A 118.0 g sample of a compound contains 72.0 g of [tex]C[/tex], 18.0 g of [tex]H[/tex], and 28.0 g of [tex]N[/tex].

Which of the following is the empirical formula of the compound?

Choose 1 answer:

A. [tex]CH_3N[/tex]

B. [tex]CH_3N_2[/tex]

C. [tex]C_3H_9N[/tex]

D. [tex]C_6H_{18}N_2[/tex]



Answer :

To determine the empirical formula of a compound, follow these steps:

1. Determine the mass of each element in the compound:
- Carbon (C): 72.0 g
- Hydrogen (H): 18.0 g
- Nitrogen (N): 28.0 g

2. Find the molar mass (atomic weight) of each element:
- Molar mass of C: 12.01 g/mol
- Molar mass of H: 1.008 g/mol
- Molar mass of N: 14.01 g/mol

3. Convert the mass of each element to moles by using the formula:
[tex]\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \][/tex]
- Moles of C:
[tex]\[ \text{Moles of C} = \frac{72.0 \text{ g}}{12.01 \text{ g/mol}} \approx 5.996 \text{ mol} \][/tex]
- Moles of H:
[tex]\[ \text{Moles of H} = \frac{18.0 \text{ g}}{1.008 \text{ g/mol}} \approx 17.857 \text{ mol} \][/tex]
- Moles of N:
[tex]\[ \text{Moles of N} = \frac{28.0 \text{ g}}{14.01 \text{ g/mol}} \approx 1.998 \text{ mol} \][/tex]

4. Determine the simplest mole ratio by dividing all moles by the smallest number of moles calculated:
- The smallest number of moles is approximately 1.998 (moles of N).

- Ratio of C:
[tex]\[ \frac{5.996}{1.998} \approx 3 \][/tex]
- Ratio of H:
[tex]\[ \frac{17.857}{1.998} \approx 8.942 \approx 9 \][/tex]
- Ratio of N:
[tex]\[ \frac{1.998}{1.998} = 1 \][/tex]

5. These ratios represent the subscripts in the empirical formula:
The ratio of C:H:N is approximately 3:9:1.

Therefore, the empirical formula of the compound is [tex]\( C_3H_9N \)[/tex].

The correct answer is:

(C) [tex]\( C_3H_9N \)[/tex]