When 100 mg sample of calcium oxalate monohydrate (Mol. wt. 146.12 g) was heated at the rate of 10°C/min in air, loss of 7.0 mg was noted at onset temperature of 140°C corresponding to water. The percentage (w/w) of calcium oxalate monohydrate in the original sample is
56.82%
In thermogravimetric analysis, CaC2O4·H2O loses its one molecule of water of crystallisation in the first step near 100–200 °C. The mass of water lost therefore fixes the amount of hydrate present: one mole of water comes from one mole of hydrate.
Step 1 — Moles of water lost: \(n = \dfrac{7.0\ \text{mg}}{18.0\ \text{mg mmol}^{-1}} = 0.3889\ \text{mmol}\).
Step 2 — Mass of hydrate: \(0.3889 \times 146.12 = 56.82\ \text{mg}\).
Step 3 — Percentage: \(\dfrac{56.82}{100} \times 100 = 56.82\%\) (w/w).
The other percentages come from rounding the ratio 146/18 differently or from using the anhydrous oxalate mass; only the stoichiometric hydrate calculation with water as 18.0 g mol−1 gives the value obtained here.
56.82% option is correct.
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