In Liebig's method, of an organic compound yielded of and of . The percentage of carbon in the compo — Practical Organic Chemistry and Purification Chemistry Question
Question
In Liebig's method, $0.2475\text{ g}$ of an organic compound yielded $0.4950\text{ g}$ of $CO_2$ and $0.2025\text{ g}$ of $H_2O$. The percentage of carbon in the compound is:
Answer: B
💡 Solution & Explanation
$\%C = \frac{12}{44} \times \frac{\text{Mass of } CO_2}{\text{Mass of cmpd}} \times 100 = \frac{12}{44} \times \frac{0.4950}{0.2475} \times 100 = 54.54\%$.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes