Import Question JSON

« Back to Question List Edit Question »

Current Question (ID: 7363)

Question:
$\text{An organic substance containing C, H, and O gave the following percentage composition:}$\n\n$\text{C = 40.687\%, H = 5.085\% and O = 54.228\%. The vapour density of this organic substance is 59.}$\n\n$\text{The molecular formula of the compound will be:}$
Options:
  • 1. $\text{C}_4\text{H}_6\text{O}_4$
  • 2. $\text{C}_4\text{H}_6\text{O}_2$
  • 3. $\text{C}_4\text{H}_4\text{O}_2$
  • 4. $\text{None of the above}$
Solution:
$\text{Hint: Molecular mass = 2} \times \text{Vapour density}$\n\n$\text{Explanation:}$\n\n$\text{STEP 1:}$\n\n$\begin{array}{|c|c|c|c|c|c|c|}\hline \text{Element} & \text{Symbol} & \text{Percentage of element} & \text{Atomic mass of element} & \text{Relative number of atoms} = \frac{\text{Percentage}}{\text{At. mass}} & \text{Simplest atomic ratio} & \text{Simplest whole number atomic ratio} \\ \hline \text{Carbon} & \text{C} & 40.687 & 12 & \frac{40.687}{12} = 3.390 & \frac{3.390}{3.389} = 1 & 2 \\ \hline \text{Hydrogen} & \text{H} & 5.085 & 1 & \frac{5.085}{1} = 5.085 & \frac{5.085}{3.389} = 1.5 & 3 \\ \hline \text{Oxygen} & \text{O} & 54.228 & 16 & \frac{54.228}{16} = 3.389 & \frac{3.389}{3.389} = 1 & 2 \\ \hline \end{array}$\n\n$\therefore \text{Empirical formula is C}_2\text{H}_3\text{O}_2$\n\n$\therefore \text{Empirical formula mass of C}_2\text{H}_3\text{O}_2 = 59$\n\n$\text{STEP 2: Also, Molecular mass = 2} \times \text{Vapour density = 2} \times 59 = 118$\n\n$\therefore n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{118}{59} = 2$\n\n$\text{Now, Molecular formula} = n \times (\text{Empirical formula}) = 2 \times (\text{C}_2\text{H}_3\text{O}_2) = \text{C}_4\text{H}_6\text{O}_4$\n\n$\therefore \text{Molecular formula is C}_4\text{H}_6\text{O}_4.$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}