1.10. Additional Practice Questions

1. A compound contains 40.00 % sulfur (S) and 60.00 % oxygen (O) by mass.
   (a) Determine its empirical formula.
   (b) If the molar mass is known to be approximately 160.06 g/mol, find the molecular formula.

To find the empirical formula from the percentage composition, we first need to assume we have 100g of the compound. That means we have 40g of sulfur and 60g of oxygen. We convert these masses to moles:

• Moles of S = 40g ÷ 32.07 g/mol (molar mass of S) = 1.25 mol
• Moles of O = 60g ÷ 16.00 g/mol (molar mass of O) = 3.75 mol
• Next, we divide by the smallest number of moles to find the ratio:
• Ratio of S: 1.25 mol ÷ 1.25 = 1
• Ratio of O: 3.75 mol ÷ 1.25 = 3
• Hence, the empirical formula is SO₃.
  For the molecular formula, you divide the molar mass of the compound by the molar mass of the empirical formula. The molar mass of SO₃ = 32.07 + (3 × 16.00) = 80.07 g/mol.
• 160.06 g/mol ÷ 80.07 g/mol = 2, indicating the molecular formula is (SO₃)₂ or S₂O₆.

Think of the empirical formula like a recipe that tells you the simplest whole number ratio of ingredients. For instance, if you're making a dish and you find that you have 1 cup of rice (S) for every 3 cups of water (O), your empirical 'recipe' is for rice, but to serve more, you multiply everything to get the total amount needed. Just like in cooking, you'll want to adjust your ingredients depending on how many people you're serving.

2. Balance the following redox reaction in acidic solution using the half‐reaction method:
   MnO4−(aq)+SO32−(aq) ⟶ Mn2+(aq)+SO42−(aq).

Balancing redox reactions involves separating the oxidation and reduction processes. First, we identify the oxidation states of each element in MnO₄⁻ and SO₃²⁻. The Mn in MnO₄⁻ is reduced to Mn²⁺ (gains electrons), while S in SO₃²⁻ is oxidized to SO₄²⁻ (loses electrons).

• Half-reaction for reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (balanced in acidic medium).
• Half-reaction for oxidation: SO₃²⁻ → SO₄²⁻ + 2e⁻.
  To combine them, multiply the oxidation half-reaction by 5 to equalize the number of electrons.
  The balanced equation then looks like this:
  MnO₄⁻ + 5SO₃²⁻ + 8H⁺ → Mn²⁺ + 5SO₄²⁻ + 4H₂O.

Imagine a bank where some people are depositing money and others are withdrawing. You have to ensure that for every person that deposits (reduces), there’s an equivalent number of withdrawals (oxidizes). Just like keeping track of money, you must keep track of the electrons to ensure everything adds up correctly in your balanced reaction.

3. Sodium carbonate (Na₂CO₃) decomposes upon heating to produce sodium oxide (Na₂O) and carbon dioxide (CO₂):
   Na2CO3(s) ⟶ Na2O(s)+CO2(g).
   (a) If 10.0 g of sodium carbonate decomposes completely, calculate the volume of CO₂ produced at 50.0 °C and 1.00 atm.
   (b) If only 7.80 g CO₂ is collected, calculate the percent yield.

To determine the volume of CO₂ produced, we start by converting grams of sodium carbonate to moles:

• Molar mass of Na₂CO₃ = (2 × 22.99) + 12.01 + (3 × 16.00) = 105.99 g/mol.
• Moles of Na₂CO₃ = 10.0 g ÷ 105.99 g/mol = 0.0943 mol.
  For the balanced equation, 1 mol of Na₂CO₃ produces 1 mol of CO₂, so:
• Moles of CO₂ produced = 0.0943 mol.
  Next, we apply the ideal gas law: PV = nRT. Rearranging gives V = nRT/P.
  Using R = 0.0821 L·atm/(K·mol), T = 323.15 K, we find the volume of CO₂ at the conditions specified. Plugging in gives:
• V = (0.0943 mol)(0.0821 L·atm/(K·mol))(323.15 K) / (1 atm) = 2.38 L.
  For percent yield, percent yield = (actual yield/theoretical yield) × 100%, which leads to
• Percent yield = (7.80 g / (0.0943 mol × 44.01 g/mol)) × 100% = 70.36%.

Think of baking a cake. The recipe yields a certain number of slices (theoretical yield), but if some slices break or are eaten by a baker before serving, the total you can serve (actual yield) is less. Calculating the percent yield is like saying, 'I expected to bake 10 slices but only served 7—the waste was 3 slices!'

4. A stock solution of 2.50 M glucose (C₆H₁₂O₆) is used to prepare 2.00 L of a 0.500 M solution.
   (a) How many milliliters of stock must be taken?
   (b) What mass of glucose is present in the final 2.00 L solution?

To find the required volume of stock for dilution, we apply the dilution formula: C₁V₁ = C₂V₂. Here, C₁ = 2.50 M, C₂ = 0.500 M, and V₂ = 2.00 L. Rearranging gives V₁ = (C₂ × V₂) / C₁ = (0.500 M × 2.00 L) / 2.50 M = 0.400 L = 400 mL.
For mass, use the final concentration: M = moles/volume. Find moles:

• Moles = C₂ × V₂ = 0.500 mol/L × 2.00 L = 1.00 mol.
• Mass of glucose = 1 mol × 180.18 g/mol = 180.18 g.

Consider making diluted juice from a concentrated syrup. If the syrup is very strong, you wouldn’t use the same quantity you'd drink—it would be too overpowering! Instead, you measure how much syrup is needed for your preferred taste, just like using the dilution formula to ensure that your final solution has the correct concentration.

5. A 0.150 M solution of hydrochloric acid is prepared by diluting a 6.00 M stock.
   (a) Calculate the volume of stock required to make 1.00 L of 0.150 M HCl.
   (b) If the student performs the dilution and obtains 1.15 L of solution by inexact technique, what is the actual concentration of HCl in the prepared solution?

Using the dilution formula C₁V₁ = C₂V₂ for part (a):
Let C₁ = 6.00 M, C₂ = 0.150 M, and V₂ = 1.00 L. Rearranging gives V₁ = (C₂ × V₂) / C₁ = (0.150 M × 1.00 L) / 6.00 M = 0.0250 L or 25.0 mL.
For part (b), calculate the actual concentration from the new total volume:

• Moles of HCl in 0.150 M × 1.00 L = 0.150 mol. Divide by the new volume 1.15 L: Actual concentration = 0.150 mol/1.15 L = 0.130 M.

Imagine you’re trying to make a punch. If the recipe calls for a specific amount of a concentrate (the stock solution), and you accidentally add more water than needed, the finished punch won’t have the same flavor as intended. The dilution formula helps you manage those proportions correctly to hit just the right level of sweetness or acidity.

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