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An organic compound contains 78 % (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formulā of this compound is: [Atomic wt. of C is H is 1]
- a)CH
- b)CH2
- c)CH3
- d)CH4
Correct answer is option 'C'. Can you explain this answer?
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An organic compound contains 78 % (by wt.) carbon and remaining percen...
Hence, the empirical formula of this compound is CH3.
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An organic compound contains 78 % (by wt.) carbon and remaining percen...
Given:
- An organic compound contains 78% (by wt.) carbon.
- The remaining percentage is hydrogen.
To find:
- The empirical formula of this compound.
Solution:
1. Determining the percentage composition:
- Let's assume we have 100 grams of the compound.
- 78% of 100 grams is carbon, which is equal to 78 grams.
- The remaining 22 grams must be hydrogen.
2. Converting grams to moles:
- The atomic weight of carbon is 12 g/mol, so 78 grams of carbon is equal to 78/12 = 6.5 moles of carbon.
- The atomic weight of hydrogen is 1 g/mol, so 22 grams of hydrogen is equal to 22/1 = 22 moles of hydrogen.
3. Determining the empirical formula:
- The ratio of carbon to hydrogen can be determined by dividing the number of moles of each element by the smallest number of moles.
- In this case, the smallest number of moles is 6.5 moles of carbon.
- Dividing the number of moles of hydrogen by 6.5 gives us 22/6.5 = 3.38 moles of hydrogen.
- Since we can't have fractional moles in a chemical formula, we need to multiply both the carbon and hydrogen by a whole number to get whole number ratios.
- The closest whole number ratio we can get is 2:7 (approximately).
- Therefore, the empirical formula of the compound is CH3.
Conclusion:
- The correct option for the empirical formula of this compound is CH3 (option C).
- An organic compound contains 78% (by wt.) carbon.
- The remaining percentage is hydrogen.
To find:
- The empirical formula of this compound.
Solution:
1. Determining the percentage composition:
- Let's assume we have 100 grams of the compound.
- 78% of 100 grams is carbon, which is equal to 78 grams.
- The remaining 22 grams must be hydrogen.
2. Converting grams to moles:
- The atomic weight of carbon is 12 g/mol, so 78 grams of carbon is equal to 78/12 = 6.5 moles of carbon.
- The atomic weight of hydrogen is 1 g/mol, so 22 grams of hydrogen is equal to 22/1 = 22 moles of hydrogen.
3. Determining the empirical formula:
- The ratio of carbon to hydrogen can be determined by dividing the number of moles of each element by the smallest number of moles.
- In this case, the smallest number of moles is 6.5 moles of carbon.
- Dividing the number of moles of hydrogen by 6.5 gives us 22/6.5 = 3.38 moles of hydrogen.
- Since we can't have fractional moles in a chemical formula, we need to multiply both the carbon and hydrogen by a whole number to get whole number ratios.
- The closest whole number ratio we can get is 2:7 (approximately).
- Therefore, the empirical formula of the compound is CH3.
Conclusion:
- The correct option for the empirical formula of this compound is CH3 (option C).
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An organic compound contains 78 % (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formul of this compound is: [Atomic wt. of C is H is 1]a)CHb)CH2c)CH3d)CH4Correct answer is option 'C'. Can you explain this answer?
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