- N: 16.67g of nitrogen
3. Convert the grams of each element to moles:
- C: 42.86g C × (1 mol C / 12.01g C) = 3.570 mol C
- H: 2.40g H × (1 mol H / 1.01g H) = 2.376 mol H
- N: 16.67g N × (1 mol N / 14.01g N) = 1.191 mol N
4. Determine the empirical formula by dividing each mole value by the smallest mole value:
- C: 3.570 mol C / 1.191 mol N = 3
- H: 2.376 mol H / 1.191 mol N = 2
- N: 1.191 mol N / 1.191 mol N = 1
The empirical formula is C3H2N.
5. Calculate the molar mass of the empirical formula:
- Molar mass C: 12.01 g/mol × 3 = 36.03 g/mol
- Molar mass H: 1.01 g/mol × 2 = 2.02 g/mol
- Molar mass N: 14.01 g/mol × 1 = 14.01 g/mol
Molar mass of C3H2N = 36.03 g/mol + 2.02 g/mol + 14.01 g/mol = 52.06 g/mol
6. Now that we know the molar mass of the empirical formula is 52.06 g/mol, we can calculate the molecular formula by dividing the molar mass of the by-product by the empirical molar mass:
- Molar mass of the by-product = 52.06 g/mol
- Experimental molar mass of the by-product = 5.5 g / (45 g/mol) = 0.1222 mol
Molecular formula molar mass = (Experimental molar mass) / (Empirical formula molar mass)
Molecular formula molar mass = 0.1222 mol / 52.06 g/mol ≈ 0.00235
Since we need a whole number for the molecular formula, we multiply the subscripts by an appropriate factor to get as close as possible to a whole number. In this case, multiplying by 1000 would yield a whole number:
Molecular formula molar mass ≈ 0.00235 × 1000 = 2.35
Therefore, the empirical formula C3H2N would be multiplied by 2, resulting in C6H4N2.
7. Finally, we can conclude that the by-product has the molecular formula C6H4N2.
To summarize, the molecular formula of the by-product obtained when benzene reacts with hot concentrated nitric acid dissolved in sulfuric acid is C6H4N2.