DP Chemistry (first assessment 2025)

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Question 23M.2.HL.TZ1.5

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Date May 2023 Marks available [Maximum mark: 9] Reference code 23M.2.HL.TZ1.5
Level HL Paper 2 Time zone TZ1
Command term Calculate, Deduce, Determine, Draw, Formulate Question number 5 Adapted from N/A
5.
[Maximum mark: 9]
23M.2.HL.TZ1.5

Double salts are substances with two cations and one anion. A hydrated sulfate containing two cations has this percentage composition.

Element

 Percentage (%)

 Nitrogen (N)

 7.09

 Hydrogen (H)

 5.11

 Sulfur (S)

 16.22

 Cobalt (Co)

 14.91

 Oxygen (O)

 —
(a.i)

Draw one Lewis (electron dot) structure of the sulfate ion.

[1]

Markscheme

✓

 

Accept any combination of dots, crosses and lines.
Double bonds do not have to be opposite each other.
Do not penalise missing square brackets.

(a.ii)

Calculate the percentage of oxygen present in the double salt.

[1]

Markscheme

«100 − (7.09 + 5.11 + 16.22 + 14.91) =» 56.67 «%» ✓

(a.iii)

Determine the empirical formula of the double salt. Use section 6 of the data booklet.

[3]

Markscheme

n(N): 7.09 g/14.01 g mol−1, n(H): 5.11 g/1.01 g mol−1, n(S): 16.22 g/32.07 g mol−1,
n(Co): 14.91 g/58.93 g mol−1 and n(O): 56.67 g/16.00 g mol−1
OR
n(N): 0.506, n(H): 5.06, n(S): 0.506, n(Co): 0.253 and n(O): 3.54 ✓

 

0.506/0.253, 5.06/0.253, 0.506/0.253, 0.253/0.253, 3.54/0.253
OR
2.00, 20.0, 2.00, 1.00 14.00 ✓

N2H20S2CoO14 ✓

 

Award [3] for the correct final formula.

Accept (NH4)2Co(SO4)2·6H2O

(a.iv)

The molar mass of the empirical formula is the same as the molar mass of the formula unit. Deduce the formula unit of the hydrated double salt.

[1]

Markscheme

(NH4)2Co(SO4)2⋅6H2O
OR
Co(NH4)2(SO4)2⋅6H2O) ✓

 

Accept (NH4)2Co(SO4)2(H2O)6.

(b)

1.20 g of the double salt was dissolved in water and an excess of aqueous barium chloride was added, precipitating all the sulfate ions as barium sulfate.

(b.i)

Formulate an ionic equation, including state symbols, for the reaction of barium ions with sulfate ions.

[1]

Markscheme

Ba2+(aq) + SO42−(aq) ⇌ BaSO4(s) ✓

 

Accept single arrow in place of equilibrium sign.

(b.ii)

Calculate the mass of barium sulfate precipitate. Use your answer to part (a)(iii) and section 6 of the data booklet. (If you did not obtain an answer for part (a)(iii), use 400.0 g mol−1 as Mr for the double salt, but this is not the correct value.)

[2]

Markscheme

«1.20g/395.29 g mol−1 salt = 2 × 3.04 ×10−3 «mol» SO42− =» 6.08 × ×10−3 «mol» ✓
«233.40 g mol−1 × 6.08 ×10−3 =» 1.42«g» ✓
OR 
«(1.20 g/400) × 2 g mol−1 =» 6.00 ×10−3 «mol» ✓
«233.40 g mol−1 × 6.00 ×10−3 =» 1.40«g» ✓

 

Award [2] for correct final answer.
Accept ×2 in any step.
Award [1] for half the answer, 0.70«g»