Find the amount of each element in a compound

Question

A sample of a new anti-malarial drug with a mass of 0.2394 g was made to undergo a series of reactions that changed all of the nitrogen in the compound into N2. This gas had a volume of 18.90 mL when collected over water at 296.95 K and a pressure of 746.0 torr. At 23.80 °C, the vapor pressure of water is 22.110 torr. When 6.478 mg of the compound was burned in pure oxygen, 17.57 mg of CO2 and 4.319 mg of H2O were obtained.

 

What are the percentages of elements in the compound?

What is the empirical formula?

What is the molecular formula?

Answer 1

Step-by-step solution:

1. Calculate percentage of nitrogen (N) in the compound:

- Correct the pressure for water vapor: $$746.0 \text{ torr} - 22.110 \text{ torr} = 723.89 \text{ torr}$$.

- Convert to atm: $$723.89 / 760 = 0.9525 \text{ atm}$$.

- Convert volume to liters: $$18.90 \text{ mL} = 0.01890 \text{ L}$$.

- Use ideal gas law to calculate moles of $$N_2$$:

$$

n = \frac{PV}{RT} = \frac{0.9525 \times 0.0189}{0.08206 \times 296.95} = 7.29 \times 10^{-4} \text{ mol}

$$

- Moles of nitrogen atoms: $$2 \times 7.29 \times 10^{-4} = 1.458 \times 10^{-3} \text{ mol}$$.

- Mass of nitrogen atoms: $$1.458 \times 10^{-3} \times 14.01 = 0.0204 \text{ g}$$.

- Percent nitrogen in sample: $$\frac{0.0204}{0.2394} \times 100 = 8.65\%$$.

2. Calculate percentage of carbon (C) and hydrogen (H) from combustion

- Mass of C from $$CO_2$$: $$17.57 \text{ mg} \times \frac{12.01}{44.01} = 4.79 \text{ mg}$$.

- Mass of H from $$H_2O$$: $$4.319 \text{ mg} \times \frac{2.02}{18.02} = 0.484 \text{ mg}$$.

- Total burned mass was 6.478 mg, so percentages:

- Carbon: $$\frac{4.79}{6.478} \times 100 = 74.0\%$$.

- Hydrogen: $$\frac{0.484}{6.478} \times 100 = 7.47\%$$.

3. Calculate oxygen percentage by difference:

- Oxygen: $$100\% - 8.65\% - 74.0\% - 7.47\% = 9.87\%$$.

4. Determine empirical formula from percentages:

- Moles of each element per 100 g:

- N: $$8.65 / 14.01 = 0.618 \text{ mol}$$.

- C: $$74.0 / 12.01 = 6.16 \text{ mol}$$.

- H: $$7.47 / 1.008 = 7.41 \text{ mol}$$.

- O: $$9.87 / 16.00 = 0.616 \text{ mol}$$.

- Normalize by smallest (0.616):

- N: $$0.618 / 0.616 = 1.00$$.

- C: $$6.16 / 0.616 = 10.0$$.

- H: $$7.41 / 0.616 = 12.0$$.

- O: $$0.616 / 0.616 = 1.00$$.

- Empirical formula is $$N_1C_{10}H_{12}O_1$$.

5. Determine molecular formula:

- Calculate empirical formula molar mass:

$$

1 \times 14.01 + 10 \times 12.01 + 12 \times 1.008 + 1 \times 16.00 = 162.2 \text{ g/mol}

$$

- Using given data and calculations, molecular formula likely matches empirical formula (ratio ~1), so molecular formula is $$N_1C_{10}H_{12}O_1$$.

Summary:

- Percent composition: N = 8.65%, C = 74.0%, H = 7.47%, O = 9.87%.

- Empirical formula: $$NC_{10}H_{12}O$$.

- Molecular formula: same as empirical, $$NC_{10}H_{12}O$$ .