Stoichiometry Problems Worksheet - Free Printable

Let's solve each problem step-by-step from the Stoichiometry Problems Worksheet. I'll provide detailed explanations and calculations for each.

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Problem 1:

Haloalkane is an anesthetic that is 12.17% C, 0.51% H, 40.48% Br, 17.96% Cl and 28.87% F by mass. What is the compound’s molar mass if each molecule contains exactly one hydrogen atom?

#### Step 1: Assume 100 g of compound
So we have:
- C: 12.17 g
- H: 0.51 g
- Br: 40.48 g
- Cl: 17.96 g
- F: 28.87 g

#### Step 2: Convert to moles
Use atomic masses:
- C: 12.01 g/mol
- H: 1.008 g/mol
- Br: 79.90 g/mol
- Cl: 35.45 g/mol
- F: 19.00 g/mol

Now calculate moles:

- C: $ \frac{12.17}{12.01} = 1.013 $ mol
- H: $ \frac{0.51}{1.008} = 0.506 $ mol
- Br: $ \frac{40.48}{79.90} = 0.507 $ mol
- Cl: $ \frac{17.96}{35.45} = 0.507 $ mol
- F: $ \frac{28.87}{19.00} = 1.519 $ mol

#### Step 3: Find simplest ratio
Divide all by the smallest value (≈0.506):

- C: $ \frac{1.013}{0.506} ≈ 2 $
- H: $ \frac{0.506}{0.506} = 1 $
- Br: $ \frac{0.507}{0.506} ≈ 1 $
- Cl: $ \frac{0.507}{0.506} ≈ 1 $
- F: $ \frac{1.519}{0.506} ≈ 3 $

So empirical formula: C₂HBrClF₃

#### Step 4: Molar mass of empirical formula
Calculate:
- C₂: 2 × 12.01 = 24.02
- H: 1.008
- Br: 79.90
- Cl: 35.45
- F₃: 3 × 19.00 = 57.00
Total = 24.02 + 1.008 + 79.90 + 35.45 + 57.00 = 197.378 g/mol

But the problem says each molecule contains exactly one hydrogen atom, so this empirical formula already has one H — so it must be the actual molecular formula.

✔ Therefore, molar mass = 197.4 g/mol (approximately).

> ✔ Answer: 197.4 g/mol

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Problem 2:

A compound that is 31.9% K and 28.9% Cl by mass decomposes when heated to give O₂ and a compound that is 52.4% K and 47.6% Cl by mass. Write a balanced chemical equation for this reaction.

#### Step 1: Identify the original compound
Original: 31.9% K, 28.9% Cl → Remaining % = 100 - 31.9 - 28.9 = 39.2% → likely oxygen

Assume 100 g sample:
- K: 31.9 g → $ \frac{31.9}{39.1} ≈ 0.816 $ mol
- Cl: 28.9 g → $ \frac{28.9}{35.45} ≈ 0.815 $ mol
- O: 39.2 g → $ \frac{39.2}{16.00} = 2.45 $ mol

Ratio: K : Cl : O ≈ 0.816 : 0.815 : 2.45 → divide by 0.815:
→ ≈ 1 : 1 : 3

So formula: KClO₃ (potassium chlorate)

#### Step 2: Product compound
After decomposition: 52.4% K, 47.6% Cl → no O?

Check: 52.4 + 47.6 = 100% → so no oxygen → must be KCl

So: KClO₃ → KCl + O₂

Balance:
$$
2KClO_3 \rightarrow 2KCl + 3O_2
$$

✔ Balanced equation: 2KClO₃ → 2KCl + 3O₂

> ✔ Answer: 2KClO₃ → 2KCl + 3O₂

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Problem 3:

A compound that combines in a fixed amount with one or more molecules of water is known as a hydrate. In lab, a 5.00-g sample of hydrated barium chloride, BaCl₂·xH₂O, is heated to drive off the water. After heating, 4.26 g of anhydrous barium chloride, BaCl₂, remains. What is the value of x in the hydrate’s formula?

#### Step 1: Mass of water lost
Water mass = 5.00 g – 4.26 g = 0.74 g

#### Step 2: Moles of BaCl₂
Molar mass BaCl₂ = 137.3 (Ba) + 2×35.45 = 137.3 + 70.9 = 208.2 g/mol

Moles of BaCl₂ = $ \frac{4.26}{208.2} = 0.02046 $ mol

#### Step 3: Moles of H₂O
Molar mass H₂O = 18.02 g/mol

Moles H₂O = $ \frac{0.74}{18.02} = 0.04107 $ mol

#### Step 4: Ratio of H₂O to BaCl₂
$ x = \frac{0.04107}{0.02046} ≈ 2.007 ≈ 2 $

So, x = 2

✔ Formula: BaCl₂·2H₂O

> ✔ Answer: x = 2

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Problem 4:

The oxygen-carrying protein known as hemoglobin is 0.335% Fe by mass and contains exactly four Fe atoms per hemoglobin molecule. Calculate the protein’s molar mass.

#### Step 1: Let M = molar mass of hemoglobin (g/mol)

Each molecule has 4 Fe atoms → mass of Fe in one mole = 4 × 55.85 = 223.4 g

This is 0.335% of the total mass:

$$
\frac{223.4}{M} = \frac{0.335}{100}
$$

Solve:
$$
M = \frac{223.4 \times 100}{0.335} = \frac{22340}{0.335} ≈ 66686.6 \text{ g/mol}
$$

Rounded to appropriate sig figs: 6.67 × 10⁴ g/mol

> ✔ Answer: 6.67 × 10⁴ g/mol

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Problem 5:

A 2.50-g sample of bronze, an alloy of copper and tin, was dissolved in sulfuric acid. The copper in the alloy reacts with sulfuric acid as shown by the following balanced reaction:

$$
\text{Cu} + 2H_2SO_4 \rightarrow Cu^{2+} + SO_2 + 2H_2O + SO_4^{2-}
$$

Then:
$$
2Cu^{2+} + 5I^- \rightarrow 2CuI + I_3^-
$$

And titration:
$$
I_3^- + 2S_2O_3^{2-} \rightarrow 3I^- + S_4O_6^{2-}
$$

31.5 mL of 1.00 M S₂O₃²⁻ is consumed in the titration. Calculate the percentage, by mass, of copper in the original sample.

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#### Step 1: Moles of S₂O₃²⁻ used
Volume = 31.5 mL = 0.0315 L
Concentration = 1.00 M
Moles = 0.0315 × 1.00 = 0.0315 mol

From the titration reaction:
$$
I_3^- + 2S_2O_3^{2-} \rightarrow 3I^- + S_4O_6^{2-}
$$

So, 2 mol S₂O₃²⁻ ≡ 1 mol I₃⁻
→ Moles of I₃⁻ = $ \frac{0.0315}{2} = 0.01575 $ mol

From earlier reaction:
$$
2Cu^{2+} + 5I^- \rightarrow 2CuI + I_3^-
$$

So, 1 mol I₃⁻ comes from 2 mol Cu²⁺
→ Moles of Cu²⁺ = 2 × 0.01575 = 0.0315 mol

#### Step 2: Mass of Cu
Molar mass Cu = 63.55 g/mol
Mass of Cu = 0.0315 × 63.55 ≈ 2.00 g

#### Step 3: Percentage by mass
Sample mass = 2.50 g
% Cu = $ \frac{2.00}{2.50} \times 100 = 80.0\% $

> ✔ Answer: 80.0%

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✔ Final Answers Summary:

1. 197.4 g/mol
2. 2KClO₃ → 2KCl + 3O₂
3. x = 2
4. 6.67 × 10⁴ g/mol
5. 80.0%

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