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Some Basic Concepts of Chemistry question

2025 · 2 Apr · Shift 1 · Q10
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Some Basic Concepts of Chemistry question

2025 · 2 Apr · Shift 1 · Q10

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
On complete combustion 1.0 g of an organic compound (X)(\mathrm{X})(X) gave 1.46 g of CO2\mathrm{CO}_2CO2​ and 0.567 g of H2O\mathrm{H}_2 \mathrm{O}H2​O. The empirical formula mass of compound (X)(\mathrm{X})(X) is ‾\underline{\hspace{2cm}}​ g. (Given molar mass in gmol−1C:12,H:1,O:16\mathrm{g} \mathrm{mol}^{-1} \mathrm{C}: 12, \mathrm{H}: 1, \mathrm{O}: 16gmol−1C:12,H:1,O:16 )
  1. A
    60
  2. B
    45
  3. C
    30
  4. D
    15
View written solutionFree

Correct answer: C

  1. Find mass of carbon from CO2\mathrm{CO_2}CO2​

Given 1.46 g1.46\,\text{g}1.46g of CO2\mathrm{CO_2}CO2​ is formed.

Mass of carbon in CO2\mathrm{CO_2}CO2​:

Mass of C=1.46×1244\text{Mass of C} = 1.46 \times \frac{12}{44}Mass of C=1.46×4412​ =1.46×311=0.398 g≈0.40 g= 1.46 \times \frac{3}{11} = 0.398\,\text{g} \approx 0.40\,\text{g}=1.46×113​=0.398g≈0.40g
  1. Find mass of hydrogen from H2O\mathrm{H_2O}H2​O

Given 0.567 g0.567\,\text{g}0.567g of H2O\mathrm{H_2O}H2​O is formed.

Mass of hydrogen in H2O\mathrm{H_2O}H2​O:

Mass of H=0.567×218\text{Mass of H} = 0.567 \times \frac{2}{18}Mass of H=0.567×182​ =0.567×19=0.063 g= 0.567 \times \frac{1}{9} = 0.063\,\text{g}=0.567×91​=0.063g
  1. Find mass of oxygen in the compound

Total mass of compound taken =1.0 g= 1.0\,\text{g}=1.0g

Mass of O=1.0−(0.398+0.063)\text{Mass of O} = 1.0 - (0.398 + 0.063)Mass of O=1.0−(0.398+0.063) =1.0−0.461=0.539 g= 1.0 - 0.461 = 0.539\,\text{g}=1.0−0.461=0.539g
  1. Convert masses into moles

For carbon:

Moles of C=0.39812=0.0332\text{Moles of C} = \frac{0.398}{12} = 0.0332Moles of C=120.398​=0.0332

For hydrogen:

Moles of H=0.0631=0.063\text{Moles of H} = \frac{0.063}{1} = 0.063Moles of H=10.063​=0.063

For oxygen:

Moles of O=0.53916=0.0337\text{Moles of O} = \frac{0.539}{16} = 0.0337Moles of O=160.539​=0.0337
  1. Find simplest mole ratio

Divide by the smallest value, approximately 0.03320.03320.0332:

C:H:O=0.03320.0332:0.0630.0332:0.03370.0332\text{C} : \text{H} : \text{O} = \frac{0.0332}{0.0332} : \frac{0.063}{0.0332} : \frac{0.0337}{0.0332}C:H:O=0.03320.0332​:0.03320.063​:0.03320.0337​ ≈1:1.90:1.02≈1:2:1\approx 1 : 1.90 : 1.02 \approx 1 : 2 : 1≈1:1.90:1.02≈1:2:1

So the empirical formula is:

CH2O\mathrm{CH_2O}CH2​O
  1. Calculate empirical formula mass
Empirical formula mass=12+2(1)+16=30\text{Empirical formula mass} = 12 + 2(1) + 16 = 30Empirical formula mass=12+2(1)+16=30
  1. Check options
  • A: 606060
  • B: 454545
  • C: 303030 ✅
  • D: 151515

Therefore, the correct answer is Option C.

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