Question

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 CO2(g) formed when 3.38 g of this hydrocarbon is completely burnt in oxygen is: (Given: Molar mass in g mol\(^{-1}\): C: 12, H: 1, O: 16)}

• A. 5.68
• B. 11.44
• C. 22.74
• D. 17.05

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We first determine the empirical formula of the hydrocarbon using the given mass ratio of Carbon to Hydrogen.
Using the fact that it contains exactly two carbon atoms, we find its exact molecular formula and molar mass.
Finally, we apply stoichiometry to the combustion reaction to find the mass of \(\text{CO}_2\) produced.
Step 2: Key Formula or Approach:
Moles = \(\frac{\text{Given Mass}}{\text{Atomic/Molar Mass}}\).
Balanced combustion equation: \(\text{C}_x\text{H}_y + (x + \frac{y}{4})\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}\).
Step 3: Detailed Explanation:
The given mass ratio of C : H is 12 : 1.
Calculate the mole ratio by dividing by their respective atomic masses:
Moles of C = \(\frac{12}{12} = 1\).
Moles of H = \(\frac{1}{1} = 1\).
The empirical formula is CH.
Since the molecule contains exactly two carbon atoms, its molecular formula must be \(\text{C}_2\text{H}_2\) (Ethyne).
The molar mass of \(\text{C}_2\text{H}_2\) is \((2 \times 12) + (2 \times 1) = 26 \text{ g/mol}\).
The amount of hydrocarbon burned is 3.38 g. Calculate the moles of \(\text{C}_2\text{H}_2\):
\[ \text{Moles of } \text{C}_2\text{H}_2 = \frac{3.38}{26} = 0.13 \text{ mol} \] Write the balanced combustion equation:
\[ \text{C}_2\text{H}_2 + \frac{5}{2}\text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O} \] From the stoichiometry, 1 mole of \(\text{C}_2\text{H}_2\) produces 2 moles of \(\text{CO}_2\).
So, 0.13 moles of \(\text{C}_2\text{H}_2\) will produce:
\[ \text{Moles of } \text{CO}_2 = 2 \times 0.13 = 0.26 \text{ mol} \] Now, calculate the mass of the produced \(\text{CO}_2\) (Molar mass of \(\text{CO}_2 = 44 \text{ g/mol}\)):
\[ \text{Mass of } \text{CO}_2 = 0.26 \text{ mol} \times 44 \text{ g/mol} = 11.44 \text{ g} \] Step 4: Final Answer:
The weight of \(\text{CO}_2\) formed is \(11.44 \text{ g}\).