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Chapter 4: Problem 69

A 1.42 -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 g. Determine the atomic mass of M, and identify M.

Short Answer

Expert verified
The atomic mass of M is 23 g/mol, and the element is sodium (Na). The compound \(\mathrm{M}_{2}\mathrm{SO}_{4}\) is sodium sulfate (Na₂SO₄).

Step by step solution

01

Calculate the moles of sulfate ions in calcium sulfate

The weight of calcium sulfate (CaSO₄) after precipitation is provided in the problem as 1.36 g. To find the moles of sulfate ions, we need to first calculate the molar mass of calcium sulfate. The molar mass of CaSO₄ = 40.08 (Ca) + 32.07 (S) + 4 × 16.00 (O) = 40.08 + 32.07 + 64.00 = 136.15 g/mol Now, we can find the moles of sulfate ions: Moles of CaSO₄ = mass / molar mass = 1.36 g / 136.15 g/mol = 0.01 mol Since there is one sulfate ion in each calcium sulfate molecule, the moles of sulfate ions will be the same as the moles of calcium sulfate, which is 0.01 mol.
02

Calculate the mass of sulfate ions in \(\mathrm{M}_{2}\mathrm{SO}_{4}\)

We know that the moles of sulfate ions in \(\mathrm{M}_{2}\mathrm{SO}_{4}\) are also 0.01 mol. We can calculate the mass of SO₄²⁻ ions in the compound by multiplying the moles with the molar mass of sulfate ions. Molar mass of SO₄²⁻ = 32.07 (S) + 4 × 16.00 (O) = 32.07 + 64.00 = 96.07 g/mol Mass of SO₄²⁻ in \(\mathrm{M}_{2}\mathrm{SO}_{4} = \text{moles} \times \text{molar mass}\) = (0.01 mol) × (96.07 g/mol) = 0.96 g
03

Calculate the mass of M in \(\mathrm{M}_{2}\mathrm{SO}_{4}\)

The given initial mass of \(\mathrm{M}_{2}\mathrm{SO}_{4}\) is 1.42 g. We have calculated the mass of SO₄²⁻ ions as 0.96 g. We can now find the mass of M in \(\mathrm{M}_{2}\mathrm{SO}_{4}\) Mass of M in \(\mathrm{M}_{2}\mathrm{SO}_{4} = \text{total mass} - \text{mass of SO₄²⁻ ions}\) = 1.42 g - 0.96 g = 0.46 g
04

Calculate the atomic mass of M and identify the element

We know that there are two M atoms in \(\mathrm{M}_{2}\mathrm{SO}_{4}\). We can find the atomic mass of M by dividing the mass of M by the moles of M atoms. Moles of M atoms = 2 × moles of SO₄²⁻ ions = 2 × 0.01 mol = 0.02 mol Atomic mass of M = mass of M / moles of M = 0.46 g / 0.02 mol = 23 g/mol The element with an atomic mass of approximately 23 g/mol is sodium (Na). Therefore, the compound \(\mathrm{M}_{2}\mathrm{SO}_{4}\) is sodium sulfate (Na₂SO₄).

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Most popular questions from this chapter

Chlorisondamine chloride $\left(\mathrm{C}_{14} \mathrm{H}_{20} \mathrm{Cl}_{6} \mathrm{N}_{2}\right)$ is a drug used in the treatment of hypertension. A 1.28-g sample of a medication containing the drug was treated to destroy the organic material and to release all the chlorine as chloride ion. When the filtered solution containing chloride ion was treated with an excess of silver nitrate, 0.104 g silver chloride was recovered. Calculate the mass percent of chlorisondamine chloride in the medication, assuming the drug is the only source of chloride.

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