GATE 2024 CY – Question 60

Inorganic Chemistry · Instrumental Methods of Analysis: UV-visible, fluorescence and FT-IR spectrophotometry, NMR and ESR spectroscopy, mass spectrometry, atomic absorption spectroscopy, Mössbauer spectroscopy (Fe and Sn) and X- ray crystallography. Electroanalytical methods – cyclic voltammetry, ion-selective electrodes. Thermo-analytical methods: TGA, DTA and DSC. · 2 marks · Numerical answer

In thermogravimetric analysis, 12.45 mg of $\mathrm{CuSO_4\cdot5H_2O}$ was heated under nitrogen. At a particular temperature the weight loss was 3.6 mg. The water molecules lost per formula unit is _____ (nearest integer). Given molar masses in g/mol: H = 1.0, O = 16.0, S = 32.0, Cu = 63.5.

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Correct answer: 4

Explanation

**Step 1: molar mass of the hydrate** $\text{CuSO}_4\cdot5\text{H}_2\text{O}$:
$$63.5+32.0+4(16.0)+5(18.0)=63.5+32+64+90=249.5\text{ g/mol}.$$

**Step 2: moles of the hydrate heated.**
$$\frac{12.45\text{ mg}}{249.5\text{ g/mol}}=0.0499\text{ mmol}.$$

**Step 3: moles of water lost.** The mass loss of 3.6 mg is water ($M=18$):
$$\frac{3.6\text{ mg}}{18\text{ g/mol}}=0.200\text{ mmol}.$$

**Step 4: water per formula unit.**
$$\frac{0.200}{0.0499}=4.008\approx\mathbf{4}.$$

(The compound has lost four of its five waters at this temperature, forming $\text{CuSO}_4\cdot\text{H}_2\text{O}$.)