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Current Question (ID: 7384)

Question:
$\text{An organic compound contains carbon, hydrogen, and oxygen. Its elemental analysis gave C, 38.71\%, and H, 9.67\%. The empirical formula of the compound would be:}$
Options:
  • 1. $\text{CH}_3\text{O}$
  • 2. $\text{CH}_2\text{O}$
  • 3. $\text{CHO}$
  • 4. $\text{CH}_4\text{O}$
Solution:
$\text{Hint: Empirical formula concept}$ $\text{Step 1:}$ $\text{The empirical formula of the given compound can be calculated in the following manner:}$ $\begin{array}{|c|c|c|c|c|} \hline \text{Element} & \text{\%} & \text{At wt} & \text{Relative number of atoms} & \text{Simplest ratio of atoms} \\ \hline \text{C} & 38.71 & 12 & 3.23 & \frac{3.23}{3.23} = 1 \\ \hline \text{H} & 9.67 & 1 & 9.67 & \frac{9.67}{3.23} = 3 \\ \hline \text{O} & 51.62 & 16 & 3.23 & \frac{3.23}{3.23} = 1 \\ \hline \end{array}$ $\text{Step 2:}$ $\text{Therefore, the empirical formula comes out to be CH}_3\text{O}.$

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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
}