Chemical Formulae, Equations, the Mole and Gases

CM + 2Cl₂ → MCl₄[1]

The big jump is after the 4th ionisation energy (3313 to 7863), so M has 4 valence electrons and is in Group 14.

M bonds to four chlorine atoms, giving MCl₄. One mole of M needs 2 mol of Cl₂.

A58.8[1]

The abundances add up to 69.3 + 26.7 + 1.20 + 3.80 + 1.00 = 102, so divide by 102, not 100.

A_r = (58 × 69.3 + 60 × 26.7 + 61 × 1.20 + 62 × 3.80 + 64 × 1.00) ÷ 102 = 5994 ÷ 102 = 58.8

A40%[1]

Let ¹²¹Sb be x%: 121x + 123(100 − x) = 12 180, so 12 300 − 2x = 12 180 and x = 60.

¹²³Sb is 100 − 60 = 40%. Option D is the percentage of ¹²¹Sb.

B2n[1]

10 g of CH₄ (M_r 16) is 0.625 mol, which is n molecules.

55 g of CO₂ (M_r 44) is 1.25 mol, twice as many moles, so it has 2n molecules.

DIt contains 3.01 × 10²² ions.[1]

n = 20.3 ÷ 406 = 0.0500 mol. Each unit has 2 Co³⁺ and 3 SO₄²⁻, so there are 0.250 mol of ions, 1.51 × 10²³, not 3.01 × 10²².

A: 0.0500 mol of gas at 293 K and 1 atm is 0.0500 × 8.31 × 293 ÷ 101 325 = 1.2 × 10⁻³ m³. B: 12 × 16.0 ÷ 406 = 47.3%. C: 3 × 0.0500 = 0.15 mol.

CThe number of ions in 68.05 g of CaSO₄.[1]

CaSO₄ has M_r = 40.1 + 32.1 + 4 × 16.0 = 136.2, so 68.05 g is 0.50 mol. Each unit is 2 ions, Ca²⁺ and SO₄²⁻, so there is 1.0 mol of ions.

A is 1 mol of O₂, which is 2 mol of atoms. B is pV/RT = 0.067 mol. D is 1 mol of C₂H₆, which has 2 mol of C atoms.

D1, 2, 3 and 4[1]

1: 1.20 × 4 = 4.80 mol of atoms. 2: 1.445 × 10²⁴ ÷ 6.02 × 10²³ = 2.40 mol of O₂, which is 4.80 mol of atoms.

3: 36.0 ÷ 180 = 0.200 mol, and 0.200 × 24 atoms = 4.80 mol. 4: 57.6 ÷ 24 = 2.40 mol of Cl₂, which is 4.80 mol of atoms. All four are equal.

CC₃H₈[1]

n_C = n_CO₂ = 0.66 ÷ 44 = 0.015 mol. n_H = 2 × 0.36 ÷ 18 = 0.040 mol.

C : H = 0.015 : 0.040 = 3 : 8, so the empirical formula is C₃H₈.

CC₈H₁₅BO₂[1]

n_C = 4.58 ÷ 44.0 = 0.104 mol, 8 per molecule (÷ 0.0130). n_H = 2 × 1.76 ÷ 18.0 = 0.196 mol, 15 per molecule.

M_r = 2.0 ÷ 0.0130 = 154. C₈H₁₅B accounts for 96 + 15 + 10.8 = 121.8, leaving 32 for 2 O atoms: C₈H₁₅BO₂.

Aphosphorus[1]

Mass of X = 106.5 − 60 = 46.5 g. n_O = 60 ÷ 16 = 3.75 mol, so n_X = 3.75 × 2/5 = 1.50 mol.

A_r(X) = 46.5 ÷ 1.50 = 31, which is phosphorus.

B43.7%[1]

50 000 g of Na₂S is 640 mol, which could give 2 × 640 = 1280 mol of NaHCO₃. Only 20 500 ÷ 84.0 = 244 mol formed, an overall yield of 19.1%.

With the same yield y in both steps, y × y = 0.191, so y = 0.437 = 43.7%. Option C forgets that each Na₂CO₃ gives 2NaHCO₃.

B37.5 cm³[1]

c₁V₁ = c₂V₂: 0.280 × 50 = 0.160 × V₂, so V₂ = 87.5 cm³.

V₂ is the final volume, so the water added is 87.5 − 50 = 37.5 cm³. Option D is the final volume.

CThe number of molecules in A < the number of molecules in B[1]

The masses are equal but A is denser, so A occupies the smaller volume (D is false).

At the same temperature and pressure a smaller volume holds fewer molecules, so A has fewer molecules than B. With the same mass, its molar mass is larger (A is false). The number of atoms depends on the formulae, which are not given.

Agraph A[1]

Boyle's law: at constant temperature p ∝ 1/V, so p against 1/V is a straight line through the origin.

p against V is a curve, not a straight line. C is Charles's law, V ∝ T, and V against 1/T matches neither law.

BC₂H₄O[1]

n = pV/RT = (50 × 10³ × 40 × 10⁻⁶) ÷ (8.31 × 300) = 8.02 × 10⁻⁴ mol, so M_r = 0.0353 ÷ 8.02 × 10⁻⁴ = 44.0.

Volume ratio = mole ratio: 40 cm³ gives 80 cm³ of CO₂ (2 C) and 80 cm³ of H₂O (4 H). C₂H₄ is 28, so one O makes 44: C₂H₄O.

D0.100 × 373 × 22.70.0250 × 273[1]

At s.t.p. (273 K and 1 bar) the vapour would occupy 0.0250 × 273/373 dm³, since V ∝ T at constant pressure.

n = that volume ÷ 22.7, and M_r = 0.100 ÷ n = (0.100 × 373 × 22.7) ÷ (0.0250 × 273).

C64.1[1]

²H₂ has M_r = 4.0, so n = 0.6 ÷ 4.0 = 0.15 mol. With the same V, n ∝ p/T: n_A = 0.15 × 2 × 293/313 = 0.281 mol.

M_r(A) = 18.0 ÷ 0.281 = 64.1

A14 127[1]

NH₃ is limiting: 0.25 mol reacts with 0.25 mol of HCl, leaving 0.42 − 0.25 = 0.17 mol of HCl. NH₄Cl is a solid, so it adds nothing to the pressure.

p = nRT/V = 0.17 × 8.31 × 500 ÷ (50 × 10⁻³) = 14 127 Pa.

D0.25P[1]

Equal amounts, so each gas starts at 0.5P. 2H₂ + O₂ → 2H₂O: the H₂ (0.5P) uses half as much O₂, 0.25P.

0.5P − 0.25P = 0.25P of O₂ is left. The water is liquid at room temperature, so it adds no pressure.

B1 and 2 only[1]

2CH₃OH + 3O₂ → 2CO₂ + 4H₂O, so 1 is correct. x dm³ of methanol uses 1.5x of O₂, leaving 2.5x, and makes x of CO₂ and 2x of steam: 5.5x in all.

p(CO₂) = x ÷ 5.5x × P = 0.18P, so 2 is correct. 3 is wrong: an ideal gas's particles have negligible volume, but the gas itself fills its container.

DXeF₄[1]

At constant volume and temperature, partial pressure is proportional to amount. Xe used: 1.6 atm. F₂ used: 7.0 − 3.8 = 3.2 atm.

Xe : F₂ = 1.6 : 3.2 = 1 : 2, so Xe + 2F₂ → XeF₄.

Brow B[1]

2H₂S + 3O₂ → 2SO₂ + 2H₂O and CS₂ + 3O₂ → CO₂ + 2SO₂. 2 mol of H₂S and 1 mol of CS₂ give 4 mol of SO₂ and 1 mol of CO₂.

CS₂ has more electrons than CO₂, so its instantaneous dipole–induced dipole forces are stronger and its molecules larger: it deviates more.