A sample of the compound MSO4 weighing 0.1131 g reacts with BaCl2 and yields 0.2193 g BaSO4. What is the elemental mass of M and its identity? Hint: All the SO42 - from the MSO4 appears in the BaSO4

The atomic mass of the given metal M is 24.3 g/mol.

Therefore, the given metal M is magnesium (Mg)

Explanation:

Reaction involved: MSO₄ + BaCl₂ → BaSO₄ + MCl₂

Atomic mass (g/mol) : oxygen (O) = 16, sulphur (S) = 32,

Molar mass of SO₄²⁻ = 96 g/mol and molar mass of BaSO₄ = 233.38 g/mol

Let the atomic mass of M be m g/mol.

Therefore, molar mass of MSO₄ = (m + 96) g/mol

If 0.1131 g of MSO₄ gives 0.2193 g of BaSO₄ on reaction

Then, (m + 96) g/mol of MSO₄ gives 233.38 g/mol of BaSO₄

Therefore, (m + 96) g/mol of MSO₄ = [ (233.38 g/mol) * (0.1131 g) ] : (0.2193 g)

⇒ (m + 96) g/mol = [26.395] : (0.2193)

⇒ (m + 96) g/mol = 120.36 g/mol

⇒m = 120.36 g/mol - 96 g/mol = 24.36 g/mol ≈ 24.3 g/mol

Since, the atomic mass of the given metal M is 24.3 g/mol.

Therefore, the given metal M is magnesium (Mg)