A) \[{{C}_{2}}{{H}_{2}}{{O}_{2}}\]
B) \[{{C}_{2}}{{H}_{4}}{{O}_{2}}\]
C) \[C{{H}_{2}}O\]
D) \[{{C}_{2}}{{H}_{4}}{{O}_{6}}\]
Correct Answer: C
Solution :
Key Idea: First calculate the percentage of elements, then find relative number of atoms. The relative number of atoms is used to find ratio of atoms, which is infact empirical formula. Given: Mass of compound = 60 g, mass of carbon = 24 g, mass of hydrogen = 4 g, mass of oxygen = 32 g. \[%\,\text{of}\,C=\frac{\text{mass}\,\text{of}\,\text{carbon}}{\text{mass}\,\text{of}\,\text{compound}}\times 100\] \[=\frac{24}{60}\times 100=40%\] \[%\,\text{of}\,\text{H}\,\text{=}\,\frac{\text{mass}\,\text{of}\,\text{hydrogen}}{\text{mass}\,\text{ofcompound}}\times 100\] \[=\frac{4}{60}\times 100=6.66%\] % of \[O=100-(%\,of\,carbon+\,%\,of\,hydrogen)\] \[=100-(40\,+\,6.66)\] \[=100-46.66=53.34%\]| Element | Percen-tage | Atomic mass | Relative no. of atoms | Ratio |
| C | 40% | 12 | \[\frac{40}{12}=3.33\] | \[\frac{3.33}{3.33}=1\] |
| H | \[6.66%\] | 1 | \[\frac{6.66}{1}=6.66\] | \[\frac{6.66}{3.33}=2\] |
| O | \[53.34%\] | 16 | \[\frac{53.34}{16}=3.33\] | \[\frac{3.33}{3.33}=1\] |
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