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Understand The Mole Concept - IB Chemistry

  1. A 0.450 g sample of a solid monoprotic acid (HA) is dissolved in water and titrated with 0.100 mol dm⁻³ NaOH solution. 25.0 cm³ of the NaOH solution is required to reach the equivalence point. What is the molar mass of the acid in g mol⁻¹?
    Answer: 180 1. The reaction is HA + NaOH → NaA + H₂O. Since the acid is monoprotic, the mole ratio of acid to base is 1:1. 2. Calculate the moles of NaOH used in the titration: n(NaOH) = C × V = 0.100 mol dm⁻³ × 0.0250 dm³ = 0.00250 mol. 3. At the equivalence point, n(HA) = n(NaOH) = 0.00250 mol. 4. Calculate the molar mass of the acid: M = mass / moles = 0.450 g / 0.00250 mol = 180 g mol⁻¹. Distractor B arises from incorrectly assuming n(HA) = 2 × n(NaOH). Distractor D arises from incorrectly assuming the acid is diprotic, H₂A, so n(H₂A) = 0.5 × n(NaOH).
  2. What volume of water, in cm³, must be added to 50.0 cm³ of 0.500 mol dm⁻³ NaOH solution to decrease its concentration to 0.200 mol dm⁻³?
    Answer: 75.0 cm³ 1. Use the dilution formula M₁V₁ = M₂V₂ to find the final volume (V₂) of the diluted solution. M₁ = 0.500 mol dm⁻³, V₁ = 50.0 cm³, M₂ = 0.200 mol dm⁻³. 2. V₂ = (M₁V₁) / M₂ = (0.500 mol dm⁻³ × 50.0 cm³) / 0.200 mol dm⁻³ = 125 cm³. 3. This is the total final volume. The question asks for the volume of water added. 4. Volume added = V₂ - V₁ = 125 cm³ - 50.0 cm³ = 75.0 cm³. Distractor C is the final volume, not the added volume.
  3. A gas mixture contains 4.4 g of carbon dioxide (CO₂), 2.8 g of carbon monoxide (CO), and 3.2 g of oxygen (O₂). What is the total number of molecules in the mixture? (L = Avogadro's constant)
    Answer: 0.30 L 1. Calculate the moles of each component gas. M(CO₂) = 44.0 g mol⁻¹, M(CO) = 28.0 g mol⁻¹, M(O₂) = 32.0 g mol⁻¹. n(CO₂) = 4.4 g / 44.0 g mol⁻¹ = 0.10 mol. n(CO) = 2.8 g / 28.0 g mol⁻¹ = 0.10 mol. n(O₂) = 3.2 g / 32.0 g mol⁻¹ = 0.10 mol. 2. Calculate the total moles of gas in the mixture: n(total) = 0.10 + 0.10 + 0.10 = 0.30 mol. 3. The total number of molecules is the total moles multiplied by Avogadro's constant (L). Total molecules = 0.30 × L. Distractor C is the result of calculating the total number of atoms, not molecules. Distractor D incorrectly uses the total mass instead of total moles.
  4. What is the atom economy for the production of ethanol by the fermentation of glucose? C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g)
    Answer: 51.2 % Atom economy = (Molar mass of desired products / Molar mass of all reactants) × 100. 1. Identify the desired product: 2 moles of ethanol (C₂H₅OH). 2. Identify the reactant: 1 mole of glucose (C₆H₁₂O₆). 3. Calculate molar masses: M(C₂H₅OH) = 46.08 g mol⁻¹. M(C₆H₁₂O₆) = 180.18 g mol⁻¹. 4. Calculate atom economy: AE = (Mass of 2C₂H₅OH / Mass of C₆H₁₂O₆) × 100 = (2 × 46.08 / 180.18) × 100 = (92.16 / 180.18) × 100 ≈ 51.2 %. Distractor B represents the atom economy for the waste product, CO₂.
  5. A 0.758 g sample of a volatile liquid, when vaporized, occupies a volume of 250 cm³ at a temperature of 100 °C and a pressure of 1.01 × 10⁵ Pa. Which compound has a molar mass consistent with this data? (R = 8.31 J K⁻¹ mol⁻¹)
    Answer: Phenylamine, C₆H₅NH₂ (Mr = 93.1) 1. Use the ideal gas law, PV = nRT, to find the number of moles (n). Ensure all units are in SI. P = 1.01 × 10⁵ Pa. V = 250 cm³ = 2.50 × 10⁻⁴ m³. T = 100 °C = 373 K. R = 8.31 J K⁻¹ mol⁻¹. 2. Rearrange to solve for n: n = PV / RT = (1.01 × 10⁵ Pa × 2.50 × 10⁻⁴ m³) / (8.31 J K⁻¹ mol⁻¹ × 373 K) ≈ 0.00814 mol. 3. Calculate the molar mass (M): M = mass / moles = 0.758 g / 0.00814 mol ≈ 93.1 g mol⁻¹. 4. Compare this molar mass to the options. The molar mass of phenylamine (C₆H₅NH₂) is 6(12.01) + 7(1.01) + 14.01 = 93.14 g mol⁻¹, which matches the calculated value.
  6. A laboratory balance can measure masses to the nearest 0.001 g. When measuring exactly 1.00 mol of different compounds, which compound would be measured with the highest relative precision (lowest relative uncertainty)?
    Answer: CaCO3\text{CaCO}_3 (molar mass = 100.1 g/mol) Relative uncertainty = absolute uncertainty ÷ measured value. The absolute uncertainty is ±0.001 g for all measurements. For 1.00 mol of each compound: LiH: 0.001/7.95 = 0.000126 (0.0126%), H₂O: 0.001/18.0 = 0.0000556 (0.00556%), NaCl: 0.001/58.4 = 0.0000171 (0.00171%), CaCO₃: 0.001/100.1 = 0.00000999 (0.000999%). The compound with the largest mass (CaCO₃) has the smallest relative uncertainty, giving the highest precision. Choices A, B, and C have progressively larger relative uncertainties.
  7. Calcium carbonate decomposes upon heating: CaCO₃(s) → CaO(s) + CO₂(g). The carbon dioxide produced is then bubbled through sodium hydroxide solution: CO₂(g) + 2NaOH(aq) → Na₂CO₃(aq) + H₂O(l). What initial mass of calcium carbonate is required to produce 5.30 g of sodium carbonate?
    Answer: 5.00 g 1. Calculate moles of the final product, Na₂CO₃. M(Na₂CO₃) ≈ 106.0 g mol⁻¹. n(Na₂CO₃) = 5.30 g / 106.0 g mol⁻¹ = 0.0500 mol. 2. Use stoichiometry to relate moles of Na₂CO₃ to moles of CO₂. From the second equation, the ratio is 1:1, so n(CO₂) = 0.0500 mol. 3. Use stoichiometry to relate moles of CO₂ to moles of CaCO₃. From the first equation, the ratio is 1:1, so n(CaCO₃) = 0.0500 mol. 4. Calculate the required mass of CaCO₃. M(CaCO₃) ≈ 100.1 g mol⁻¹. Mass = n × M = 0.0500 mol × 100.1 g mol⁻¹ ≈ 5.00 g. Distractor D incorrectly applies the 1:2 ratio from NaOH to the mole calculation.
  8. The density of a certain gas at STP is 1.96 g L−11.96 \text{ g L}^{-1}. If 6.02×10236.02 \times 10^{23} molecules of this gas occupy 22.4 L at STP, what is the molecular formula of the gas, assuming it contains only carbon and oxygen?
    Answer: CO2\text{CO}_2 At STP, 1 mole of any gas occupies 22.4 L. The mass of 22.4 L = 1.96 g/L × 22.4 L = 43.9 g. This represents the molar mass of the gas (≈44.0 g/mol). For the given options: CO = 28.0 g/mol, CO₂ = 44.0 g/mol, C₂O = 40.0 g/mol, C₃O₂ = 68.0 g/mol. The calculated molar mass of 44.0 g/mol matches CO₂. Choice A has too low molar mass. Choice C is close but still incorrect. Choice D has too high molar mass.
  9. A hydrated salt has the formula MgSO₄·xH₂O. When a 5.00 g sample of the hydrated salt is heated to a constant mass, 2.44 g of anhydrous MgSO₄ remains. What is the value of x?
    Answer: 7 1. Find the mass of water lost during heating: Mass of H₂O = Initial mass - Final mass = 5.00 g - 2.44 g = 2.56 g. 2. Calculate the moles of anhydrous MgSO₄ and water. M(MgSO₄) = 24.31 + 32.07 + 4(16.00) = 120.38 g mol⁻¹. M(H₂O) = 2(1.01) + 16.00 = 18.02 g mol⁻¹. n(MgSO₄) = 2.44 g / 120.38 g mol⁻¹ ≈ 0.0203 mol. n(H₂O) = 2.56 g / 18.02 g mol⁻¹ ≈ 0.142 mol. 3. Determine the ratio of moles of water to moles of MgSO₄ to find x: x = n(H₂O) / n(MgSO₄) = 0.142 mol / 0.0203 mol ≈ 6.99. 4. The value of x is the nearest whole number, which is 7.
  10. A 1.80 g sample of an unknown organic compound containing only carbon, hydrogen, and oxygen was combusted completely to produce 3.52 g of CO₂ and 1.80 g of H₂O. The molar mass of the compound is approximately 90 g mol⁻¹. What is the molecular formula of the compound?
    Answer: C₄H₁₀O₂ 1. Calculate moles of C and H from products: n(C) = n(CO₂) = 3.52 g / 44.01 g mol⁻¹ = 0.080 mol. n(H) = 2 × n(H₂O) = 2 × (1.80 g / 18.02 g mol⁻¹) = 0.200 mol. 2. Calculate mass of C and H: m(C) = 0.080 mol × 12.01 g mol⁻¹ = 0.96 g. m(H) = 0.200 mol × 1.01 g mol⁻¹ = 0.20 g. 3. Calculate mass of O by difference: m(O) = 1.80 g - 0.96 g - 0.20 g = 0.64 g. 4. Calculate moles of O: n(O) = 0.64 g / 16.00 g mol⁻¹ = 0.040 mol. 5. Determine empirical formula by finding the simplest whole number ratio of moles C:H:O = 0.080:0.200:0.040. Divide by the smallest value (0.040) to get the ratio 2:5:1. The empirical formula is C₂H₅O. 6. Calculate empirical formula mass: 2(12.01) + 5(1.01) + 16.00 = 45.07 g mol⁻¹. 7. Find the multiplier: Molar Mass / Empirical Mass = 90 / 45.07 ≈ 2. 8. The molecular formula is (C₂H₅O)₂ = C₄H₁₀O₂.
  11. Which sample contains the greatest number of oxygen atoms?
    Answer: 0.30 mol of dinitrogen pentoxide, N₂O₅ To find the number of oxygen atoms, multiply the moles of the compound by the number of oxygen atoms in its formula. A: 0.20 mol KClO₃ × 3 O/molecule = 0.60 mol O atoms. B: 0.30 mol N₂O₅ × 5 O/molecule = 1.50 mol O atoms. C: 0.40 mol SO₂ × 2 O/molecule = 0.80 mol O atoms. D: 0.50 mol C₂H₅OH × 1 O/molecule = 0.50 mol O atoms. Comparing the results, 1.50 mol is the largest amount.
  12. 10 cm³ of a gaseous hydrocarbon, CₓHᵧ, is completely combusted with 70 cm³ of oxygen. After cooling to the initial temperature and pressure, the resulting gas mixture has a volume of 50 cm³. When this mixture is passed through aqueous potassium hydroxide, the volume reduces to 20 cm³. What is the formula of the hydrocarbon?
    Answer: C₃H₈ 1. Potassium hydroxide absorbs acidic gases, so the volume of CO₂ produced is the reduction in volume: V(CO₂) = 50 cm³ - 20 cm³ = 30 cm³. 2. The remaining 20 cm³ is excess, unreacted oxygen. 3. The volume of oxygen that reacted is the initial volume minus the excess volume: V(O₂ reacted) = 70 cm³ - 20 cm³ = 50 cm³. 4. By Avogadro's law, volume ratios equal mole ratios for gases at the same temperature and pressure. The reaction ratio is V(CₓHᵧ) : V(O₂ reacted) : V(CO₂) = 10 : 50 : 30, which simplifies to 1 : 5 : 3. 5. From CₓHᵧ → xCO₂, the ratio is 1:x. So, x = 3. 6. The general combustion equation is CₓHᵧ + (x + y/4)O₂ → xCO₂ + (y/2)H₂O. The ratio of CₓHᵧ to O₂ is 1 : (x + y/4). So, x + y/4 = 5. 7. Substitute x=3: 3 + y/4 = 5, which gives y/4 = 2, so y = 8. The formula is C₃H₈.
  13. Excess aqueous silver nitrate is added to 50.0 cm³ of a potassium chloride solution, and the silver chloride precipitate is filtered, dried, and weighed. The mass of the precipitate is 0.717 g. What was the concentration of the potassium chloride solution? (Mr of AgCl = 143.3)
    Answer: 0.100 mol dm⁻³ 1. The reaction is KCl(aq) + AgNO₃(aq) → AgCl(s) + KNO₃(aq). The mole ratio of KCl to AgCl is 1:1. 2. Calculate the moles of AgCl precipitated: n(AgCl) = mass / Molar mass = 0.717 g / 143.3 g mol⁻¹ = 0.00500 mol. 3. Due to the 1:1 stoichiometry, the moles of KCl in the original solution is also 0.00500 mol. 4. Calculate the concentration of the KCl solution. Volume V = 50.0 cm³ = 0.0500 dm³. Concentration C = n / V = 0.00500 mol / 0.0500 dm³ = 0.100 mol dm⁻³.
  14. A sample of an element exists as two isotopes, ⁶⁹X and ⁷¹X. The relative atomic mass of the element is 69.8. What is the percentage abundance of the lighter isotope, ⁶⁹X?
    Answer: 60 % Let the fractional abundance of ⁶⁹X be 'a'. Then the fractional abundance of ⁷¹X is (1-a). The relative atomic mass is the weighted average: Ar = (mass₁ × abundance₁) + (mass₂ × abundance₂). 69.8 = (69 × a) + (71 × (1 - a)). 69.8 = 69a + 71 - 71a. 69.8 = 71 - 2a. 2a = 71 - 69.8. 2a = 1.2. a = 0.6. The percentage abundance is 0.6 × 100 = 60 %. Distractor B represents the abundance of the heavier isotope.
  15. Consider the following data for three different samples of compounds containing only nitrogen and oxygen. Use the information in the table below to determine which samples could represent the same compound. Sample X: 1.40 g contains 3.01×10223.01 \times 10^{22} molecules Sample Y: 2.80 g contains 6.02×10226.02 \times 10^{22} molecules Sample Z: 4.20 g contains 6.02×10226.02 \times 10^{22} molecules Which conclusion is correct?
    Answer: Samples X and Y represent the same compound; Sample Z represents a different compound When you encounter molecular composition problems like this, the key insight is that samples of the same compound must have identical molar masses, regardless of the amount present. You need to calculate the molar mass for each sample using the relationship: molar mass = (mass × Avogadro's number) ÷ number of molecules. For Sample X: Molar mass = (1.40 g×6.022×1023)÷(3.01×1022)=28.0 g/mol(1.40 \text{ g} \times 6.022 \times 10^{23}) ÷ (3.01 \times 10^{22}) = 28.0 \text{ g/mol} For Sample Y: Molar mass = (2.80 g×6.022×1023)÷(6.02×1022)=28.0 g/mol(2.80 \text{ g} \times 6.022 \times 10^{23}) ÷ (6.02 \times 10^{22}) = 28.0 \text{ g/mol} For Sample Z: Molar mass = (4.20 g×6.022×1023)÷(6.02×1022)=42.0 g/mol(4.20 \text{ g} \times 6.022 \times 10^{23}) ÷ (6.02 \times 10^{22}) = 42.0 \text{ g/mol} Since Samples X and Y both have molar masses of 28.0 g/mol, they represent the same compound (likely N₂O). Sample Z has a different molar mass of 42.0 g/mol, indicating a different compound (likely N₂O₃). Answer A is wrong because Sample Z has a different molar mass than X and Y. Answer B incorrectly pairs X with Z when they have different molar masses (28.0 vs 42.0 g/mol). Answer D is wrong because X and Y clearly have identical molar masses, proving they're the same compound. Remember: identical compounds always have identical molar masses. When analyzing molecular data, always calculate molar mass first to identify which samples represent the same substance.
  16. A stock solution of hydrochloric acid is 12.0 mol dm⁻³. What is the final concentration if 10.0 cm³ of this stock solution is diluted to 100.0 cm³ in a volumetric flask, and then 25.0 cm³ of this new solution is further diluted to 250.0 cm³?
    Answer: 0.120 mol dm⁻³ This is a two-step serial dilution. 1. First dilution: Use M₁V₁ = M₂V₂. (12.0 mol dm⁻³) × (10.0 cm³) = C_intermediate × (100.0 cm³). C_intermediate = 120 / 100 = 1.20 mol dm⁻³. 2. Second dilution: This intermediate solution is now diluted. (1.20 mol dm⁻³) × (25.0 cm³) = C_final × (250.0 cm³). C_final = (1.20 × 25.0) / 250.0 = 1.20 / 10 = 0.120 mol dm⁻³. Distractor D is the concentration after only the first dilution.
  17. A student measures 25.0 mL of a solution containing 1.20×10231.20 \times 10^{23} formula units of NaCl\text{NaCl}. After evaporating the water, the student wants to dissolve the remaining solid in enough water to create a new solution where each 10.0 mL contains exactly 6.02×10226.02 \times 10^{22} Na+\text{Na}^+ ions. What volume of the new solution should be prepared?
    Answer: 20.0 mL The solid contains 1.20×10²³ NaCl formula units, which means 1.20×10²³ Na⁺ ions (since each NaCl produces one Na⁺). In the new solution, each 10.0 mL should contain 6.02×10²² Na⁺ ions. Total volume needed = (1.20×10²³ Na⁺ ions) ÷ (6.02×10²² Na⁺ ions per 10.0 mL) × 10.0 mL = 2.0 × 10.0 mL = 20.0 mL. Choice B doubles the answer incorrectly. Choice C assumes equal concentrations. Choice D represents a ten-fold dilution error.
  18. 25.0 cm³ of a 0.100 mol dm⁻³ solution of a metal nitrate, M(NO₃)₂, is mixed with 50.0 cm³ of a 0.150 mol dm⁻³ solution of sodium phosphate, Na₃PO₄. A precipitate of the metal phosphate is formed. Based on the principles of ionic compound formation, what is the chemical formula of the precipitate?
    Answer: M₃(PO₄)₂ The formula of the precipitate is determined by the charges of the reacting ions, not the specific amounts mixed (which may be non-stoichiometric). From M(NO₃)₂, the metal ion is M²⁺. From Na₃PO₄, the phosphate ion is PO₄³⁻. To form a neutral ionic compound, the total positive charge must equal the total negative charge. The lowest common multiple of the charges (+2 and -3) is 6. This requires three M²⁺ ions (3 × +2 = +6) and two PO₄³⁻ ions (2 × -3 = -6). Therefore, the formula of the precipitate is M₃(PO₄)₂. The provided volumes and concentrations represent a limiting reactant problem and do not alter the fundamental stoichiometry of the compound formed.
  19. 14.0 g of nitrogen gas is reacted with 4.0 g of hydrogen gas to produce ammonia according to the equation: N₂(g) + 3H₂(g) → 2NH₃(g). What is the maximum mass of ammonia that can be formed?
    Answer: 17.0 g 1. Calculate the moles of each reactant. M(N₂) ≈ 28.0 g mol⁻¹, M(H₂) ≈ 2.0 g mol⁻¹. n(N₂) = 14.0 g / 28.0 g mol⁻¹ = 0.500 mol. n(H₂) = 4.0 g / 2.0 g mol⁻¹ = 2.0 mol. 2. Identify the limiting reactant. The stoichiometric ratio is 1 mol N₂ to 3 mol H₂. 0.500 mol N₂ requires 3 × 0.500 = 1.50 mol H₂. Since 2.0 mol H₂ is available, H₂ is in excess and N₂ is the limiting reactant. 3. Calculate the moles of product based on the limiting reactant. The ratio is 1 mol N₂ to 2 mol NH₃. n(NH₃) = 2 × n(N₂) = 2 × 0.500 mol = 1.00 mol. 4. Calculate the mass of ammonia produced. M(NH₃) ≈ 17.0 g mol⁻¹. Mass = n × M = 1.00 mol × 17.0 g mol⁻¹ = 17.0 g.
  20. A solution is prepared by dissolving 10.0 g of glucose (C₆H₁₂O₆, Mr = 180.2) in water to make a final volume of 200 cm³. The density of the resulting solution is 1.02 g cm⁻³. What is the molar concentration (in mol dm⁻³) of the glucose solution?
    Answer: 0.278 mol dm⁻³ Molar concentration (molarity) is defined as moles of solute per liter (dm³) of solution. The density of the solution is extraneous information if the final volume is known. 1. Calculate moles of glucose: n = mass / Molar mass = 10.0 g / 180.2 g mol⁻¹ ≈ 0.0555 mol. 2. Convert the solution volume to dm³: V = 200 cm³ = 0.200 dm³. 3. Calculate concentration: C = n / V = 0.0555 mol / 0.200 dm³ ≈ 0.278 mol dm⁻³. Distractor A is obtained by incorrectly using the mass of the solution (V × d = 200 × 1.02 = 204 g = 0.204 kg) in the volume term of the concentration calculation.

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