Stoichiometry

Balancing & Using Chemical Equations

5th Year · 6th Year (Leaving Cert)

✓Balance chemical equations by inspection to satisfy the Law of Conservation of Mass.
✓Use mole ratios derived from balanced equations to perform stoichiometric calculations.
✓Identify the limiting reagent in a chemical reaction and use it to determine the theoretical yield.
✓Calculate the percentage yield of a reaction given the actual yield and theoretical yield.
✓Apply stoichiometric principles to solve quantitative problems involving mass and moles.

Key concepts

Balancing Chemical Equations

Chemical equations represent chemical reactions. According to the Law of Conservation of Mass, atoms are neither created nor destroyed in a chemical reaction; they are merely rearranged. Therefore, the number of atoms of each element must be the same on both sides of a chemical equation (reactants and products). Balancing by inspection involves adjusting the stoichiometric coefficients (the numbers in front of the chemical formulae) until the number of atoms of each element is equal on both sides.

Mole Ratio

The stoichiometric coefficients in a balanced chemical equation not only represent the relative number of molecules or formula units but also the relative number of moles of each reactant and product involved in the reaction. This ratio is crucial for converting between moles of different substances in a reaction. For example, in the reaction 2H₂(g) + O₂(g) → 2H₂O(l), the mole ratio of H₂ to O₂ is 2:1, and the mole ratio of H₂ to H₂O is 2:2 (or 1:1).

Limiting Reagent (or Limiting Reactant)

In most chemical reactions, reactants are not present in exactly the stoichiometric amounts required by the balanced equation. The limiting reagent is the reactant that is completely consumed first in a chemical reaction, thereby stopping the reaction and limiting the amount of product that can be formed. All calculations for the theoretical yield of products must be based on the amount of the limiting reagent. The other reactant(s) are said to be in excess.

Theoretical Yield

The theoretical yield is the maximum amount of product that can be formed from a given amount of reactants, assuming the reaction goes to completion and there are no losses. It is calculated using the stoichiometry of the balanced chemical equation and the amount of the limiting reagent.

Actual Yield

The actual yield is the amount of product that is actually obtained from a chemical reaction when carried out in the laboratory or industrially. It is almost always less than the theoretical yield due to various factors such as incomplete reactions, side reactions, loss of product during purification, or experimental errors.

Percentage Yield

The percentage yield is a measure of the efficiency of a chemical reaction. It compares the actual yield obtained in an experiment to the theoretical yield calculated from stoichiometry, expressed as a percentage. A higher percentage yield indicates a more efficient reaction.

Percentage Yield = (Actual Yield / Theoretical Yield) × 100%

Key facts to remember

1Chemical equations must be balanced to obey the Law of Conservation of Mass.
2Stoichiometric coefficients in a balanced equation represent mole ratios.
3The limiting reagent determines the maximum amount of product that can be formed.
4Theoretical yield is the calculated maximum product; actual yield is the experimentally obtained product.
5Percentage yield quantifies the efficiency of a reaction: (Actual Yield / Theoretical Yield) × 100%.
6Molar mass (M) is the mass of one mole of a substance, expressed in g/mol.
7Moles = Mass / Molar Mass.

Worked examples

Example 1

Balance the following chemical equation and determine the moles of oxygen gas required to react completely with 0.50 mol of propane (C₃H₈). C₃H₈(g) + O₂(g) → CO₂(g) + H₂O(l)

I1. Balance Carbon atoms: There are 3 C atoms on the left, so place a '3' in front of CO₂ on the right. C₃H₈(g) + O₂(g) → 3CO₂(g) + H₂O(l)

II2. Balance Hydrogen atoms: There are 8 H atoms on the left, so place a '4' in front of H₂O on the right (4 × 2 = 8). C₃H₈(g) + O₂(g) → 3CO₂(g) + 4H₂O(l)

III3. Balance Oxygen atoms: Count O atoms on the right: (3 × 2) + (4 × 1) = 6 + 4 = 10 O atoms. So, place a '5' in front of O₂ on the left (5 × 2 = 10). C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

IV4. Verify all atoms are balanced: Left: C=3, H=8, O=10 Right: C=3, H=8, O=(3×2)+(4×1)=10. The equation is balanced.

V5. Determine the mole ratio from the balanced equation: From the balanced equation, 1 mole of C₃H₈ reacts with 5 moles of O₂. Mole ratio C₃H₈ : O₂ = 1 : 5

VI6. Calculate moles of O₂ required: Moles of O₂ = Moles of C₃H₈ × (5 mol O₂ / 1 mol C₃H₈) Moles of O₂ = 0.50 mol C₃H₈ × 5 = 2.50 mol O₂

Answer

The balanced equation is C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l). 2.50 mol of oxygen gas is required.

Always balance C and H first, then O, and finally any other elements.

Example 2

If 10.0 g of hydrogen gas (H₂) reacts with 70.0 g of oxygen gas (O₂) to produce water (H₂O), what is the limiting reagent and what is the theoretical yield of water in grams? 2H₂(g) + O₂(g) → 2H₂O(l)

I1. Calculate the molar masses: M(H₂) = 2 × 1.0 g/mol = 2.0 g/mol M(O₂) = 2 × 16.0 g/mol = 32.0 g/mol M(H₂O) = (2 × 1.0) + 16.0 = 18.0 g/mol

II2. Convert masses of reactants to moles: Moles of H₂ = Mass / Molar Mass = 10.0 g / 2.0 g/mol = 5.0 mol Moles of O₂ = Mass / Molar Mass = 70.0 g / 32.0 g/mol = 2.1875 mol

III3. Identify the limiting reagent using the mole ratio from the balanced equation (2H₂ : 1O₂): * Method: Calculate how much O₂ is needed for 5.0 mol H₂. Moles of O₂ needed = 5.0 mol H₂ × (1 mol O₂ / 2 mol H₂) = 2.5 mol O₂ We only have 2.1875 mol O₂, which is less than 2.5 mol O₂ needed. Therefore, O₂ is the limiting reagent.

IV4. Calculate the theoretical yield of H₂O based on the limiting reagent (O₂): From the balanced equation, 1 mol O₂ produces 2 mol H₂O. Moles of H₂O produced = Moles of O₂ (limiting) × (2 mol H₂O / 1 mol O₂) Moles of H₂O produced = 2.1875 mol O₂ × 2 = 4.375 mol H₂O

V5. Convert moles of H₂O to mass: Mass of H₂O = Moles × Molar Mass = 4.375 mol × 18.0 g/mol = 78.75 g

Answer

The limiting reagent is oxygen (O₂). The theoretical yield of water is 78.8 g (to 3 significant figures).

Always base theoretical yield calculations on the limiting reagent, not the reactant in excess.

Example 3

In an experiment, 13.0 g of ethyne (C₂H₂) reacts with excess hydrogen (H₂) to produce ethane (C₂H₆). If the actual yield of ethane obtained was 14.5 g, calculate the percentage yield. C₂H₂(g) + 2H₂(g) → C₂H₆(g)

I1. Calculate molar masses: M(C₂H₂) = (2 × 12.0) + (2 × 1.0) = 26.0 g/mol M(C₂H₆) = (2 × 12.0) + (6 × 1.0) = 30.0 g/mol

II2. Since H₂ is in excess, C₂H₂ is the limiting reagent. Calculate moles of C₂H₂: Moles of C₂H₂ = Mass / Molar Mass = 13.0 g / 26.0 g/mol = 0.50 mol

III3. Calculate the theoretical yield of C₂H₆ in moles using the mole ratio from the balanced equation (1C₂H₂ : 1C₂H₆): Moles of C₂H₆ (theoretical) = Moles of C₂H₂ × (1 mol C₂H₆ / 1 mol C₂H₂) Moles of C₂H₆ (theoretical) = 0.50 mol × 1 = 0.50 mol

IV4. Convert theoretical moles of C₂H₆ to mass (theoretical yield): Theoretical Yield of C₂H₆ = Moles × Molar Mass = 0.50 mol × 30.0 g/mol = 15.0 g

V5. Calculate the percentage yield: Percentage Yield = (Actual Yield / Theoretical Yield) × 100% Percentage Yield = (14.5 g / 15.0 g) × 100% = 96.666...%

Answer

The percentage yield is 96.7% (to 3 significant figures).

Percentage yield can never be greater than 100%. If it is, there's likely an error in measurement or calculation, or impurities are present.

Common mistakes

✗Failing to balance the chemical equation correctly before performing calculations.
✗Not identifying the limiting reagent when given amounts of more than one reactant, leading to incorrect theoretical yield calculations.
✗Confusing actual yield with theoretical yield.
✗Using incorrect molar masses in calculations.
✗Not paying attention to significant figures in final answers.

Exam tips

★Always start by writing down the balanced chemical equation for the reaction.
★Clearly show all steps in your calculations, including units, to maximise marks, even if you make a numerical error.
★When identifying the limiting reagent, calculate the amount of product that *could* be formed from *each* reactant, or calculate how much of one reactant is *needed* for the other. The reactant that produces less product (or is fully consumed) is the limiting reagent.
★Remember that percentage yield is always based on the theoretical yield, which is calculated from the limiting reagent.

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