1. Critically assess why the percentage yield in a reaction is rarely 100% and discuss the factors that might contribute to a lower yield.
Answer: In chemical synthesis, the actual amount of product obtained (actual yield) is almost always less than the maximum amount predicted by stoichiometry (theoretical yield). The percentage yield is calculated as (Actual Yield / Theoretical Yield) × 100. A yield of 100% is rarely achieved due to several practical and theoretical factors:
1. Reversible Reactions and Equilibrium:
Many chemical reactions are reversible. As the reaction proceeds, the products can react to reform the reactants. Eventually, a state of dynamic equilibrium is reached where both forward and reverse reactions occur at equal rates. Unless the equilibrium is shifted (e.g., by removing product or adding excess reactant), the reaction will not go to completion, limiting the yield.
2. Side Reactions and By-products:
Reactants may undergo competing side reactions to form unwanted by-products instead of the desired product. For example, in organic synthesis, a reactant might react at a different site on the molecule, or further reactions might occur
(over-reaction). These side reactions consume reactants and reduce the amount available for the main reaction.
3. Incomplete Reactions:
Even in irreversible reactions, the reaction may not proceed to completion due to kinetic limitations. If the reaction time is insufficient, or if the temperature and pressure are not optimal, some reactants may remain unreacted.
4. Loss of Product During Isolation and Purification:
The process of separating the product from the reaction mixture (e.g., filtration, distillation, extraction) often results in some loss of product. For instance, some liquid may remain in the distillation flask, or some solid may be lost during filtration or recrystallization. Transferring liquids between containers inevitably leaves residue behind.
5. Purity of Reactants:
If the starting materials are impure, the actual amount of the reactive substance is less than assumed. This means the limiting reactant is present in a smaller amount than calculated, leading to a lower theoretical yield based on the impure mass.
6. Human Error:
Errors in measurement (weighing, volume), inaccurate temperature control, or contamination can also lead to lower yields.
In summary, factors such as equilibrium, side reactions, incomplete reactions, and losses during purification mean that the actual yield is almost always lower than the theoretical yield.
2. A solution has a concentration of 0.5 mol/dm³. Propose a method to double this concentration to 1.0 mol/dm³.
Answer:
To double the concentration of a solution from 0.5 mol/dm³ to 1.0 mol/dm³, we need to increase the amount of solute per unit volume of solution. There are two main methods to achieve this:
Method 1: Evaporation of Solvent (Removing Water):
Principle: Concentration is the amount of solute divided by the volume of the solution. If we remove solvent (water) without removing solute, the volume decreases, and the concentration increases. Procedure:
A. Place the 0.5 mol/dm³ solution in an open container.
B. Heat gently to evaporate the water. The solute (e.g., NaCl) will not evaporate.
C. Monitor the volume carefully. To double the concentration, the final volume must be exactly half of the initial volume.
D. For example, if you start with 1 dm³ of 0.5 mol/dm³ solution, you must evaporate water until the volume is reduced to 0.5 dm³.
Verification: New Concentration = Initial Moles / New Volume = 0.5 moles / 0.5 dm³ = 1.0 mol/dm³.
Method 2: Addition of More Solute (Dissolving more solute):
Principle: If we add more solute to the existing solution without significantly changing the volume, the concentration increases. Procedure:
A. Take a known volume of the 0.5 mol/dm³ solution (e.g., 1 dm³).
B. Calculate the mass of solute needed to increase the moles from 0.5 to 1.0.
C. If the solute is NaCl (Molar mass = 58.5 g/mol), the additional moles needed = 1.0 - 0.5 = 0.5 moles. Mass to add = 0.5 ×
58.5 = 29.25 g.
D. Add the 29.25 g of NaCl to the 1 dm³ solution and stir until completely dissolved.
Verification: Total moles = Initial moles + Added moles = 0.5 + 0.5 = 1.0 moles. Volume remains approximately 1 dm³. New
Concentration = 1.0 mol/dm³.
Conclusion: The simplest method depends on the required accuracy and the nature of the solute. Evaporation is often used if the solute is heat-stable and the volume can be accurately monitored. Adding more solute is simpler but requires knowledge of the solute's molar mass and assumes the volume does not change significantly upon dissolution.
3. Calculate the percentage yield of a sample of sodium carbonate (Na₂CO₃) if 1.0 g of the sample requires 25.0 cm³ of 0.1 mol/dm³ hydrochloric acid for complete reaction. Na₂CO₃ + 2HCl [source symbol unavailable] 2NaCl + CO₂ + H₂O
Answer:
Step 1: Calculate Moles of HCl used.
Concentration of HCl = 0.1 mol/dm³
Volume of HCl = 25.0 cm³ = 0.025 dm³
Moles of HCl = 0.1 × 0.025 = 0.0025 moles.
Step 2: Calculate Moles of Na₂CO₃ that reacted.
From the equation, 2 moles of HCl react with 1 mole of Na₂CO₃.
Moles of Na₂CO₃ = 0.00252 = 0.00125 moles.
Step 3: Calculate the Mass of Pure Na₂CO₃ that reacted.
Molar mass of Na₂CO₃ = 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106 g/mol.
Mass = Moles × Molar Mass = 0.00125 × 106 = 0.1325 g.
Step 4: Calculate the Percentage Purity of the sample.
Percentage Purity = (Mass of Pure Na₂CO₃ / Mass of Impure Sample) × 100
Percentage Purity = (0.13251.0) × 100 = 13.25%.
(Note: The calculated purity is quite low. If the question intended the percentage yield of a reaction producing Na₂CO₃, the calculation would be different. However, based on the wording "percentage yield of a sample," percentage purity is the correct interpretation.)
4. If a solution has a concentration of 2 mol/dm³, what is its concentration in g/dm³ if the solute is H₂SO₄? (Answer: 196 g/dm³)
Answer:
Given: Concentration = 2 mol/dm³
Molar Mass of H₂SO₄: 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 g/mol.
Conversion Formula: Concentration (g/dm³) = Concentration (mol/dm³) × Molar Mass (g/mol)
Calculation: 2 mol/dm³ × 98 g/mol = 196 g/dm³.
5. Design a method to determine the concentration of acetic acid in vinegar using titration.
Answer: Objective: To determine the concentration of acetic acid (CH₃COOH) in a commercial vinegar sample using titration with a standard sodium hydroxide (NaOH) solution.
Materials:
Vinegar sample Standard NaOH solution (e.g., 0.1 mol/dm³) Phenolphthalein indicator Burette, pipette, conical flask, retort stand.
Procedure:
1. Preparation: Rinse the burette with distilled water and then with the standard NaOH solution. Fill the burette with NaOH and record the initial reading.
2. Pipetting: Rinse a clean pipette with the vinegar sample. Pipette exactly 25.0 cm³ of vinegar into a clean conical flask.
3. Adding Indicator: Add 2-3 drops of phenolphthalein indicator to the flask. The solution remains colorless.
4. Titration: Slowly add NaOH from the burette into the flask while swirling. The endpoint is reached when a faint pink color persists for at least 30 seconds.
5. Recording: Record the final burette reading. The volume of NaOH used is the difference between final and initial readings.
6. Repeat: Repeat the titration at least three times to obtain concordant results (within 0.1 cm³).
Calculation:
Equation: CH₃COOH + NaOH [source symbol unavailable] CH₃COONa + H₂O Moles of NaOH: Concentration of NaOH × Volume of NaOH used.
Moles of CH₃COOH: From the 1:1 stoichiometry, moles of CH₃COOH = moles of NaOH.
Concentration of Vinegar: Concentration = Moles of CH₃COOH / Volume of vinegar used (in dm³).
Conclusion: This method accurately determines the concentration of acetic acid in vinegar by neutralizing it with a base of known concentration.
CONCEPT ASSESSMENT EXERCISE 3.1
NH₃ is an important raw material in the manufacture of fertilisers. It is obtained by the combination of N₂ and H₂ as shown by the following balanced equation. N₂(g) + 3H₂(g) [source symbol unavailable] 2NH₃(g) How many moles of the following are required to manufacture 8.0 moles of NH₃?
(a) Nitrogen (b) Hydrogen
Answer:
From the balanced equation, 1 mole of N₂ produces 2 moles of NH₃, and 3 moles of H₂ produce 2 moles of NH₃.
(a) Nitrogen: To produce 8.0 moles of NH₃, moles of N₂ required = (12) × 8.0 = 4.0 moles.
(b) Hydrogen: To produce 8.0 moles of NH₃, moles of H₂ required = (32) × 8.0 = 12.0 moles.
CONCEPT ASSESSMENT EXERCISE 3.2
1. Hydrogen burns with oxygen according to the following equation. 2H₂(g) + O₂(g) [source symbol unavailable] 2H₂O(g) How many grams of oxygen are needed to burn 100 g of hydrogen completely?
Answer:
Molar mass of H₂ = 2 g/mol.
Moles of H₂ = 1002 = 50 moles. From equation, 2 moles H₂ need 1 mole O₂.
Moles of O₂ needed = 502 = 25 moles.
Molar mass of O₂ = 32 g/mol.
Mass of O₂ = 25 × 32 = 800 g.
CONCEPT ASSESSMENT EXERCISE 3.3
(a) How many moles of oxygen molecule are there in 50.0 dm³ of oxygen gas at RTP?
Answer: At RTP, 1 mole of gas occupies 24 dm³.
Moles = 50.024 = 2.08 moles.
(b) What volume does 0.80 mole of N₂ gas occupy at RTP?
Answer: Volume = 0.80 × 24 = 19.2 dm³.
CONCEPT ASSESSMENT EXERCISE 3.4
Determine the percentage composition of the following compounds.
1. Water H₂O
Answer: Molar mass = 18 g/mol. %H = (218)×100 = 11.11%; %O = (1618)×100 = 88.89%.
2. Sulphuric acid H₂SO₄
Answer: Molar mass = 98 g/mol. %H = (298)×100 = 2.04%; %S = (3298)×100 = 32.65%; %O = (6498)×100 = 65.31%.
3. Glucose C₆H₁₂O₆
Answer: Molar mass = 180 g/mol. %C = (72180)×100 = 40%; %H = (12180)×100 = 6.67%; %O = (96180)×100 = 53.33%.
4. Ammonium nitrate NH₄NO₃
Answer: Molar mass = 80 g/mol. %N = (2880)×100 = 35%; %H = (480)×100 = 5%; %O = (4880)×100 = 60%.
CONCEPT ASSESSMENT EXERCISE 3.5
1. Magnesium metal reacts with Sulphur to produce MgS. How many grams of magnesium sulphide (MgS) can be made from 1.50g of Mg and 1.50g of sulphur by the reaction? Mg(s) + S(s) [source symbol unavailable] MgS(s)
Answer:
Moles of Mg = 1.5024 = 0.0625 moles.
Moles of S = 1.5032 = 0.0469 moles. From equation, 1 mole Mg needs 1 mole S. S is the limiting reactant.
Moles of MgS = Moles of S = 0.0469 moles.
Molar mass of MgS = 24 + 32 = 56 g/mol.
Mass of MgS = 0.0469 × 56 = 2.63 g.
2. Zinc and Sulphur react to form Zinc Sulphide according to the following balanced chemical equation: Zn(s) + S(s) [source symbol unavailable] ZnS(s). If 10.0g of Zinc and 10.0g of Sulphur are available for reaction, then determine (a) The limiting reactant. (b) The mass of Zinc Sulphide produced.
Answer:
Moles of Zn = 10.065.4 = 0.153 moles.
Moles of S = 10.032 = 0.3125 moles. From equation, 1 mole Zn needs 1 mole S. Zn is the limiting reactant.
(a) Limiting Reactant = Zinc (Zn).
Moles of ZnS = Moles of Zn = 0.153 moles.
Molar mass of ZnS = 65.4 + 32 = 97.4 g/mol.
(b) Mass of ZnS = 0.153 × 97.4 = 14.9 g.
3. Aluminium reacts with bromine to form Aluminium bromide, as shown by the balanced chemical equation, 2Al(s) + 3Br₂(l) [source symbol unavailable] 2AlBr₃(s). If 54g of Al and 200g of Br₂ are available for reaction, then determine (a) The limiting reactant (b) The mass of AlBr₃ produced
Answer:
Moles of Al = 5427 = 2 moles.
Moles of Br₂ = 200160 = 1.25 moles.
From equation, 2 moles Al need 3 moles Br₂. For 2 moles Al, Br₂ needed = 3 moles. Available Br₂ = 1.25 moles, so Br₂ is the limiting reactant.
(a) Limiting Reactant = Bromine (Br₂).
Moles of AlBr₃ produced = (23) × 1.25 = 0.833 moles.
Molar mass of AlBr₃ = 27 + 3(80) = 267 g/mol.
(b) Mass of AlBr₃ = 0.833 × 267 = 222.4 g.
CONCEPT ASSESSMENT EXERCISE 3.6
1. In a reaction, 2.00 moles of CH₄ were reacted with an excess of Cl₂. As a result, 177.0 g of CCl₄ is obtained. What is the (a) theoretical yield (b) actual yield (c) % yield of this reaction? CH₄(g) + 4Cl₂(g) [source symbol unavailable] CCl₄(g) + 4HCl(g)
Answer:
From equation, 1 mole CH₄ produces 1 mole CCl₄.
Moles of CCl₄ = 2.00 moles.
Molar mass of CCl₄ = 12 + 4(35.5) = 154 g/mol.
(a) Theoretical Yield = 2.00 × 154 = 308 g.
(b) Actual Yield = 177.0 g.
(c) % Yield = (177.0308) × 100 = 57.5%.
2. Phosphorus reacts with bromine to form phosphorus tribromide. When 35g of bromine is reacted with 27.9g of phosphorus tribromide is produced. What is the (a) theoretical yield (b) actual yield (c) % yield of this reaction? 2P(s) + 3Br₂(l) [source symbol unavailable] 2PBr₃(l)
Answer:
Moles of Br₂ = 35160 = 0.219 moles. From equation, 3 moles Br₂ produce 2 moles PBr₃.
Moles of PBr₃ = (23) × 0.219 = 0.146 moles.
Molar mass of PBr₃ = 31 + 3(80) = 271 g/mol.
(a) Theoretical Yield = 0.146 × 271 = 39.6 g.
(b) Actual Yield = 27.9 g.
(c) % Yield = (27.939.6) × 100 = 70.45%.
CONCEPT ASSESSMENT EXERCISE 3.7
1. Determine the empirical and molecular of a compound from the following data. 26.4 % Carbon 3.3 % Hydrogen 70.3 % Oxygen, Molar Mass: 91.0 g/mol
Answer:
Moles: C = 26.412 = 2.2; H = 3.31 = 3.3; O = 70.316 = 4.39.
Divide by smallest (2.2): C = 1; H = 1.5; O = 2.
Multiply by 2 to get whole numbers: C = 2; H = 3; O = 4.
Empirical Formula = C₂H₃O₄.
Empirical Mass = 2(12) + 3(1) + 4(16) = 24 + 3 + 64 = 91 g/mol.
n = 9191 = 1.
Molecular Formula = C₂H₃O₄.
2. Determine the empirical and molecular formula of a compound composed of 18.24 g Carbon, 0.51 g Hydrogen, and 16.91 g Fluorine, which has a molar mass 562.0 g/mol.
Answer:
Moles: C = 18.2412 = 1.52; H = 0.511 = 0.51; F = 16.9119 = 0.89.
Divide by smallest (0.51): C = 3; H = 1; F = 1.75.
Multiply by 4 to get whole numbers: C = 12; H = 4; F = 7.
Empirical Formula = C₁₂H₄F₇.
Empirical Mass = 12(12) + 4(1) + 7(19) = 144 + 4 + 133 = 281 g/mol.
n = 562.0281 = 2.
Molecular Formula = (C₁₂H₄F₇)₂ = C₂₄H₈F₁₄.
CONCEPT ASSESSMENT EXERCISE 3.8
An impure sample of calcium carbonate has a mass of 70g. After purification, 56g of calcium carbonate is recovered. What is the percentage purity of calcium carbonate?
Answer:
Percentage Purity = (Mass of Pure Substance / Mass of Impure Sample) × 100
Percentage Purity = (5670) × 100 = 80%.
CONCEPT ASSESSMENT EXERCISE 3.9
Potassium chlorate (KClO₃) is a white solid. It is used in making matches and dyes. Calculate the molarity of the solution that contains.
(a) 1.5 moles of this compound dissolved in 250cm³ of solution
Answer: Volume = 250 cm³ = 0.25 dm³. Molarity = 1.50.25 = 6.0 mol/dm³.
(b) 75g of this compound dissolved to produce 1.25dm³ of solution
Answer: Molar mass of KClO₃ = 39 + 35.5 + 48 = 122.5 g/mol. Moles = 75122.5 = 0.612 moles. Molarity = 0.6121.25 = 0.49 mol/ dm³.
(c) What is the molarity of a 50cm³ sample of potassium chlorate solution that yields 0.25g residue after evaporation of the water.
Answer: Moles = 0.25122.5 = 0.00204 moles. Volume = 0.05 dm³. Molarity = 0.002040.05 = 0.0408 mol/dm³.
CONCEPT ASSESSMENT EXERCISE 3.10
1. Sodium hydroxide solutions are used to neutralize acids and in the preparation of soaps and rayon. If you dissolve 25g of NaOH to make 1 dm³ of solution, what is the molarity of this solution?
Answer: Molar mass of NaOH = 40 g/mol. Moles = 2540 = 0.625 moles. Molarity = 0.6251 = 0.625 mol/dm³.
2. A solution of NaOH has a concentration of 1.2M. Calculate the mass of NaOH in g/dm³ and g/cm³ in this solution.
Answer: Mass in g/dm³ = 1.2 mol/dm³ × 40 g/mol = 48 g/dm³. Mass in g/cm³ = 481000 = 0.048 g/cm³.
3. A solution is prepared by dissolving 10g of haemoglobin in enough water to make up 1dm³ in volume. Calculate molarity of this solution. Molar mass of haemoglobin is 6.51x10⁴ g/mol.
Answer: Moles = 1065100 = 1.54 × 10⁻⁴ moles. Molarity = 1.54 × 10⁻⁴ / 1 = 1.54 × 10⁻⁴ mol/dm³.
CONCEPT ASSESSMENT EXERCISE 3.11
1. When 10 cm³ solution of NaOH solution is titrated against 9.8 cm³ of 0.1M HCl solution for complete neutralization. HCl(aq) + NaOH(aq) [source symbol unavailable] NaCl(aq) + H₂O(l) Calculate (i) the molarity of NaOH solution (ii) the amount of NaOH present per cm³ of solution.
Answer:
Moles of HCl = 0.1 × (9.81000) = 0.00098 moles.
Moles of NaOH = Moles of HCl = 0.00098 moles.
Volume of NaOH = 10 cm³ = 0.01 dm³.
(i) Molarity of NaOH = 0.000980.01 = 0.098 mol/dm³.
(ii) Mass of NaOH per cm³: Mass in 1 dm³ = 0.098 × 40 = 3.92 g. Mass per cm³ = 3.921000 = 0.00392 g/cm³.
2. A chemist has a 200 cm³ solution of hydrochloric acid with a concentration of 6 M (molar). He wants to dilute it to a concentration of 2 M. How much of the original solution should be mixed with water to achieve the desired concentration?
Answer:
Using dilution formula: M₁V₁ = M₂V₂
6 × V₁ = 2 × 200
V₁ = 4006 = 66.67 cm³. The chemist should use 66.67 cm³ of the original solution and add water to make the final volume 200 cm³.
