Quantitative chemistry and the mole

Which row correctly describes the mole and the value of Avogadro's constant?

one mole of a substance is equal to the substance's relative atomic or molecular mass in grams; one mole of a substance contains 6.02 × 10²³ atoms, molecules or formula units One mole of any substance has a mass equal to its relative atomic or molecular mass in grams, and contains 6.02 × 10²³ particles (Avogadro's constant).

What is the mass of one mole of carbon dioxide (CO₂)? (Ar: C = 12, O = 16)

44 g Mr of CO₂ = 12 + (2 × 16) = 44, so one mole has a mass of 44 g.

How many molecules are there in one mole of hydrogen gas (H₂)?

6.02 × 10²³ molecules One mole of any substance contains 6.02 × 10²³ particles, so one mole of H₂ contains 6.02 × 10²³ molecules.

The complete combustion of methane produces carbon dioxide and steam. CH₄(g) + 2O₂(g) → 2H₂O(g) + CO₂(g). Which statements about the reaction are correct? 1. The empirical formula of methane is CH₄ 2. The number of atoms in 1 mole of methane is 4 × Avogadro's constant 3. 1 mole of methane produces 72 dm³ of gaseous products at r.t.p. 4. 1 mole of methane occupies a volume of 12 dm³ at r.t.p.

1, 2 and 3 1 is correct (CH₄ is empirical). 2: each CH₄ has 5 atoms, so 1 mole has 5 × 6.02 × 10²³ atoms, not 4. 3: 1 mole CH₄ gives 3 moles gas (2H₂O+CO₂), volume = 3 × 24 = 72 dm³. 4: 1 mole CH₄ occupies 24 dm³, not 12.

Magnesium carbonate and hydrochloric acid react: MgCO₃(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l) + CO₂(g). What is the volume of CO₂ produced when 21 g of magnesium carbonate (Mr = 84) reacts completely with excess hydrochloric acid? (One mole of gas occupies 24.0 dm³ at r.t.p.)

6 dm³ Moles MgCO₃ = 21/84 = 0.25 mol. From equation, 1 mol MgCO₃ produces 1 mol CO₂, so 0.25 mol CO₂. Volume = 0.25 × 24 = 6 dm³.

Calcium carbonate undergoes thermal decomposition: CaCO₃ → CaO + CO₂. What mass of calcium oxide (Mr = 56) is formed when 60 g of calcium carbonate (Mr = 100) is completely decomposed?

33.6 g Moles CaCO₃ = 60/100 = 0.6 mol. 1:1 ratio, so moles CaO = 0.6 mol. Mass CaO = 0.6 × 56 = 33.6 g.

Substance X was analysed and found to contain 40.00% carbon, 6.67% hydrogen and 53.33% oxygen by mass. What is the empirical formula of X?

CH₂O Assume 100 g: C = 40 g (40/12 = 3.33 mol), H = 6.67 g (6.67/1 = 6.67 mol), O = 53.33 g (53.33/16 = 3.33 mol). Divide by smallest (3.33): C=1, H=2, O=1 → CH₂O.

What is the concentration in mol dm⁻³ of a solution of sodium hydroxide that contains 16 g of NaOH in 200 cm³ of distilled water? (Mr NaOH = 40)

2 mol dm⁻³ Moles NaOH = 16/40 = 0.4 mol. Volume = 200 cm³ = 0.2 dm³. Concentration = 0.4/0.2 = 2 mol dm⁻³.

A student titrated 25 cm³ of hydrochloric acid against a standard solution of sodium hydroxide of 0.1 mol dm⁻³. The average titre of sodium hydroxide was 30 cm³. What is the concentration of the acid? (HCl + NaOH → NaCl + H₂O)

0.12 mol dm⁻³ Moles NaOH = 0.1 × (30/1000) = 0.003 mol. 1:1 ratio, so moles HCl = 0.003 mol. Concentration HCl = 0.003 / (25/1000) = 0.12 mol dm⁻³.

How many oxygen atoms are there in 10.0 g of magnesium nitrate, Mg(NO₃)₂? (Avogadro constant L = 6.02 × 10²³ mol⁻¹, Ar: Mg=24, N=14, O=16)

2.44 × 10²³ Mr Mg(NO₃)₂ = 24 + 2×(14+48) = 148. Moles = 10/148 = 0.0676 mol. Each formula unit has 6 O atoms, so moles O = 0.0676 × 6 = 0.4056 mol. Number O atoms = 0.4056 × 6.02×10²³ = 2.44×10²³.

What is the volume of hydrogen gas produced, measured at r.t.p., when 5.0 g of magnesium reacts with an excess of dilute hydrochloric acid? (Mg + 2HCl → MgCl₂ + H₂, Ar Mg = 24, molar gas volume = 24 dm³ at r.t.p.)

5.0 dm³ Moles Mg = 5/24 = 0.2083 mol. 1:1 ratio, so moles H₂ = 0.2083 mol. Volume = 0.2083 × 24 = 5.0 dm³.

A sample of chalk (limestone) with a mass of 1.70 g was reacted with excess HCl and 0.36 dm³ of CO₂ gas was collected at r.t.p. CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O. What is the percentage purity of the chalk? (Mr CaCO₃ = 100, molar gas volume = 24 dm³ at r.t.p.)

88% Moles CO₂ = 0.36/24 = 0.015 mol. 1:1 ratio, so moles CaCO₃ = 0.015 mol. Mass pure CaCO₃ = 0.015 × 100 = 1.5 g. Percentage purity = (1.5/1.70)×100 = 88%.

A student investigates the reaction of zinc with hydrochloric acid, using a gas syringe to collect the hydrogen gas produced. Several volume readings are taken as the reaction proceeds, and one reading falls noticeably off the general trend. Which of the following is the most likely reason for that one inaccurate reading?

The gas syringe had a leak A gas syringe leak would cause loss of gas, leading to a lower measured volume and an inaccurate point on the graph. Other options affect the overall trend but not necessarily a single inaccurate point.

Calcium carbide and water react: CaC₂(s) + 2H₂O(l) → C₂H₂(g) + Ca(OH)₂(s). What volume of ethyne (C₂H₂) is produced at r.t.p. when 45 g of water reacts completely? (Molar volume at r.t.p. = 24 dm³ mol⁻¹; Ar: H=1, O=16, Ca=40, C=12)

30 dm³ Moles of water = 45/18 = 2.5 mol. From equation, 2 mol H₂O produce 1 mol C₂H₂, so moles C₂H₂ = 1.25 mol. Volume = 1.25 × 24 = 30 dm³.

A student thermally decomposes 24.8 g of copper(II) carbonate and obtains 11.2 g of copper(II) oxide. CuCO₃ → CuO + CO₂. What is the percentage yield of CuO? (Ar: Cu=63.5, C=12, O=16)

82% Moles CuCO₃ = 24.8 / (63.5+12+48) = 24.8/123.5 = 0.2008 mol. Theoretical yield CuO = 0.2008 × (63.5+16) = 0.2008 × 79.5 = 15.96 g. Percentage yield = (11.2/15.96) × 100 = 70.2% ≈ 70%.

A student was asked to find the concentration in g dm⁻³ of a solution of sulfuric acid. She transferred 25 cm³ of the acid to a 250 cm³ volumetric flask and made it up to the mark. She titrated 25 cm³ of the diluted acid against 0.100 mol dm⁻³ NaOH. The average titre of NaOH was 18.0 cm³. What is the concentration of the original acid in g dm⁻³? (Mr H₂SO₄ = 98)

35.28 g dm⁻³ Moles NaOH = 0.100 × 0.018 = 0.00180 mol. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, so moles H₂SO₄ in 25 cm³ diluted = 0.00180/2 = 0.000900 mol. Concentration diluted = 0.000900 / 0.025 = 0.0360 mol dm⁻³. Dilution factor = 250/25 = 10, so original concentration = 0.360 mol dm⁻³. Mass conc = 0.360 × 98 = 35.28 g dm⁻³.

How would you design the inquiry for quantitative chemistry and the mole?

Use measured mass data to compare amounts of substance for samples of different compounds. Compare the compound being considered; record the amount of substance in moles; use the same specified sample mass.

What limitation matters when applying quantitative chemistry and the mole?

Theoretical yield assumes the specified reaction and limiting reactant; incomplete reaction and losses reduce practical yield.