teaching-notes — Chemistry (THE MOLE CONCEPT)

Topic: THE MOLE CONCEPT

Subtopic: The mole

AND CALCULATE THE PERCENTAGE YIELD IN A REACTION AND THE PERCENTAGE PURITY OF A SUBSTANCE

Meaning of percentage yield

Percentage yield tells us how much product was actually obtained in a reaction compared with the amount that should be obtained if the reaction went perfectly.

Formula: Percentage yield = actual amount obtainedtheoretical amount × 100%

To use this formula you often first calculate the theoretical amount from moles and relative formula mass (M_r).

Example 1 (simple percentage yield): A chemist expects to make 25.0 g of product but actually obtains 20.0 g. Find the percentage yield.

Substitution: Percentage yield = 20.0 g25.0 g × 100%

Answer: 80.0%

Worked example (calculating theoretical yield then % yield): 10.0 g magnesium (Mg) reacts with excess hydrochloric acid to form magnesium chloride (MgCl₂). The experimental mass of MgCl₂ collected is 35.0 g. Calculate the percentage yield. (M_r(Mg) = 24.3, M_r(MgCl₂) = 95.2)

Step 1: Calculate moles of Mg used.

Formula: moles of Mg = mass of MgM_r of Mg

Substitution: moles of Mg = 10.0 g24.3

Answer: 0.412 mol (to three significant figures)

Step 2: Use equation and mole ratio. Word equation: Magnesium + Hydrochloric acid → Magnesium chloride + Hydrogen. Balanced chemical equation: Mg + 2HCl → MgCl₂ + H₂. From the equation, 1 mol Mg gives 1 mol MgCl₂, so moles of MgCl₂ theoretical = 0.412 mol.

Step 3: Calculate theoretical mass of MgCl₂.

Formula: mass = moles × M_r

Substitution: mass = 0.412 × 95.2

Answer (theoretical mass): 39.2 g

Step 4: Percentage yield.

Formula: Percentage yield = actual amount obtainedtheoretical amount × 100%

Substitution: Percentage yield = 35.0 g39.2 g × 100%

Answer: 89.3%

Meaning of percentage purity

Percentage purity tells how much of a sample is the pure substance and how much is impurity.

Formula: Percentage purity = mass of pure substance in sampletotal mass of sample × 100%

Example: A 10.0 g salt sample contains 8.0 g of pure NaCl. Percentage purity = 8.0 g10.0 g × 100% = 80.0%

simple labelled pupil-notebook diagram of MOLE CONVERSION FL...

LIMITING REAGENT IN A GIVEN REACTION

Meaning

The limiting reagent (limiting reactant) is the reactant that runs out first and so stops the reaction. The other reactant(s) are in excess.

Method (using stoichiometric masses and given quantities)

1. Write a balanced chemical equation. 2. Convert masses of reactants to moles using moles = mass ÷ M_r. 3. Use mole ratios to find which reactant is used up first. 4. The reactant that gives the smaller amount of product is the limiting reagent.

Worked example 1 (using moles): 3.0 g H₂ is mixed with 32.0 g O₂. Which is limiting in the reaction 2H₂ + O₂ → 2H₂O? (M_r(H₂) = 2.0, M_r(O₂) = 32.0, M_r(H₂O)=18.0)

Step 1: Moles of H₂.

Formula: moles = massM_r

Substitution: moles H₂ = 3.0 g2.0

Answer: 1.50 mol H₂

Step 2: Moles of O₂.

Substitution: moles O₂ = 32.0 g32.0

Answer: 1.00 mol O₂

Step 3: Use stoichiometric ratio. Equation requires 2 mol H₂ per 1 mol O₂. For 1.00 mol O₂ we would need 2.00 mol H₂, but only 1.50 mol H₂ is present. Therefore H₂ is the limiting reagent and O₂ is in excess.

Step 4: Calculate mass of H₂O produced. Mole ratio H₂ : H₂O is 1 : 1 (from 2H₂ → 2H₂O). So moles H₂O = 1.50 mol.

Mass of H₂O:

Formula: mass = moles × M_r

Substitution: mass = 1.50 × 18.0

Answer: 27.0 g H₂O

Worked example 2 (using proportional stoichiometric masses): Use the same reaction 2H₂ + O₂ → 2H₂O. We know 4.0 g H₂ would react with 32.0 g O₂ (because 2 mol H₂ = 4.0 g and 1 mol O₂ = 32.0 g). If we actually have 3.0 g H₂, the O₂ required would be:

Calculation: required O₂ = 32.0 × 3.04.0 = 24.0 g

Given 32.0 g O₂ is available, O₂ is in excess and H₂ is limiting (same conclusion as before).

simple labelled pupil-notebook diagram of LIMITING REAGENT C...

CALCULATIONS INVOLVING DIFFERENT TYPES OF ACID–BASE TITRATION REACTIONS

Titration law (general form)

For acid–base titrations where different numbers of protons are exchanged, use the general titration law:

n₁M₁V₁ = n₂M₂V₂

Where n is the number of H⁺ or OH^- ions each molecule provides (for example n = 1 for HCl, n = 2 for H₂SO₄), M is concentration (mol dm^-3), and V is volume (use the same units for V₁ and V₂, e.g. cm³ or dm³).

Monoprotic acid with monoprotic base (simple case)

Example 1: 25.0 cm³ of HCl is titrated with 20.0 cm³ of 0.100 mol dm^-3 NaOH. Find the concentration of HCl. Both HCl and NaOH are monoprotic so n₁=n₂=1 and the formula becomes M₁V₁ = M₂V₂.

Formula: M₁ = M₂V₂V₁

Substitution: M₁ = 0.100 × 20.025.0

Answer: 0.0800 mol dm^-3

Diprotic acid with monoprotic base

Example 2: 25.0 cm³ of 0.100 mol dm^-3 H₂SO₄ (diprotic, n₁=2) is titrated with NaOH (monoprotic, n₂=1). The volume of NaOH used is 40.0 cm³. Find the concentration of NaOH (M₂).

Formula: M₂ = n₁M₁V₁n₂V₂

Substitution: M₂ = 2 × 0.100 × 25.01 × 40.0

Answer: 0.125 mol dm^-3

Note on units: If volumes are in cm³ you may use them directly in the formula because the same units cancel on both sides. Always give final concentration in mol dm^-3.

simple labelled pupil-notebook diagram of TITRATION APPARATU...

Worked example (finding amount of acid from titration result): 20.0 cm³ of 0.200 mol dm^-3 HCl is neutralised by 25.0 cm³ of NaOH. Both are monoprotic. Find the concentration of NaOH.

Formula: M₁V₁ = M₂V₂

Substitution: 0.200 × 20.0 = M₂ × 25.0

Rearrange: M₂ = 0.200 × 20.025.0

Answer: 0.160 mol dm^-3

APPLICATION TIP: Always identify n values before substituting. For acids: HCl, HNO₃ have n = 1; H₂SO₄ has n = 2. For bases: NaOH has n = 1, Ca(OH)₂ has n = 2.

SUMMARY

• Percentage yield = actual ÷ theoretical × 100%. You often calculate theoretical from moles and M_r.
• Percentage purity = mass of pure substance ÷ total mass of sample × 100%.
• Limiting reagent is found by converting reactant masses to moles and comparing with stoichiometric ratios; the smaller amount of product indicates the limiting reagent.
• Titration law (general): n₁M₁V₁ = n₂M₂V₂. Use correct n values for mono- or polyprotic acids/bases.

REVISION QUESTIONS

1. Define percentage yield and write its formula.
2. How is percentage purity calculated? Give the formula.
3. Explain how to find the limiting reagent when given masses of two reactants.
4. Calculate the percentage yield if the theoretical yield is 50.0 g and the actual yield is 42.0 g.
5. 20.0 g of A reacts with excess B to give 30.0 g of product. If the theoretical mass of product is 35.0 g, find the percentage yield.
6. State the titration law for a diprotic acid reacting with a monoprotic base.
7. 25.0 cm³ of acid required 30.0 cm³ of 0.150 mol dm^-3 NaOH for neutralisation. If the acid is monoprotic, find its concentration.
8. Why is it important to know the limiting reagent when calculating yield?

PRACTICE EXERCISE

1. A sample of impure CaCO₃ weighing 12.0 g contains 9.60 g pure CaCO₃. Calculate the percentage purity.
2. In a reaction, the theoretical yield of a product is 80.0 g but the actual yield is 62.4 g. Calculate the percentage yield.
3. Given 5.00 g of Al reacts with 14.0 g of O₂ in the reaction 4Al + 3O₂ → 2Al₂O₃, determine the limiting reagent. (M_r(Al)=27.0, M_r(O₂)=32.0)
4. 25.0 cm³ of unknown HCl required 18.0 cm³ of 0.100 mol dm^-3 NaOH for neutralisation. Find the concentration of HCl.
5. 20.0 cm³ of 0.0500 mol dm^-3 H₂SO₄ is titrated with NaOH. Volume of NaOH used is 40.0 cm³. Find the concentration of NaOH.
6. A chemist reacted 15.0 g Fe with excess HCl and obtained 20.0 g of FeCl₂. Theoretical mass of FeCl₂ is 24.5 g. Calculate the percentage yield.

ANSWERS TO PRACTICE EXERCISE

1. Work: Percentage purity = 9.60 g12.0 g × 100% = 80.0%
2. Work: Percentage yield = 62.4 g80.0 g × 100% = 78.0%
3. Work: moles Al = 5.00 g27.0

   = 0.185 mol Al. moles O₂ = 14.0 g32.0 = 0.438 mol O₂.

   Equation ratio: 4 mol Al react with 3 mol O₂. For 0.185 mol Al required O₂ = 3 × 0.1854 = 0.139 mol O₂. Available O₂ = 0.438 mol, so Al is limiting reagent.

4. Work: M₁V₁ = M₂V₂. M(HCl) = 0.100 × 18.025.0 = 0.0720 mol dm^-3.
5. Work: n₁=2 for H₂SO₄, n₂=1 for NaOH. M₂ = 2 × 0.0500 × 20.01 × 40.0 = 2.0040.0 = 0.0500 mol dm^-3.
6. Work: Percentage yield = 20.0 g24.5 g × 100% = 81.6%

END OF NOTES