As a student of chemistry analyse and interpret the available data to address management concerns.

First, calculate the percentage of Nitrogen by subtracting the percentages of Carbon and Hydrogen from 100%.

$$\% \text{N} = 100\% - 77.3\% - 7.58\% = 15.12\%$$

Next, assume a 100 g sample to convert these percentages into masses, and then into moles using their respective atomic masses (C=12.01, H=1.008, N=14.01 g/mol).

$$\text{Moles of C} = \frac{77.3 \text{ g}}{12.01 \text{ g/mol}} = 6.436 \text{ mol}$$ $$\text{Moles of H} = \frac{7.58 \text{ g}}{1.008 \text{ g/mol}} = 7.520 \text{ mol}$$ $$\text{Moles of N} = \frac{15.12 \text{ g}}{14.01 \text{ g/mol}} = 1.079 \text{ mol}$$

Divide each mole value by the smallest number of moles (1.079 mol for Nitrogen) to find the simplest whole-number ratio of atoms.

$$\text{C}: \frac{6.436}{1.079} \approx 5.96 \approx 6$$ $$\text{H}: \frac{7.520}{1.079} \approx 6.97 \approx 7$$ $$\text{N}: \frac{1.079}{1.079} = 1$$

The empirical formula of X is $\text{C}_6\text{H}_7\text{N}$.

In steam distillation, the total pressure ($P_{\text{total}}$) is the sum of the partial pressures of the immiscible components (compound X and water).

$$P_{\text{total}} = P_{\text{X}} + P_{\text{water}}$$

Given $P_{\text{total}} = 760 \text{ mmHg}$ and the saturated vapor pressure of water ($P_{\text{water}}$) at $98^\circ\text{C}$ is $734.9 \text{ mmHg}$.

$$P_{\text{X}} = P_{\text{total}} - P_{\text{water}} = 760 \text{ mmHg} - 734.9 \text{ mmHg} = 25.1 \text{ mmHg}$$

The ratio of the masses of the components in the distillate is proportional to the product of their partial pressures and molar masses.

$$\frac{m_{\text{X}}}{m_{\text{water}}} = \frac{P_{\text{X}} \times M_{\text{X}}}{P_{\text{water}} \times M_{\text{water}}}$$

The distillate contains 85% by mass water, meaning it contains 15% by mass compound X. So, for every 15 g of X, there are 85 g of water. The molar mass of water ($M_{\text{water}}$) is approximately $18.015 \text{ g/mol}$.

$$\frac{15 \text{ g}}{85 \text{ g}} = \frac{25.1 \text{ mmHg} \times M_{\text{X}}}{734.9 \text{ mmHg} \times 18.015 \text{ g/mol}}$$

Rearrange the equation to solve for $M_{\text{X}}$:

$$M_{\text{X}} = \frac{15}{85} \times \frac{734.9}{25.1} \times 18.015 \text{ g/mol}$$ $$M_{\text{X}} = 0.17647 \times 29.27888 \times 18.015 \text{ g/mol}$$ $$M_{\text{X}} \approx 93.07 \text{ g/mol}$$

The molar mass of X is approximately $\boxed{\text{93.1 g/mol}}$.

Calculate the empirical formula mass (EFM) for $\text{C}_6\text{H}_7\text{N}$:

$$\text{EFM} = (6 \times 12.01) + (7 \times 1.008) + (1 \times 14.01) = 72.06 + 7.056 + 14.01 = 93.126 \text{ g/mol}$$

Compare the calculated molar mass ($M_{\text{X}} \approx 93.07 \text{ g/mol}$) with the empirical formula mass.

$$n = \frac{\text{Molar Mass}}{\text{Empirical Formula Mass}} = \frac{93.07 \text{ g/mol}}{93.126 \text{ g/mol}} \approx 1$$

Since $n$ is approximately 1, the molecular formula is the same as the empirical formula.
The molecular formula of X is $\boxed{\text{C}_6\text{H}_7\text{N}}$.

Steam distillation is a valuable purification technique for organic compounds that are immiscible with water, volatile in steam, and often sensitive to high temperatures. It allows these compounds to be distilled at a temperature lower than their normal boiling points, preventing decomposition.

• Purification of high-boiling organic compounds: It is used to purify compounds like aniline (compound X), nitrobenzene, and other organic substances that would decompose if distilled at their normal boiling points.
• Extraction of natural products: This method is widely employed in industries such as perfume and food to extract essential oils from plants (e.g., eucalyptus, lemon, lavender) without causing thermal degradation of the delicate compounds.
• Separation of volatile compounds from non-volatile impurities: Steam distillation effectively separates volatile organic compounds from non-volatile impurities or inorganic salts, providing a clean product.