Understand Amount of Chemical Change - IB Chemistry

1. Methanol can be produced by two different processes. Process 1: $\text{CO(g) + 2H}_2\text{(g)} \rightarrow \text{CH}_3\text{OH(l)}$ Process 2: $\text{CO}_2\text{(g) + 3H}_2\text{(g)} \rightarrow \text{CH}_3\text{OH(l) + H}_2\text{O(l)}$ Which statement correctly compares the atom economy of the two processes for producing methanol?
   Answer: Process 1 has an atom economy of 100% while Process 2 has an atom economy less than 100%. Atom economy = (Molar mass of desired product / Total molar mass of all reactants) × 100. Process 1 is an addition reaction with only one product, so all atoms from the reactants end up in the desired product, giving it a 100% atom economy. Process 2 produces water (H₂O) as a by-product, so some of the reactant atoms do not end up in the desired product (CH₃OH). Therefore, the atom economy of Process 2 is less than 100%. Specifically, Atom Economy(2) = M(CH₃OH) / [M(CO₂) + 3M(H₂)] * 100 = 32.05 / (44.01 + 3*2.02) * 100 = 64%.
2. Which of the following would result in an experimentally determined percentage yield greater than 100% for the precipitation of a solid product?
   Answer: The solid product was not completely dried before its final mass was measured. Percentage yield is calculated as (actual mass / theoretical mass) × 100. A value over 100% means the measured actual mass is greater than the maximum possible mass of pure product. If the solid product is not completely dried, its measured mass will include the mass of the remaining solvent (e.g., water), making it artificially high. Losing product (A) or incomplete reaction (C) would lower the yield. Using an excess of a reactant (D) ensures the other is limiting but does not affect the theoretical yield calculation or make the actual yield higher than theoretical.
3. A 4.58 g sample of a mixture of sodium chloride (NaCl) and anhydrous sodium carbonate (Na₂CO₃) was dissolved in water and made up to 250.0 cm³. A 25.00 cm³ aliquot of this solution required 22.80 cm³ of 0.150 mol dm⁻³ HCl for complete reaction with the carbonate. What is the percentage by mass of Na₂CO₃ in the original mixture? ($M_r$: Na₂CO₃ = 105.99)
   Answer: 79.1% The reaction is Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. 1. Calculate moles of HCl used: n(HCl) = 0.02280 dm³ × 0.150 mol dm⁻³ = 0.00342 mol. 2. Calculate moles of Na₂CO₃ in the 25.00 cm³ aliquot: n(Na₂CO₃) = 0.00342 mol HCl × (1 mol Na₂CO₃ / 2 mol HCl) = 0.00171 mol. 3. Calculate moles of Na₂CO₃ in the original 250.0 cm³ solution: 0.00171 mol × (250.0 / 25.00) = 0.0171 mol. 4. Calculate mass of Na₂CO₃ in the original sample: 0.0171 mol × 105.99 g mol⁻¹ = 3.625 g. 5. Calculate percentage by mass: (3.625 g / 4.58 g) × 100 = 79.1%.
4. What is the minimum mass of TiO₂ required to produce 100.0 g of pure titanium, assuming the yields are as stated? ($A_r$: Ti = 47.87, O = 16.00)
   Answer: 195.2 g Work backwards from the final product. 1. Moles of Ti produced = 100.0 g / 47.87 g mol⁻¹ = 2.089 mol. This is the actual yield. 2. Theoretical moles of Ti for Step 2 = Actual moles / Yield = 2.089 mol / 0.95 = 2.199 mol. 3. Stoichiometry of Step 2 is 1:1 for TiCl₄:Ti, so 2.199 mol of TiCl₄ must be produced in Step 1. This is the actual yield for Step 1. 4. Theoretical moles of TiCl₄ needed from Step 1 = 2.199 mol / 0.90 = 2.443 mol. 5. Stoichiometry of Step 1 is 1:1 for TiO₂:TiCl₄, so 2.443 mol of TiO₂ is required. 6. Mass of TiO₂ = 2.443 mol × 79.87 g mol⁻¹ = 195.2 g.
5. Given the reaction $\text{2SO}_2\text{(g) + O}_2\text{(g)} \rightarrow \text{2SO}_3\text{(g)}$, what is the theoretical yield of SO₃, if the reaction starts with 50.0 g of SO₂ and 50.0 g of O₂? ($M_r$: SO₂ = 64.07, O₂ = 32.00, SO₃ = 80.07)
   Answer: 62.5 g 1. Calculate initial moles: n(SO₂) = 50.0 g / 64.07 g mol⁻¹ = 0.780 mol. n(O₂) = 50.0 g / 32.00 g mol⁻¹ = 1.563 mol. 2. Determine the limiting reactant: The ratio is 2 SO₂ : 1 O₂. Moles of O₂ needed for 0.780 mol SO₂ = 0.780 / 2 = 0.390 mol. We have 1.563 mol of O₂, so O₂ is in excess and SO₂ is the limiting reactant. 3. Calculate the theoretical yield of SO₃ based on the limiting reactant (SO₂): The ratio is 2 SO₂ : 2 SO₃, or 1:1. So, n(SO₃) = n(SO₂) = 0.780 mol. 4. Calculate the mass of SO₃: Mass = 0.780 mol × 80.07 g mol⁻¹ = 62.5 g.
6. A student performs a titration using 25.00 mL of 0.1050 M $\text{NaOH}$ to neutralize 30.00 mL of $\text{HCl}$ solution. If the student accidentally reads the burette incorrectly and records the volume as 24.50 mL instead of 25.00 mL, what is the percent error in the calculated concentration of the $\text{HCl}$?
   Answer: The calculated concentration is 2.0% lower than the actual value True concentration: $\text{HCl}$ molarity = (0.02500 L × 0.1050 M) / 0.03000 L = 0.08750 M. Calculated concentration with error: $\text{HCl}$ molarity = (0.02450 L × 0.1050 M) / 0.03000 L = 0.08575 M. Percent error = (0.08575 - 0.08750) / 0.08750 × 100% = -2.0%. The calculated value is 2.0% lower than the actual value. Choice B incorrectly assumes higher concentration. Choices C and D use 2.04% which would result from calculation error (50/2450 ≈ 0.0204).
7. In the reaction $2\text{Al} + 3\text{CuSO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3\text{Cu}$, if 5.40 g of aluminum reacts with excess copper(II) sulfate, what mass of copper is produced and how many moles of electrons are transferred?
   Answer: 19.1 g Cu produced; 0.600 mol electrons transferred When you encounter stoichiometry problems involving redox reactions, you need to track both mass relationships and electron transfer. This question tests your ability to convert between moles, mass, and electrons in a single calculation. Start by finding moles of aluminum: $\frac{5.40 \text{ g Al}}{26.98 \text{ g/mol}} = 0.200 \text{ mol Al}$ Using the balanced equation's molar ratio, 2 mol Al produces 3 mol Cu, so: $0.200 \text{ mol Al} \times \frac{3 \text{ mol Cu}}{2 \text{ mol Al}} = 0.300 \text{ mol Cu}$ Converting to mass: $0.300 \text{ mol Cu} \times 63.55 \text{ g/mol} = 19.1 \text{ g Cu}$ For electron transfer, examine the half-reactions: Al goes from 0 to +3 (loses 3 electrons), while Cu²⁺ goes from +2 to 0 (gains 2 electrons). Each aluminum atom transfers 3 electrons, so: $0.200 \text{ mol Al} \times 3 \text{ electrons/Al atom} = 0.600 \text{ mol electrons}$ Answer A incorrectly calculates copper mass as if the molar ratio were 1:1 instead of 2:3, giving 12.7 g instead of 19.1 g. Answer B gets the copper mass right but miscalculates electrons by using the moles of copper produced rather than considering that each Al transfers 3 electrons. Answer D combines both errors from A and further reduces the electron count. Remember: in stoichiometry problems, always use the balanced equation's coefficients for molar ratios, and for electron transfer, focus on the oxidation state changes of the element you're given, not the products.
8. When 4.80 g of $\text{CaCO}_3$ is heated, it decomposes according to: $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$. If the reaction goes to 85% completion, what volume of $\text{CO}_2$ gas is produced at STP?
   Answer: 0.915 L of $\text{CO}_2$ is produced at STP conditions This problem tests your ability to perform stoichiometric calculations with percent completion and gas volume at STP. When you see decomposition reactions with incomplete conversion, you need to account for the actual yield, not just the theoretical yield. Start by finding the moles of $\text{CaCO}_3$: Using the molar mass of 100.09 g/mol, you have $\frac{4.80 \text{ g}}{100.09 \text{ g/mol}} = 0.04796 \text{ mol}$. From the balanced equation, the mole ratio is 1:1, so theoretically 0.04796 mol of $\text{CO}_2$ would be produced at 100% completion. However, the reaction only goes to 85% completion, so the actual moles of $\text{CO}_2$ produced is: $0.04796 \text{ mol} \times 0.85 = 0.04077 \text{ mol}$. At STP, one mole of any gas occupies 22.4 L, so the volume is: $0.04077 \text{ mol} \times 22.4 \text{ L/mol} = 0.913 \text{ L}$, which rounds to 0.915 L. Choice A (1.080 L) represents a calculation error where someone might have used incorrect molar mass or STP conditions. Choice B (1.076 L) likely comes from forgetting to apply the 85% completion factor entirely. Choice C (0.918 L) is close but suggests rounding errors or slight miscalculations in the molar mass or percent completion step. Remember that percent completion problems require you to multiply your theoretical yield by the completion percentage. Always double-check that you're using the correct molar mass and STP volume (22.4 L/mol) in your calculations.
9. In the Ostwald process for producing nitric acid, ammonia is oxidized: $\text{4NH}_3\text{(g) + 5O}_2\text{(g)} \rightarrow \text{4NO(g) + 6H}_2\text{O(g)}$. If 10.0 g of NH₃ is reacted with 20.0 g of O₂, what mass of the excess reactant remains after the reaction is complete? ($M_r$: NH₃ = 17.03, O₂ = 32.00)
   Answer: 1.48 g 1. Calculate initial moles: n(NH₃) = 10.0 g / 17.03 g mol⁻¹ = 0.587 mol; n(O₂) = 20.0 g / 32.00 g mol⁻¹ = 0.625 mol. 2. Determine limiting reactant: The ratio is 4 NH₃ : 5 O₂. Moles of O₂ needed for 0.587 mol NH₃ = 0.587 × (5/4) = 0.734 mol. We only have 0.625 mol, so O₂ is limiting. NH₃ is in excess. 3. Calculate moles of excess reactant used: n(NH₃) used = 0.625 mol O₂ × (4 mol NH₃ / 5 mol O₂) = 0.500 mol. 4. Calculate moles of excess reactant remaining: 0.587 mol - 0.500 mol = 0.087 mol. 5. Calculate mass of excess reactant remaining: 0.087 mol × 17.03 g mol⁻¹ = 1.48 g.
10. A chemical process is described as having a high percentage yield but a low atom economy. Which statement provides the best explanation for this outcome?
    Answer: The reaction went almost to completion, but it also produced a large mass of undesirable by-products. High percentage yield means the reaction was efficient in converting reactants to products (it went almost to completion). Low atom economy means that a large proportion of the atoms from the reactants ended up in waste or by-products, rather than the desired product. Therefore, the reaction must have produced the desired product along with a significant mass of other substances.
11. Which type of reaction is most likely to have a theoretical atom economy of 100%?
    Answer: An addition reaction, such as the hydrogenation of ethene. Atom economy is a measure of how many atoms from the reactants are incorporated into the desired product. An addition reaction, like C₂H₄ + H₂ → C₂H₆, combines all reactant atoms into a single product, resulting in a 100% atom economy. Substitution, elimination, and neutralization reactions all produce at least one by-product in addition to the desired product, meaning their atom economies are always less than 100%.
12. The atom economy for a reaction is 80.0% and the percentage yield is 75.0%. For every 100.0 g of reactants consumed, what is the mass of waste products generated?
    Answer: 40.0 g 1. Atom economy of 80.0% means that for every 100 g of reactants that are fully converted to products, 80.0 g is the desired product and 20.0 g are by-products (waste). 2. Percentage yield of 75.0% means that only 75.0% of the reactants are converted to products. So, for 100.0 g of reactants, 75.0 g are converted to products and 25.0 g remain unreacted (this is also waste). 3. From the 75.0 g of reactants that are converted, 80.0% of the resulting mass is desired product and 20.0% is by-product waste. Mass of by-products = 75.0 g × 0.200 = 15.0 g. 4. Total waste = unreacted reactants + by-products = 25.0 g + 15.0 g = 40.0 g.
13. A hydrated salt has the formula $\text{MgSO}_4 \cdot x\text{H}_2\text{O}$. When 2.466 g of the hydrated salt is heated to remove all water, 1.204 g of anhydrous $\text{MgSO}_4$ remains. Determine the value of x and the percentage of water in the original hydrated salt.
    Answer: x = 7; water percentage is 51.2% by mass When you encounter hydrated salt problems, you're dealing with stoichiometry and mass relationships. The key is finding how many water molecules are associated with each formula unit of the anhydrous salt. Start by calculating the mass of water lost during heating: 2.466 g (hydrated) - 1.204 g (anhydrous) = 1.262 g of water. Next, convert both masses to moles. For anhydrous $\text{MgSO}_4$: molar mass = 24.3 + 32.1 + 4(16.0) = 120.4 g/mol, so 1.204 g ÷ 120.4 g/mol = 0.01000 mol. For water: 1.262 g ÷ 18.02 g/mol = 0.07005 mol. The ratio of water to anhydrous salt gives you x: 0.07005 mol $\text{H}_2\text{O}$ ÷ 0.01000 mol $\text{MgSO}_4$ = 7.005 ≈ 7. So x = 7. The water percentage is: (1.262 g ÷ 2.466 g) × 100% = 51.2%. Option A incorrectly rounds the ratio to 6 instead of 7, leading to the wrong formula. Option C correctly identifies x = 7 but miscalculates the water percentage as 48.8% - this would be the percentage of anhydrous salt, not water. Option D combines both errors from A and C. The correct answer is B: x = 7 and water percentage is 51.2%. Study tip: Always double-check your mole ratio calculations by rounding to the nearest whole number, and remember that water percentage means the mass of water divided by the total mass of the hydrated compound.
14. In the thermite reaction, 10.8 g of aluminium is reacted with 30.0 g of iron(III) oxide. The equation for the reaction is: $\text{2Al(s) + Fe}_2\text{O}_3\text{(s)} \rightarrow \text{Al}_2\text{O}_3\text{(s) + 2Fe(l)}$. What is the maximum mass of iron that can be produced? ($A_r$: Al = 26.98, Fe = 55.85, O = 16.00)
    Answer: 21.0 g First, calculate the moles of each reactant. Moles Al = 10.8 g / 26.98 g mol⁻¹ = 0.400 mol. Moles Fe₂O₃ = 30.0 g / 159.70 g mol⁻¹ = 0.188 mol. Next, determine the limiting reactant using the stoichiometry (2:1 ratio). Moles of Fe₂O₃ needed to react with 0.400 mol Al is 0.400 / 2 = 0.200 mol. Since only 0.188 mol of Fe₂O₃ is available, Fe₂O₃ is the limiting reactant. The amount of product is determined by the limiting reactant. Moles of Fe produced = 0.188 mol Fe₂O₃ × (2 mol Fe / 1 mol Fe₂O₃) = 0.376 mol Fe. Mass of Fe = 0.376 mol × 55.85 g mol⁻¹ = 21.0 g.
15. A student aims to prepare 12.8 g of copper by reacting excess zinc with copper(II) sulfate solution. $\text{Zn(s) + CuSO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq) + Cu(s)}$. After collecting and drying the product, the student finds the actual mass of copper to be 11.5 g. Which statement correctly identifies a possible reason for the percentage yield being less than 100% and a way to improve it?
    Answer: Some copper was lost during filtration; using a finer filter paper could increase the amount of product collected. A percentage yield less than 100% means the actual mass of product collected is less than the theoretical maximum. Option A describes a situation that would lead to a yield >100% and the 'improvement' would lower the measured mass, not increase the true yield. Option B correctly identifies a source of product loss (mechanical loss during transfer/filtration) and a plausible method to mitigate it. Option C is incorrect because the problem states excess zinc was already used, so making it more excess won't change the theoretical yield determined by the copper(II) sulfate. Option D is incorrect as concentration affects rate, but not necessarily the final yield in this type of reaction.
16. A 12.5 g sample of impure limestone is heated strongly, causing the calcium carbonate within it to decompose. After the reaction, 5.50 g of calcium oxide is collected. $\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s) + CO}_2\text{(g)}$. If the limestone sample is known to be 92.0% CaCO₃ by mass, what is the percentage yield of calcium oxide? ($M_r$: CaCO₃ = 100.09, CaO = 56.08)
    Answer: 85.4% First, find the mass of pure CaCO₃ in the sample: 12.5 g × 0.920 = 11.5 g. Then, calculate the theoretical yield of CaO. Moles CaCO₃ = 11.5 g / 100.09 g mol⁻¹ = 0.1149 mol. From the 1:1 stoichiometry, the theoretical moles of CaO is also 0.1149 mol. Theoretical mass of CaO = 0.1149 mol × 56.08 g mol⁻¹ = 6.44 g. The percentage yield is (actual yield / theoretical yield) × 100 = (5.50 g / 6.44 g) × 100 = 85.4%.
17. Equal volumes of all gases at the same temperature and pressure contain equal numbers of particles. Consider the reaction: $\text{N}_2\text{(g) + 3H}_2\text{(g)} \rightarrow \text{2NH}_3\text{(g)}$. If 20 cm³ of nitrogen is reacted with 40 cm³ of hydrogen under constant conditions, what is the total volume of gas remaining in the container?
    Answer: 33.3 cm³ According to Avogadro's law, volume ratios equal mole ratios for gases at constant T and P. The stoichiometric ratio is 1 N₂ : 3 H₂. 1. Determine the limiting reactant: For 20 cm³ of N₂, 60 cm³ of H₂ is required. Since only 40 cm³ of H₂ is available, H₂ is the limiting reactant. 2. Calculate volumes based on limiting reactant: 40 cm³ of H₂ will react with 40/3 = 13.33 cm³ of N₂, producing 40 × (2/3) = 26.67 cm³ of NH₃. 3. Calculate remaining volumes: N₂ remaining = 20.0 - 13.33 = 6.67 cm³; NH₃ produced = 26.67 cm³. 4. Total final volume = 6.67 + 26.67 = 33.3 cm³.
18. When hydrated copper(II) sulfate, CuSO₄·5H₂O, is heated, it decomposes: $\text{CuSO}_4\cdot\text{5H}_2\text{O(s)} \rightarrow \text{CuSO}_4\text{(s) + 5H}_2\text{O(g)}$. A student heats 10.0 g of the hydrated salt and collects 5.80 g of anhydrous copper(II) sulfate. What is the percentage yield for this reaction? ($M_r$: CuSO₄·5H₂O = 249.72, CuSO₄ = 159.62)
    Answer: 90.8% First, calculate the theoretical yield of anhydrous CuSO₄ from 10.0 g of the hydrated salt. Moles of CuSO₄·5H₂O = 10.0 g / 249.72 g mol⁻¹ = 0.0401 mol. The mole ratio between the hydrated salt and the anhydrous salt is 1:1, so the theoretical moles of CuSO₄ is 0.0401 mol. Theoretical mass of CuSO₄ = 0.0401 mol × 159.62 g mol⁻¹ = 6.39 g. The actual yield was 5.80 g. Percentage yield = (actual / theoretical) × 100 = (5.80 g / 6.39 g) × 100 = 90.8%.
19. The decomposition of sodium azide, NaN₃, is used in some automobile airbags. $\text{2NaN}_3\text{(s)} \rightarrow \text{2Na(s) + 3N}_2\text{(g)}$. What volume of nitrogen gas, in dm³, is produced from the decomposition of 13.0 g of NaN₃ if the gas is collected at 25 °C and 101 kPa? ($M_r$: NaN₃ = 65.02; R = 8.31 J K⁻¹ mol⁻¹)
    Answer: 7.36 dm³ First, calculate the moles of NaN₃: n(NaN₃) = 13.0 g / 65.02 g mol⁻¹ = 0.200 mol. Use the mole ratio to find moles of N₂: n(N₂) = 0.200 mol NaN₃ × (3 mol N₂ / 2 mol NaN₃) = 0.300 mol. Use the ideal gas law, PV = nRT, to find the volume. Convert units: T = 25 + 273 = 298 K, P = 101 kPa = 101000 Pa. V = nRT/P = (0.300 mol × 8.31 J K⁻¹ mol⁻¹ × 298 K) / 101000 Pa = 0.00736 m³. Convert m³ to dm³: 0.00736 m³ × 1000 dm³ m⁻³ = 7.36 dm³.
20. What is the percentage purity of the calcium carbonate sample? ($M_r$: CaCO₃ = 100.09)
    Answer: 83.6% This is a back titration problem. 1. Initial moles HCl = 0.0500 dm³ × 1.00 mol dm⁻³ = 0.0500 mol. 2. Moles NaOH used = 0.0165 dm³ × 0.500 mol dm⁻³ = 0.00825 mol. 3. Moles of excess HCl = moles NaOH = 0.00825 mol. 4. Moles HCl reacted with CaCO₃ = Initial moles - Excess moles = 0.0500 - 0.00825 = 0.04175 mol. 5. Moles CaCO₃ = 0.04175 mol HCl × (1 mol CaCO₃ / 2 mol HCl) = 0.020875 mol. 6. Mass of pure CaCO₃ = 0.020875 mol × 100.09 g mol⁻¹ = 2.089 g. 7. Percentage purity = (2.089 g / 2.50 g) × 100 = 83.6%.

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