Quantitative analysis of an organic compound (X) shows following % composition.

C : $14.5 \%$

Cl : 64.46%

H: 1.8 %

(Empirical formula mass of the compound $(\mathrm{X})$ is _________ $\times 10^{-1}$

(Given molar mass in $\mathrm{g} \mathrm{~mol}^{-1}$ of $\mathrm{C}: 12, \mathrm{H}: 1, \mathrm{O}: 16, \mathrm{Cl}: 35.5$)

<p>Identify the given weight percentages from a 100 g sample:</p> <p><p>Carbon (C): 14.5 g</p></p> <p><p>Chlorine (Cl): 64.46 g</p></p> <p><p>Hydrogen (H): 1.8 g</p></p> <p><p>Since the sample must sum to 100 g, the remaining mass is from oxygen (O):</p> <p>$\text{O: } 100 - (14.5 + 64.46 + 1.8) = 100 - 80.76 = 19.24\text{ g}.$</p></p> <p><p>Next, convert these masses to moles using the given atomic masses:</p></p> <p><p>Moles of C: </p> <p>$\frac{14.5\text{ g}}{12\text{ g/mol}} \approx 1.2083\text{ mol}.$</p></p> <p><p>Moles of Cl: </p> <p>$\frac{64.46\text{ g}}{35.5\text{ g/mol}} \approx 1.8171\text{ mol}.$</p></p> <p><p>Moles of H:</p> <p>$\frac{1.8\text{ g}}{1\text{ g/mol}} = 1.8\text{ mol}.$</p></p> <p><p>Moles of O:</p> <p>$\frac{19.24\text{ g}}{16\text{ g/mol}} \approx 1.2025\text{ mol}.$</p></p> <p>Now, find the simplest whole-number ratio by dividing each by the smallest number of moles (approximately 1.2025 mol):</p> <p><p>Ratio for C: </p> <p>$1.2083 / 1.2025 \approx 1.00$</p></p> <p><p>Ratio for O:</p> <p>$1.2025 / 1.2025 = 1.00$</p></p> <p><p>Ratio for H:</p> <p>$1.8 / 1.2025 \approx 1.50$</p></p> <p><p>Ratio for Cl:</p> <p>$1.8171 / 1.2025 \approx 1.51 \approx 1.50$</p> <p>Since the ratios for H and Cl are about 1.5, multiplying all ratios by 2 will give whole numbers:</p></p> <p><p>C: 1 × 2 = 2</p></p> <p><p>O: 1 × 2 = 2</p></p> <p><p>H: 1.5 × 2 = 3</p></p> <p><p>Cl: 1.5 × 2 = 3</p></p> <p><p>Thus, the empirical formula of the compound is:</p> <p>$\mathrm{C_2H_3O_2Cl_3}.$</p></p> <p><p>Calculate the empirical formula mass by summing the contributions:</p></p> <p><p>C: $2 \times 12 = 24\text{ g/mol}$</p></p> <p><p>H: $3 \times 1 = 3\text{ g/mol}$</p></p> <p><p>O: $2 \times 16 = 32\text{ g/mol}$</p></p> <p><p>Cl: $3 \times 35.5 = 106.5\text{ g/mol}$</p> <p>Total mass:</p> <p>$24 + 3 + 32 + 106.5 = 165.5\text{ g/mol}.$</p></p> <p><p>The problem asks for the empirical formula mass in the form “<strong><em>__</em></strong> × 10⁻¹”. We can express 165.5 g/mol as:</p> <p>$165.5 = 1655 \times 10^{-1}\text{ g/mol}.$</p></p> <p>Thus, the answer is: $1655 \times 10^{-1}\text{ g/mol}.$</p>