In a laboratory quantitative analysis, exactly of an organic compound is subjected to complete combu — Qualitative and Quantitative Analysis Chemistry Question
Question
In a laboratory quantitative analysis, exactly $0.200 \text{ g}$ of an organic compound is subjected to complete combustion. The resulting products are passed through an anhydrous $CaCl_2$ U-tube, which shows a mass increase of exactly $0.144 \text{ g}$. Calculate the percentage of Hydrogen in the organic compound.
Answer: 8
💡 Solution & Explanation
The mass increase of the $CaCl_2$ tube corresponds to the mass of $H_2O$ formed. Mass of $H_2O = 0.144 \text{ g}$. The formula for $\%H$ is: $\%H = \frac{2}{18} \times \frac{\text{Mass of } H_2O}{\text{Mass of compound}} \times 100$. Substituting the values: $\%H = \frac{2}{18} \times \frac{0.144}{0.200} \times 100 = \frac{1}{9} \times 0.72 \times 100 = 8\%$.
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