Chemistry Q51 – JEE-MAIN 2026

Home Chemistry Basic ChemistryQuestion #38

The ratio of mass percentage (w/w) of C : H in a hydrocarbon is 12 : 1. It has two carbon atoms. The weight (in g) of

C O_{2} (g)

formed when 3.38 g of this hydrocarbon is completely burnt in oxygen is : (Given : Molar mass in g

m o l^{- 1}

C : 12, H : 1, O : 16)

Clue 1 of 3: Determine the molecular formula

Use the given mass ratio of carbon to hydrogen and the number of carbon atoms to find the exact chemical formula of the hydrocarbon.

🧠 Full Solution Path

Step 1: Determine the Molecular Formula of the Hydrocarbon✦ Active

The mass ratio of C:H is $12 : 1$ . Given molar mass of C is 12 g/mol and H is 1 g/mol. The mole ratio of C:H is $\frac{12}{12} : \frac{1}{1} = 1 : 1$ . Thus, the empirical formula is $C H$ . Since the hydrocarbon has two carbon atoms, its molecular formula must be $(C H)_{2} = C_{2} H_{2}$ . The molar mass of $C_{2} H_{2}$ is $(2 \times 12) + (2 \times 1) = 26$ g/mol.

Step 2: Write and Balance the Combustion Reaction○ Expand

The complete combustion reaction for $C_{2} H_{2}$ is:

2 C_{2} H_{2} (g) + 5 O_{2} (g) \to 4 C O_{2} (g) + 2 H_{2} O (l)

From the balanced equation, 2 moles of $C_{2} H_{2}$ produce 4 moles of $C O_{2}$ , which simplifies to 1 mole of $C_{2} H_{2}$ producing 2 moles of $C O_{2}$ .

Step 3: Calculate the Mass of

C O_{2}

Formed○ Expand

First, calculate the moles of $C_{2} H_{2}$ burnt:

Moles of C_{2} H_{2} = \frac{mass}{molar mass} = \frac{3.38 g}{26 g/mol} = 0.13 mol

Using the stoichiometric ratio from Step 2, calculate the moles of $C O_{2}$ produced:

Moles of C O_{2} = 0.13 mol C_{2} H_{2} \times \frac{2 mol C O_{2}}{1 mol C_{2} H_{2}} = 0.26 mol C O_{2}

Finally, calculate the mass of $C O_{2}$ formed. The molar mass of $C O_{2} = 12 + (2 \times 16) = 44$ g/mol.

Mass of C O_{2} = 0.26 mol \times 44 g/mol = 11.44 g

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Chemistry Question 51 – JEE-MAIN 2026

Clue 1 of 3: Determine the molecular formula

🧠 Full Solution Path