4.74 g of an inorganic compound contains 0.39 g of K, 0.27 g of Al, 1.92 g of $SO_4$ radicals and 2.16 g of wa

Question

4.74 g of an inorganic compound contains 0.39 g of K, 0.27 g of Al, 1.92 g of $SO_4$ radicals and 2.16 g of water. If molar mass of the compound is $948 	ext{ g mol}^{-1}$, the molecular formula of the inorganic compound is:

Accepted Answer

1. **Calculate Moles of Constituents:**
   - Moles of K = $\frac{0.39}{39} = 0.01 	ext{ mol}$
   - Moles of Al = $\frac{0.27}{27} = 0.01 	ext{ mol}$
   - Moles of $SO_4$ = $\frac{1.92}{96} = 0.02 	ext{ mol}$
   - Moles of $H_2O$ = $\frac{2.16}{18} = 0.12 	ext{ mol}$

2. **Determine Empirical Formula:**
   Divide by the smallest number of moles (0.01):
   - K : Al : $SO_4$ : $H_2O$ = $1 : 1 : 2 : 12$
   - Empirical Formula = $KAl(SO_4)_2 \cdot 12H_2O$

3. **Calculate Empirical Formula Mass:**
   $$ 	ext{Mass} = 39 + 27 + 2(96) + 12(18) = 66 + 192 + 216 = 474 	ext{ g mol}^{-1} $$

4. **Determine Molecular Formula:**
   - $n = \frac{	ext{Molar Mass}}{	ext{Empirical Mass}} = \frac{948}{474} = 2$
   - Molecular Formula = $2 	imes [KAl(SO_4)_2 \cdot 12H_2O] = K_2Al_2(SO_4)_4 \cdot 24H_2O$
   - This can be rewritten in the standard double salt notation as: $K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$ (Potash Alum).

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