Three students were asked to find the identity of the metal in a particular sulfate salt. They dissolved a \(0.1472-\mathrm{g}\) sample of the salt in water and treated it with excess barium chloride, resulting in the precipitation of barium sulfate. After the precipitate had been filtered and dried, it weighed \(0.2327 \mathrm{~g}\). Each student analyzed the data independently and came to different conclusions. Pat decided that the metal was titanium. Chris thought it was sodium. Randy reported that it was gallium. What formula did each student assign to the sulfate salt? Look for information on the sulfates of gallium, sodium, and titanium in this text and reference books such as the \(C R C\) Handbook of Chemistry and Physics. What further tests would you suggest to determine which student is most likely correct?

Step by step solution

01

Identify the balanced chemical reactions

Given that Titanium sulfate is Ti(SO₄)2, Sodium sulfate is Na2SO4, and Gallium sulfate is Ga2(SO4)3, let's write the balanced chemical reactions for each sulfate reacting with barium chloride (BaCl2): (a) Ti(SO₄)₂ + 2 BaCl₂ → 2 BaSO₄ + TiCl₄ (b) Na₂SO₄ + BaCl₂ → 2 NaCl + BaSO₄ (c) Ga₂(SO₄)₃ + 3 BaCl₂ → 2 GaCl₃ + 3 BaSO₄

02

Calculate the molar mass of the metal sulfates

Using the periodic table, we can calculate the molar mass of each metal sulfate: (a) Ti(SO₄)₂: Ti = 47.87 g/mol, S = 32.07 g/mol, O = 16.00 g/mol Molar mass = 47.87 + 2(32.07 + 4(16.00)) = 159.91 g/mol (b) Na₂SO₄: Na = 22.99 g/mol, S = 32.07 g/mol, O = 16.00 g/mol Molar mass = 2(22.99) + 32.07 + 4(16.00) = 142.06 g/mol (c) Ga₂(SO₄)₃: Ga = 69.72 g/mol, S = 32.07 g/mol, O = 16.00 g/mol Molar mass = 2(69.72) + 3(32.07 + 4(16.00)) = 427.64 g/mol

03

Calculate moles of sulfate ions

We know the mass of barium sulfate formed: 0.2327 g The molar mass of BaSO₄ is: Ba = 137.33 g/mol, S = 32.07 g/mol, O = 16.00 g/mol Molar mass (BaSO₄) = 137.33 + 32.07 + 4(16.00) = 233.44 g/mol Now we can calculate the moles of sulfate ions present in the precipitation: Moles of sulfate ions = mass of BaSO₄ / molar mass of BaSO₄ = 0.2327 g / 233.44 g/mol = 0.0010 mol

04

Find the moles of the metal ions and their mass

Now let's find the moles of each metal ion in the sulfate salt sample and their corresponding mass: (a) For titanium sulfate: According to the balanced equation, 1 mol of Ti(SO₄)₂ requires 2 mol of sulfate ions. Moles of Ti(SO₄)₂ = 0.0010 mol / 2 = 0.0005 mol Mass of Ti(SO₄)₂ = moles × molar mass = 0.0005 mol × 159.91 g/mol = 0.07996 g (b) For sodium sulfate: According to the balanced equation, 1 mol of Na₂SO₄ requires 1 mol of sulfate ions. Moles of Na₂SO₄ = 0.0010 mol Mass of Na₂SO₄ = moles × molar mass = 0.0010 mol × 142.06 g/mol = 0.14206 g (c) For gallium sulfate: According to the balanced equation, 1 mol of Ga₂(SO₄)₃ requires 3 mol of sulfate ions. Moles of Ga₂(SO₄)₃ = 0.0010 mol / 3 = 0.000333 mol Mass of Ga₂(SO₄)₃ = moles × molar mass = 0.000333 mol × 427.64 g/mol = 0.14254 g

05

Compare the results and suggest further tests

Comparing the calculated mass for each metal sulfate: (a) Mass of Ti(SO₄)₂ = 0.07996 g (b) Mass of Na₂SO₄ = 0.14206 g (c) Mass of Ga₂(SO₄)₃ = 0.14254 g The mass of the original sample was 0.1472 g. The closest match to that mass is the mass of Na₂SO₄ and Ga₂(SO₄)₃. However, we should perform further tests to determine which student is most likely correct since the masses of Na₂SO₄ and Ga₂(SO₄)₃ are too close. A possible test would be to perform a flame test, which would show a distinct color for each metal ion. Sodium ions give a yellow flame, while gallium ions give a colorless/white flame. This test would allow us to determine, with higher certainty, which metal ion is present in the sulfate salt.

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