Question

A \(1.42-\mathrm{g}\) sample of a pure compound, with formula \(\mathrm{M}_{2} \mathrm{SO}_{4}\), was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulfate ions as calcium sulfate. The precipitate was collected, dried, and found to weigh \(1.36 \mathrm{~g}\). Determine the atomic mass of \(\mathrm{M}\), and identify \(\mathrm{M}\).

Short answer

The atomic mass of M in the compound M2SO4 is approximately 71 g/mol. By comparing this value with the periodic table, we identify M as Gallium (Ga) with an atomic mass of approximately 69.7 g/mol.

Step-by-step solution

Calculate the moles of sulfate ions in the calcium sulfate precipitate

Since the precipitate is dried, we have anhydrous calcium sulfate (CaSO4), and its molar mass is the sum of the molar mass of calcium, sulfur, and oxygen: Molar mass of CaSO4 = M(Ca) + M(S) + 4 * M(O) = 40.08 + 32.07 + 4 * 16.00 = 136.15 g/mol Given the mass of the calcium sulfate precipitate (1.36 g), we can find the moles of sulfate ions within it by: moles of CaSO4 = mass of CaSO4 / molar mass of CaSO4 = 1.36 g / 136.15 g/mol ≈ 0.01 mol Since there's a 1:1 mole ratio between CaSO4 and SO4^2-, we have 0.01 mol of sulfate ions in the precipitate.

Calculate the moles of M in M2SO4

Each formula unit of M2SO4 corresponds to one sulfate ion, so: moles of M2SO4 = moles of SO4^2- = 0.01 mol Since each M2SO4 formula unit contains two atoms of M, we have: moles of M = 2 * moles of M2SO4 = 2 * 0.01 mol = 0.02 mol

Calculate the atomic mass of M

Now, we can find the atomic mass of M by using the given mass of the compound (1.42 g) and the moles calculated above: atomic mass of M = mass of M / moles of M = 1.42 g / 0.02 mol ≈ 71 g/mol

Identify M

Finally, by comparing the calculated atomic mass of M (71 g/mol) with values from the periodic table, we identify M as Gallium (Ga) with an atomic mass of approximately 69.7 g/mol. Therefore, the unknown metal in the compound M2SO4 is Gallium (Ga) with an atomic mass of approximately 69.7 g/mol.