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Chemistry Question 63 – JEE-MAIN 2025

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On complete combustion 1.0 g of an organic compound (X) gave 1.46 g of CO2 and 0.567 g of H2O. The empirical formula mass of compound (X) is _______ g. (Given molar mass in g mol−1 C : 12, H : 1, O : 16)

Determine the mass of carbon and hydrogen from the products of combustion (CO2 and H2O).

● Video Walkthrough
Step 1: Calculate mass of C and H from products✦ Active

First, calculate the mass of carbon (C) from the CO2 produced and the mass of hydrogen (H) from the H2O produced. The molar mass of C is 12 g/mol, CO2 is 44 g/mol, H is 1 g/mol, and H2O is 18 g/mol.

Mass of C=1.46 g CO2×12 g C44 g CO2=0.398 g C Mass of H=0.567 g H2O×2 g H18 g H2O=0.063 g H
Step 2: Determine mass of O and moles of each element○ Expand

Assuming the organic compound (X) contains only C, H, and O, the mass of oxygen can be found by subtracting the masses of C and H from the total mass of the compound. Then, convert the masses of each element to moles.

Mass of O=1.0 g (X)−(0.398 g C+0.063 g H)=0.539 g O Moles of C=0.398 g12 g/mol=0.03317 mol Moles of H=0.063 g1 g/mol=0.063 mol Moles of O=0.539 g16 g/mol=0.03369 mol
💡 Teacher's Secret Hint

Ensure to use the correct molar masses for each element when converting to moles.

Step 3: Find the empirical formula and its mass○ Expand

Divide the moles of each element by the smallest number of moles to find the simplest whole-number ratio, which gives the empirical formula. Then, calculate the empirical formula mass.

Ratio C=0.033170.03317≈1 Ratio H=0.0630.03317≈1.9≈2 Ratio O=0.033690.03317≈1.01≈1 Empirical Formula=CH2O Empirical Formula Mass=(1×12)+(2×1)+(1×16)=12+2+16=30 g/mol
💡 Teacher's Secret Hint

Remember to round the mole ratios to the nearest whole number to obtain the empirical formula.

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