JEE Main 2025 — Some Basic Concepts Of Chemistry Question with Solution

Identify the given weight percentages from a 100 g sample:

Carbon (C): 14.5 g

Chlorine (Cl): 64.46 g

Hydrogen (H): 1.8 g

Since the sample must sum to 100 g, the remaining mass is from oxygen (O):

$$\text{O: }100-(14.5+64.46+1.8)=100-80.76=19.24\text{ g}.$$

Next, convert these masses to moles using the given atomic masses:

Moles of C:

$$\frac{14.5\text{ g}}{12\text{ g/mol}}\approx 1.2083\text{ mol}.$$

Moles of Cl:

$$\frac{64.46\text{ g}}{35.5\text{ g/mol}}\approx 1.8171\text{ mol}.$$

Moles of H:

$$\frac{1.8\text{ g}}{1\text{ g/mol}}=1.8\text{ mol}.$$

Moles of O:

$$\frac{19.24\text{ g}}{16\text{ g/mol}}\approx 1.2025\text{ mol}.$$

Now, find the simplest whole-number ratio by dividing each by the smallest number of moles (approximately 1.2025 mol):

Ratio for C:

$$1.2083/1.2025\approx 1.00$$

Ratio for O:

$$1.2025/1.2025=1.00$$

Ratio for H:

$$1.8/1.2025\approx 1.50$$

Ratio for Cl:

$$1.8171/1.2025\approx 1.51\approx 1.50$$

Since the ratios for H and Cl are about 1.5, multiplying all ratios by 2 will give whole numbers:

C: 1 × 2 = 2

O: 1 × 2 = 2

H: 1.5 × 2 = 3

Cl: 1.5 × 2 = 3

Thus, the empirical formula of the compound is:

$${\mathrm{C}}_2{\mathrm{H}}_3{\mathrm{O}}_2\mathrm{C}{\mathrm{l}}_3.$$

Calculate the empirical formula mass by summing the contributions:

C: $$2\times 12=24\text{ g/mol}$$

H: $$3\times 1=3\text{ g/mol}$$

O: $$2\times 16=32\text{ g/mol}$$

Cl: $$3\times 35.5=106.5\text{ g/mol}$$

Total mass:

$$24+3+32+106.5=165.5\text{ g/mol}.$$

The problem asks for the empirical formula mass in the form “__ × 10⁻¹”. We can express 165.5 g/mol as:

$$165.5=1655\times 10^{-1}\text{ g/mol}.$$

Thus, the answer is: $$1655\times 10^{-1}\text{ g/mol}.$$