Empirical Formula From Percentage Composition
Elemental analysis of an organic compound gives carbon 40.0 percent, hydrogen 6.7 percent, and oxygen 53.3 percent by mass. What is the empirical formula of this compound?
Select the correct option:
Solution
CH2O
Deriving an empirical formula from percentage composition requires converting each mass percentage into moles by dividing by the atomic mass, then finding the simplest whole-number ratio. Taking 100 g of compound, the moles are: carbon 40.0/12 = 3.33, hydrogen 6.7/1 = 6.7, and oxygen 53.3/16 = 3.33. Dividing each by the smallest value, 3.33, gives carbon 1, hydrogen 2, and oxygen 1, so the empirical formula is CH_2O. Option CH_3O is wrong because the hydrogen-to-carbon mole ratio is 2, not 3, when computed correctly. Option C_2H_4O is incorrect because the oxygen ratio relative to carbon must remain 1:1, not 1:2. Option CHO is wrong because it understates the hydrogen content, which the data clearly fix at twice the carbon. This mole-ratio procedure is the standard NCERT route from composition to formula. It is worth noting that the empirical formula gives only the simplest ratio of atoms, not the actual molecular formula; the molecular formula is a whole-number multiple of the empirical unit and can be found only if the molar mass of the compound is also known. Here CH_2O is the empirical unit of carbohydrates such as glucose, whose molecular formula C_6H_12O_6 is six times this unit. Plausibility check: the empirical formula CH_2O has a formula mass of 30, and the back-calculated composition, carbon 40%, hydrogen 6.7%, oxygen 53.3%, matches the given data exactly, confirming the result.
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About This Question
- Subject
- chemistry
- Chapter
- principles related to practical chemistry
- Topic
- empirical formula from percentage composition
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
CH2O
Deriving an empirical formula from percentage composition requires converting each mass percentage into moles by dividing by the atomic mass, then finding the simplest whole-number ratio. Taking 100 g of compound, the moles are: carbon 40.0/12 = 3.33, hydrogen 6.7/1 = 6.7, and oxygen 53.3/16 = 3.33. Dividing each by the smallest value, 3.33, gives carbon 1, hydrogen 2, and oxygen 1, so the empirical formula is CH_2O. Option CH_3O is wrong because the hydrogen-to-carbon mole ratio is 2, not 3, when computed correctly. Option C_2H_4O is incorrect because the oxygen ratio relative to carbon must remain 1:1, not 1:2. Option CHO is wrong because it understates the hydrogen content, which the data clearly fix at twice the carbon. This mole-ratio procedure is the standard NCERT route from composition to formula. It is worth noting that the empirical formula gives only the simplest ratio of atoms, not the actual molecular formula; the molecular formula is a whole-number multiple of the empirical unit and can be found only if the molar mass of the compound is also known. Here CH_2O is the empirical unit of carbohydrates such as glucose, whose molecular formula C_6H_12O_6 is six times this unit. Plausibility check: the empirical formula CH_2O has a formula mass of 30, and the back-calculated composition, carbon 40%, hydrogen 6.7%, oxygen 53.3%, matches the given data exactly, confirming the result.
This medium difficulty chemistry question is from the chapter principles related to practical chemistry, covering the topic of empirical formula from percentage composition. It appeared in the 2025 exam.
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