Question

An organic compound A that contains 31.9% by mass of cart,6.8% hydrogen,42.79% of oxygen and 18.51% nitrogen has a molecular mass 75/mol
Detert
I) the impirical for A
II) molecular formula for A.( C=12,o=16,N=14,H=1)

Answer 1

To determine the empirical and molecular formula of the compound A with the given percentages by mass:

### Step 1: Convert percentages to grams
Assuming we have 100 g of the compound:
- Carbon (C): 31.9 g
- Hydrogen (H): 6.8 g
- Oxygen (O): 42.79 g
- Nitrogen (N): 18.51 g

### Step 2: Convert grams to moles
Next, we convert grams to moles using the molar masses:
- Molar mass of C = 12 g/mol
- Molar mass of H = 1 g/mol
- Molar mass of O = 16 g/mol
- Molar mass of N = 14 g/mol

Calculating moles for each element:
1. Moles of Carbon (C):
\[
\text{Moles of C} = \frac{31.9 \text{ g}}{12 \text{ g/mol}} = 2.658 \text{ mol}
\]
2. Moles of Hydrogen (H):
\[
\text{Moles of H} = \frac{6.8 \text{ g}}{1 \text{ g/mol}} = 6.8 \text{ mol}
\]
3. Moles of Oxygen (O):
\[
\text{Moles of O} = \frac{42.79 \text{ g}}{16 \text{ g/mol}} = 2.673 \text{ mol}
\]
4. Moles of Nitrogen (N):
\[
\text{Moles of N} = \frac{18.51 \text{ g}}{14 \text{ g/mol}} = 1.321 \text{ mol}
\]

### Step 3: Find the simplest ratio of moles
Now we divide each mole quantity by the smallest number of moles calculated (which is for nitrogen, 1.321 mol):
1. Moles of C:
\[
\frac{2.658}{1.321} \approx 2.01 \approx 2
\]
2. Moles of H:
\[
\frac{6.8}{1.321} \approx 5.14 \approx 5
\]
3. Moles of O:
\[
\frac{2.673}{1.321} \approx 2.02 \approx 2
\]
4. Moles of N:
\[
\frac{1.321}{1.321} \approx 1
\]

### Step 4: Write the empirical formula
The ratio of moles gives us approximately:
- C: 2
- H: 5
- O: 2
- N: 1

Thus, the empirical formula for compound A is:
\[
\text{C}_2\text{H}_5\text{O}_2\text{N}
\]

### Step 5: Calculate the empirical formula mass
Calculating the molar mass of the empirical formula \(C_2H_5O_2N\):
\[
= (2 \times 12) + (5 \times 1) + (2 \times 16) + (1 \times 14)\\
= 24 + 5 + 32 + 14 = 75 \text{ g/mol}
\]

### Step 6: Determine the molecular formula
The empirical formula mass is equal to the molecular mass given (75 g/mol), which means that the molecular formula is the same as the empirical formula.

### Final Answers
1. **Empirical Formula for A**: \(C_2H_5O_2N\)
2. **Molecular Formula for A**: \(C_2H_5O_2N\)