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Standard XII

Chemistry

Empirical & Molecular Formula

A 0.60 g samp...

Question

A $0.60 g$ sample consisting of only $C a C_{2} O_{4}$ and $M g C_{2} O_{4}$ is heated at $500^{o} C$ converting the two salts to $C a C O_{3}$ and $M g C O_{3}$ . The sample then weight $0.465 g$ . If the sample had been heated to $900^{o} C$ , where the products are $C a O$ and $M g O$ , what would the mixtures of oxide have weighed (in grams)?
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Solution

Let $x g$ is the weight of $C a C_{2} O_{4}$ and $(0.6 - x) g$ is the weight of $M g C_{2} O_{4}$
$C a C_{2} O_{4} Δ - --- \to C a C O_{3} + C O$
$M g C_{2} O_{4} Δ - --- \to M g C O_{3} + C O$
$Weight of C a C O_{3} produced = \frac{x}{128} \times 100$

$Weight of M g C O_{3} produced = \frac{(0.6 - x)}{112} \times 84$
$∴ \frac{x}{128} \times 100 + \frac{(0.6 - x)}{112} \times 84 = 0.465$
$x = 0.48 g$
Due to further heating,
$\begin{matrix} C a C O_{3} & Δ - ---- \to & C a O & + & C O_{2} \frac{x}{128} \end{matrix}$
$\begin{matrix} M g C O_{3} & Δ - ----- \to & M g O & + & C O_{2} (\frac{0.6 - x}{112}) \end{matrix}$
Weight of $C a O$ and $M g O$ produced,
$= \frac{0.48}{128} \times 56 + \frac{0.12}{112} \times 40 = 0.252 g$

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0.60 g

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M g C_{2} O_{4}

is heated at

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converting the two salts of

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