Chem Units 2

Molarity Calculation

• Problem Statement: Calculate the molarity of a solution prepared by dissolving 15.0 g of Potassium Permanganate (KMnO₄) in enough water to make a final volume of 250.0 mL.

• Step 1: Calculate Moles of KMnO₄

  • Molar mass of KMnO₄ is calculated as follows:

  • K = 39.10 g/mol

  • Mn = 54.94 g/mol

  • O = 16.00 g/mol × 4

  • Molar mass = 39.10 + 54.94 + (16.00 × 4) = 158.04 g/mol

  • Moles of KMnO₄ = \frac{15.0 ext{ g}}{158.04 ext{ g/mol}} ≈ 0.0950 mol

• Step 2: Calculate Molarity

  • Volume of solution in liters = \frac{250.0 ext{ mL}}{1000 ext{ mL/L}} = 0.250 L

  • Molarity (M) = \frac{moles\; of\; solute}{liters\; of\; solution} = \frac{0.0950 \text{ mol}}{0.250 ext{ L}} = 0.380 M

Mass Percent Calculation

• Problem Statement: Calculate the mass percent of nitrogen in Ammonium Carbonate (NH₄)₂CO₃.

• Step 1: Calculate Molar Mass of (NH₄)₂CO₃

  • N = 14.01 g/mol × 2 = 28.02 g/mol

  • H = 1.01 g/mol × 8 = 8.08 g/mol

  • C = 12.01 g/mol × 1 = 12.01 g/mol

  • O = 16.00 g/mol × 3 = 48.00 g/mol

  • Total molar mass = 28.02 + 8.08 + 12.01 + 48.00 = 96.11 g/mol

• Step 2: Calculate Mass Percent of Nitrogen

  • Mass percent = \frac{mass\; of\; nitrogen}{total\; mass\; of\; compound} \times 100

  • Mass percent = \frac{28.02}{96.11} \times 100 ≈ 29.14%

Percent by Mass of Water in Chromium(III) Nitrate

• Problem Statement: A 5.00 g sample of Chromium(III) Nitrate Nonahydrate [Cr(NO₃)₃·9H₂O] is heated until all the water is removed. The mass of anhydrous Chromium(III) Nitrate remaining is 3.42 g. Calculate the percent by mass of water in the hydrate.

• Step 1: Calculate Mass of Water Lost

  • Mass of water lost = initial mass - mass of anhydrous compound

  • Mass of water lost = 5.00 g - 3.42 g = 1.58 g

• Step 2: Calculate Percent by Mass of Water

  • Percent water = \frac{mass\; of\; water}{mass\; of\; hydrate} \times 100

  • Percent water = \frac{1.58}{5.00} \times 100 = 31.60%

Empirical and Molecular Formula Calculation

• Problem Statement: A compound is found to be 40.00% carbon, 6.71% hydrogen, and 53.29% oxygen. Its molecular mass is approximately three times its empirical mass. Determine both the empirical and molecular formulas.

• Step 1: Convert Percent to Grams (Assume 100 g sample)

  • C: 40.00 g, H: 6.71 g, O: 53.29 g

• Step 2: Convert to Moles

  • Moles of C = \frac{40.00}{12.01} ≈ 3.32 mol

  • Moles of H = \frac{6.71}{1.008} ≈ 6.64 mol

  • Moles of O = \frac{53.29}{16.00} ≈ 3.33 mol

• Step 3: Find Mole Ratio

  • Divide by the smallest number of moles (≈ 3.32)

  • C = 1, H ≈ 2, O ≈ 1

  • Empirical formula = C₁H₂O₁ or CH₂O

• Step 4: Calculate Empirical Formula Mass

  • Molar mass of CH₂O = 12.01 + (2 × 1.008) + 16.00 = 30.03 g/mol

• Step 5: Calculate Molecular Formula

  • Molecular mass = 3 × empirical mass = 3 × 30.03 g/mol = 90.09 g/mol

  • Moles of the compound = \frac{90.09}{30.03} ≈ 3

  • Molecular formula = (CH₂O)₃ = C₃H₆O₃

Balanced Chemical Equation

• Problem Statement: Write a balanced chemical equation, including physical states, for the reaction between aqueous solutions of Ammonium Dichromate and Lead(II) Nitrate.

• Reactants:

  • Ammonium Dichromate (NH₄)₂Cr₂O₇ (aq)

  • Lead(II) Nitrate Pb(NO₃)₂ (aq)

• Products:

  • Lead(II) Dichromate PbCr₂O₇ (s)

  • Aqueous Ammonium Nitrate NH₄NO₃ (aq)

• Balanced Equation:

  • 2(NH₄)₂Cr₂O₇ (aq) + 3Pb(NO₃)₂ (aq) \rightarrow 3PbCr₂O₇ (s) + 6NH₄NO₃ (aq)

Mass of Sulfur Dioxide Produced

• Problem Statement: Given the balanced reaction, if 20.0 g of Iron(II) Sulfide is reacted with excess Oxygen, what mass of Sulfur Dioxide is produced?

• Balanced Reaction:

  • 2FeS + 3O₂ \rightarrow 2Fe + 2SO₂

• Step 1: Calculate Moles of FeS

  • Molar mass of FeS = 55.85 + 32.07 = 87.92 g/mol

  • Moles of FeS = \frac{20.0}{87.92} ≈ 0.227 mol

• Step 2: Using Stoichiometry to Find Moles of SO₂

  • From the balanced equation, 2 moles of FeS produce 2 moles of SO₂

  • Moles of SO₂ = 0.227 mol (same as FeS due to 1:1 ratio)

• Step 3: Calculate Mass of SO₂

  • Molar mass of SO₂ = 32.07 + (16.00 × 2) = 64.07 g/mol

  • Mass of SO₂ = 0.227 \text{ mol} \times 64.07 \text{ g/mol} ≈ 14.53 ext{ g}

Limiting Reactant Identification

• Problem Statement: For the reaction between Barium Hydroxide and Iron(III) Sulfate. If 12.0 g of Barium Hydroxide reacts with 18.0 g of Iron(III) Sulfate, identify the limiting reactant.

• Balanced Reaction:

  • 3Ba(OH)₂ + Fe₂(SO₄)₃ \rightarrow 3BaSO₄ + 2Fe(OH)₃

• Step 1: Calculate Moles of Reactants

  • Molar mass of Ba(OH)₂ = 137.33 + (16.00 × 2) + 1.01 × 2 = 171.34 g/mol

  • Moles of Ba(OH)₂ = \frac{12.0}{171.34} ≈ 0.0700 mol

  • Molar mass of Fe₂(SO₄)₃ = (55.85 × 2) + (32.07) + (16.00 × 12) = 399.88 g/mol

  • Moles of Fe₂(SO₄)₃ = \frac{18.0}{399.88} ≈ 0.0450 mol

• Step 2: Determine Limiting Reactant

  • From the balanced equation:

  • 3 moles of Ba(OH)₂ are required for every 1 mole of Fe₂(SO₄)₃.

  • Moles of Ba(OH)₂ required = 3 × 0.0450 = 0.135 mol

  • Since we only have 0.0700 mol of Ba(OH)₂, it is the limiting reactant.

Net Ionic Equation

• Problem Statement: Write the net ionic equation for the reaction that occurs when aqueous Potassium Phosphate is mixed with aqueous Iron(III) Perchlorate.

• Balanced Molecular Equation:

  • 3K₃PO₄ + 2Fe(ClO₄)₃ \rightarrow Fe₃(PO₄)₂ (s) + 6KClO₄

• Step 1: Write Out the Complete Ionic Equation

  • 3K^+ + 3PO4^{3-} + 2Fe^{3+} + 6ClO4^{-} \rightarrow Fe₃(PO₄)₂ (s) + 6K^+ + 6ClO₄^{-}

• Step 2: Identify Spectator Ions

  • Spectator ions are K⁺ and ClO₄⁻

• Net Ionic Equation:

  • 2Fe^{3+} + 3PO_4^{3-} \rightarrow Fe₃(PO₄)₂ (s)

Percent Yield Calculation

• Problem Statement: For the high-temperature synthesis of Calcium Phosphide. If 15.0 g of Calcium Phosphate is reacted with excess Carbon and 3.50 g of Calcium Phosphide is produced experimentally, calculate the percent yield for this reaction.

• Balanced Reaction:

  • 2Ca₃(PO₄)₂ + 6C \rightarrow 6CaP + 6CO

• Step 1: Calculate Theoretical Yield

  • Calculate moles of Calcium Phosphate

  • Molar mass of Ca₃(PO₄)₂ = 3 × 40.08 + 2 × (30.97 + (16.00 × 4)) = 310.18 g/mol

  • Moles of Calcium Phosphate = \frac{15.0}{310.18} ≈ 0.0485 mol

  • From stoichiometry of the balanced equation, 1 mol of Calcium Phosphate produces 6 mol of Calcium Phosphide:

  • Moles of Calcium Phosphide = 0.0485 mol × 6 ≈ 0.291 mol

• Step 2: Calculate Theoretical Mass of Calcium Phosphide

  • Molar mass of Calcium Phosphide = 40.08 + 30.97 = 71.05 g/mol

  • Theoretical mass of Calcium Phosphide = 0.291 mol × 71.05 g/mol ≈ 20.69 g

• Step 3: Calculate Percent Yield

  • Percent yield = \frac{actual\; yield}{theoretical\; yield} \times 100

  • Percent yield = \frac{3.50 g}{20.69 g} \times 100 ≈ 16.92%

Oxidation Number Determination

• Problem Statement: Determine the oxidation number of bromine in the compound Perbromic Acid and the average oxidation number of sulfur in the polyatomic ion Thiosulfate.

• Perbromic Acid: HBrO₄

  • Assign oxidation numbers:

  • H = +1, O = -2 (4 O atoms contribute -8),

  • Let oxidation number of Br = x:

  • 1 + x + (-8) = 0 \Rightarrow x = +7

• Thiosulfate Ion (S₂O₃²⁻)

  • Assign oxidation numbers:

  • O = -2 (3 O atoms contribute -6), let oxidation number of S (two S) = 2y:

  • 2y + (-6) = -2 \Rightarrow 2y = 4 \Rightarrow y = +2

  • Average oxidation number of sulfur = \frac{+2 + +2}{2} = +2

Identifying Oxidized and Reduced Elements

• Problem Statement: For the unbalanced reaction between Dichromate ion and Iron(II) ion, which produces Chromium(III) ion and Iron(III) ion, identify the element that is oxidized and the element that is reduced.

• Step 1: Assign Oxidation Numbers

  • Dichromate ion: Cr = +6

  • Iron(II) ion: Fe = +2

  • Chromium(III) ion: Cr = +3

  • Iron(III) ion: Fe = +3

• Step 2: Identify Changes in Oxidation Numbers

  • Chromium goes from +6 to +3 (Reduced)

  • Iron goes from +2 to +3 (Oxidized)

• Identified Elements:

  • Oxidized: Iron

  • Reduced: Chromium