Question

A 1.42-g sample of a pure compound with formula m2so4 was dissolved in water and treated with an ex- cess of aqueous calcium chloride, resulting in the precipi- tation of all the sulfate ions as calcium sulfate. the pre- cipitate was collected, dried, and found to weigh 1.36 g. determine the atomic mass of m and identify m.

Answer 1

The reaction between $m_{2}SO_{4}$ with calcium chloride can be shown as-$m_{2}SO_{4}$+$CaCl_{2}$→$CaSO_{4}$↓+2mCl. The molecular weight of $CaSO_{4}$ is 136.14g. The weight of sulfate ion is 96.06g. The molecular weight of $m_{2}SO_{4}$ = (2×m + 96.06). From the reaction we can see that 1 mole of calcium chloride reacts with 1 moles$m_{2}SO_{4}$ to produce 1 mole of calcium sulfate. Now 1.36g of calcium sulfate is equivalent to 1.36/136.14=9.989×$10^{-3}$ moles of calcium sulfate.

Thus, 9.989×$10^{-3}$ moles of $m_{2}SO_{4}$ reacts in this reaction.

Let assume the atomic mass of m is x thus the molecular weight of $m_{2}SO_{4}$ is 2x+96.

So we may write 9.989×$10^{-3}$× (2x+96) =1.42

Or, 2x + 96 = 142.146

Or, 2x = 46.146

Or, x = 23.073

Thus the atomic mass of m is 23.073. The atom (m) is sodium (Na).