Unit 7: Stoichiometry

Moles · Molar Mass · Limiting Reagents · Percent Yield · Molarity · Dilution · Solution Stoichiometry

⚗️ Concept Overview

The Mole Map

The mole is the central unit connecting everything in stoichiometry. Use the mole map to convert between any two quantities:

Particles
(atoms, molecules, ions)
÷ 6.022×10²³
────
× 6.022×10²³
Moles
× molar mass
────
÷ molar mass
Grams
use mole ratio
from equation
Moles of
other substance
  • Avogadro's number: 6.022 × 10²³ particles/mol
  • Molar mass = sum of atomic masses from periodic table (g/mol)
  • Example: H₂O = 2(1.008) + 16.00 = 18.02 g/mol

7.4–7.5 — Mole Ratios & Stoichiometry

  • A mole ratio comes from the coefficients of a balanced equation.
  • Example: 4 Al + 3 O₂ → 2 Al₂O₃ → mole ratio Al:Al₂O₃ = 4:2 = 2:1
  • Stoichiometry road map: given grams → moles given → moles wanted → grams wanted

7.6–7.7 — Limiting Reagent & Excess

  • The limiting reagent is used up first and determines the maximum product formed.
  • The excess reactant is the one that remains after reaction is complete.
  • To find limiting reagent: convert each reactant to moles of product — whichever gives less product is limiting.
  • Leftover excess = (initial moles excess) − (moles excess actually used)

7.8 — Percent Yield

% Yield = (actual yield / theoretical yield) × 100%
  • Theoretical yield: calculated from stoichiometry (maximum possible)
  • Actual (experimental) yield: what you measured in the lab
  • % Yield ≤ 100% in real experiments (losses due to side reactions, spilling, incomplete reaction)

7.9–7.13 — Molarity & Solutions

Molarity (M) = moles of solute / liters of solution
Dilution: C₁V₁ = C₂V₂
  • Solute: substance dissolved; Solvent: substance doing the dissolving (usually water)
  • To prepare a solution: weigh out calculated grams, dissolve, then dilute to final volume
  • Ion concentration: multiply molarity by the subscript of the ion in the formula
    Example: 0.5 M CaCl₂ → [Ca²⁺] = 0.5 M; [Cl⁻] = 2 × 0.5 = 1.0 M
  • Dilution (C₁V₁ = C₂V₂): moles of solute stay constant — you just add more solvent

7.14 — Solution Stoichiometry

Combine molarity with stoichiometry: use M × V (in liters) to find moles, then use mole ratios as usual.

moles = M × V(L)

🃏 Flashcards

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Mole
A counting unit equal to 6.022 × 10²³ particles (atoms, molecules, ions). 1 mole of any substance contains Avogadro's number of particles.
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Molar Mass
The mass of one mole of a substance in grams. Found by summing atomic masses from the periodic table. Units: g/mol. Example: NaCl = 22.99 + 35.45 = 58.44 g/mol.
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Avogadro's Number
6.022 × 10²³ — the number of particles in exactly one mole. Used to convert between moles and individual particles (atoms, molecules, ions).
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Mole Ratio
The ratio of moles of one substance to moles of another from a balanced chemical equation. Derived from coefficients. Used to convert moles of one substance to moles of another.
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Limiting Reagent
The reactant that is completely consumed first and limits how much product can form. Identified by calculating moles of product from each reactant — the one giving LESS product is limiting.
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Excess Reactant
The reactant that remains after the limiting reagent is used up. Some of it is left over. Leftover = initial moles − moles actually consumed (via mole ratio with limiting reagent).
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Theoretical Yield
The maximum amount of product calculated from stoichiometry assuming complete reaction of the limiting reagent. The "best-case" result.
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Actual (Experimental) Yield
The amount of product actually collected in a lab experiment. Always ≤ theoretical yield due to losses, side reactions, or incomplete reaction.
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Percent Yield
(actual yield ÷ theoretical yield) × 100%. Measures efficiency of a reaction. Always ≤ 100% in real lab settings.
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Molarity
Concentration unit: moles of solute per liter of solution. M = mol/L. A 1.0 M solution has 1.0 mol of solute dissolved in enough solvent to make 1.0 L of solution.
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Solute vs. Solvent
Solute: the substance dissolved (smaller amount). Solvent: the substance that does the dissolving (larger amount, usually water). Together they form a solution.
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Dilution (C₁V₁ = C₂V₂)
Adding solvent to decrease concentration. Moles of solute are conserved: C₁V₁ = C₂V₂. C = molarity, V = volume (same units on both sides).
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Ion Concentration
In ionic solutions, multiply molarity by the subscript of each ion. Example: 0.30 M Al(NO₃)₃ → [Al³⁺] = 0.30 M; [NO₃⁻] = 3 × 0.30 = 0.90 M.
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Solution Stoichiometry
Using molarity in stoichiometric calculations. Key step: moles = M × V(L). Then use mole ratios from the balanced equation as usual.
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Stoichiometry Road Map
Given grams → ÷ molar mass → moles given → × mole ratio → moles wanted → × molar mass → grams wanted. Every step requires the balanced equation.
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Coefficient vs. Subscript
Coefficient (in front of formula): mole ratio for stoichiometry. Subscript (inside formula): atom count for molar mass. In 3H₂O — coefficient=3 (3 mol water); subscript=2 (2 H per molecule).
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Preparing a Solution
1) Calculate grams needed using M × V × molar mass. 2) Weigh out grams of solute. 3) Dissolve in small amount of solvent. 4) Transfer to volumetric flask. 5) Add solvent to the calibration mark.

⚡ Quick Review

📝 Practice Quiz

20 multiple choice + 3 short answer.

1. How many moles are in 58.44 g of NaCl? (Molar mass NaCl = 58.44 g/mol)

2. What is the molar mass of Ca(OH)₂? (Ca = 40.08, O = 16.00, H = 1.008)

3. Given: 4 Al + 3 O₂ → 2 Al₂O₃. How many moles of Al₂O₃ are produced from 12 moles of Al (excess O₂)?

4. For the reaction S₈ + 8 O₂ → 8 SO₂, how many grams of SO₂ can be produced from 2.00 g of S₈? (MM: S₈ = 256.5 g/mol, SO₂ = 64.07 g/mol)

5. For 2 H₂ + O₂ → 2 H₂O, you have 2.00 mol H₂ and 2.30 mol O₂. Which is the limiting reagent?

6. A student obtains 4.50 g of a product. Stoichiometry predicts 5.00 g. What is the percent yield?

7. How many molecules are in 3.00 moles of CO₂?

8. 60.0 g of NaOH (MM = 40.00 g/mol) is dissolved in enough water to make 500.0 mL of solution. What is the molarity?

9. What volume of 2.00 M NaCl is needed to prepare 500. mL of 0.500 M NaCl via dilution?

10. What is the concentration of NO₃⁻ in a 0.40 M Mg(NO₃)₂ solution?

11. One mole of CO₂ contains:

12. If 200 mL of 1.60 M NaOH is diluted to 350 mL, the new concentration is:

13. In an experiment, 3 moles of X reacted with 2 moles of Y to produce 5 moles of L and 1 mole of B. The correctly balanced equation is:

14. What mass of water is produced when 4.00 mol of H₂ reacts with excess O₂? (2H₂ + O₂ → 2H₂O; MM H₂O = 18.02 g/mol)

15. Which species is NOT present in a 0.1 M NaCl solution?

16. The correct method for preparing 0.500 L of 0.20 M Na₂SO₄ (MM = 142.0 g/mol) is:

17. For the reaction: Zn + 2HCl → ZnCl₂ + H₂. If you react 25.0 g Zn (MM = 65.38) with 0.25 mol HCl, what is the limiting reagent and max grams of H₂ produced? (MM H₂ = 2.016)

18. When 1 mole of Al(NO₃)₃ dissolves, the concentration of NO₃⁻ in 1 L of solution is:

19. How many moles of Ca²⁺ are in 250.0 mL of 0.300 M CaCl₂?

20. For the reaction H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, how many grams of Na₂SO₄ (MM = 142.0 g/mol) are produced when 25.0 mL of 0.250 M H₂SO₄ reacts completely?

Short Answer Questions

SA1. For the reaction N₂ + 3H₂ → 2NH₃: if you start with 5.00 g of N₂ (MM = 28.02 g/mol) and 2.00 g of H₂ (MM = 2.016 g/mol), (a) identify the limiting reagent, (b) calculate the theoretical yield of NH₃ in grams (MM NH₃ = 17.03 g/mol), and (c) if 2.80 g of NH₃ is actually collected, what is the percent yield?

(a) mol N₂ = 5.00/28.02 = 0.1785 mol; mol H₂ = 2.00/2.016 = 0.9921 mol
From N₂: 0.1785 mol N₂ × (2 mol NH₃/1 mol N₂) = 0.3570 mol NH₃
From H₂: 0.9921 mol H₂ × (2 mol NH₃/3 mol H₂) = 0.6614 mol NH₃
N₂ is the limiting reagent (produces less product).

(b) Theoretical yield = 0.3570 mol × 17.03 g/mol = 6.08 g NH₃

(c) % Yield = 2.80/6.08 × 100% = 46.1%

SA2. You have a stock solution of 2.50 M NaOH. (a) How many mL of this stock solution do you need to prepare 500. mL of 0.100 M NaOH? (b) Describe the physical steps to make this solution in the lab.

(a) C₁V₁ = C₂V₂ → (2.50 M)(V₁) = (0.100 M)(500. mL)
V₁ = 50.0/2.50 = 20.0 mL

(b) Measure out 20.0 mL of the 2.50 M NaOH stock solution using a pipette or graduated cylinder. Add it to a 500. mL volumetric flask that already contains about 250 mL of distilled water. Swirl to mix. Then add distilled water up to the 500. mL calibration mark, stopper, and invert several times to mix thoroughly.

SA3. A solution contains 0.250 M H₂SO₄. If 25.0 mL of this solution reacts with excess NaOH: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. (a) How many moles of H₂SO₄ are present? (b) How many grams of Na₂SO₄ (MM = 142.0 g/mol) are produced?

(a) mol H₂SO₄ = M × V = 0.250 mol/L × 0.0250 L = 6.25 × 10⁻³ mol

(b) 6.25 × 10⁻³ mol H₂SO₄ × (1 mol Na₂SO₄ / 1 mol H₂SO₄) = 6.25 × 10⁻³ mol Na₂SO₄
g = 6.25 × 10⁻³ mol × 142.0 g/mol = 0.888 g Na₂SO₄