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Current Question (ID: 20841)

Question:
$\text{Compound A contains 8.7\% Hydrogen, 74\% Carbon and 17.3\% Nitrogen. The molecular formula of the compound is:}$ $\text{(Given: Atomic masses of C, H and N are 12, 1 and 14 amu respectively. The molar mass of the compound A is 162 g mol}^{-1})$
Options:
  • 1. $\text{C}_4\text{H}_6\text{N}_2$
  • 2. $\text{C}_2\text{H}_3\text{N}$
  • 3. $\text{C}_5\text{H}_7\text{N}$
  • 4. $\text{C}_{10}\text{H}_{14}\text{N}_2$
Solution:
$\text{Empirical formula} \times n = \text{Molecular formula}$ $\begin{array}{ccc} \text{Element} & \% \text{ mass} & \text{Moles} & \text{Whole number ratio} \\ \text{C} & 74 & 6.17 & 5 \\ \text{H} & 8.7 & 8.7 & 7 \\ \text{N} & 17.3 & 1.236 & 1 \end{array}$ $\text{Empirical Formula} = \text{C}_5\text{H}_7\text{N}$ $\text{Empirical formula mass} = 81 \text{ g}$ $n \times 81 = 162$ $n = 2$ $\text{Hence molecular formula is } \text{C}_{10}\text{H}_{14}\text{N}_2$

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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
}