Edexcel A-Level Chemistry AS Paper 2, June 2017: Question 8

11 marks · Medium difficulty · Calculations

Calculate the moles, mass, and percentage yield of propanedioic acid produced from the oxidation of propane-1,3-diol, including reagent selection, colour change, and structural isomer representation.

Practise this question

Question

Question text

8 This question is about the chemistry of propane-1,3-diol and propanedioic acid.

(a) Give the structures of propane-1,3-diol and another diol which is an

isomer of propane-1,3-diol.

(2)

(b) Propane-1,3-diol can be oxidised to propanedioic acid in the same way as other

primary alcohols.

(i) Suitable reagents and conditions are

(1)

Reagents Conditions

A sodium dichromate(VI) + sulfuric acid heating under reflux

B sodium dichromate(VI) + hydrochloric acid heating under reflux

C potassium dichromate(VI) + sulfuric acid room temperature

D potassium dichromate(VI) + hydrochloric acid room temperature

(ii) The colour change in this reaction is

(1)

A green to orange

B orange to green

C orange to colourless

D colourless to orange

which is a solid dibasic acid. This acid may be represented as H2X. 21

250 3cmof a solution was prepared from all of the acid in a volumetric flask.*P49857A02124*

10.0 3cmportions of this solution were then titrated with 0.400 mol dm–3

sodium hydroxide solution. The mean titre was 18.45 cm3.

H2X + 2NaOH o Na2X + 2H2O

[Relative formula masses: propane-1,3-diol = 76.0; propanedioic acid = 104.0]

(i) Calculate the moles of propanedioic acid in 10.0 cm3 of the acid solution.

(2)

(ii) Calculate the mass of propanedioic acid in the 250 cm3 solution.

(2)

(iii) Calculate the percentage yield for the oxidation of propane-1,3-diol to

propanedioic acid.

(2)

(iv) Giveonereason why the yield calculated in (iii) is less than 100%.*P49857A02224*

(1)

(Total for Question 8 = 11 marks)

Mark scheme

Show the mark scheme

Question Acceptable Answer

Additional Guidance Mark

Number

8(a) CH2OHCH2CH2OH/ H H H (1) allow displayed / skeletal formula (2)

l l l allow OH undisplayed

H C C C H If more than one formula given for a

l l l molecule, both must be correct

O H O

l l Penalise C–H–O only once

H H Do not penalise bond to H of

pendent OH

Do not award C

CH2OHCHOHCH3/ H H H (1) l

l l l H

H C C C H l

l l l O

O O H

l l

H H Ignore names even if incorrect

Penalise missing alkane H once only

Do not award missing H from OH

Allow formulae of propane-1,1-diol

or propane-2,2-diol

Do not award for other diols

Question Acceptable Answer Mark

Number

8(b)(i) 8(b)(i). The only correct answer is A (1)

B is not correct because hydrogen chloride would be lost during heating

C is not correct because reflux is required to ensure complete oxidation

D is not correct because reflux is required to ensure complete oxidation

Question Acceptable Answer Mark

Number

8(b)(ii) 8(b)(i). The only correct answer is B (1)

A is not correct because the correct colour change is reversed

C is not correct because the orange dichromate(VI) ions are reduced to green chromium(III)

ions

D is not correct because the orange dichromate (VI) ions are reduced to green

chromium(III) ions

Question Acceptable Answer Additional Guidance Mark

Number

8(c)(i) moles of NaOH (2)

18.45 x 0.400/1000 = 7.38 x 10-3/0.00738 (1)

moles of propanedioic acid

7.38 x 10-3/2 = 3.69 x 10-3/0.00369 (1) TE: moles of NaOH/2

Question Acceptable Answer Additional Guidance Mark

Number

8(c)(ii) example of calculation (2)

moles of propanedioic acid in 250 cm3 (1) moles of propanedioic acid

25 x answer to (c)(i) = 25 x 3.69 x

10-3 = 0.09225

mass of propanedioic acid in 250 cm3 (1) 0.09225 x 104

= 9.6/9.59/9.594 (g)

Allow calculation in either order e.g.

calculate mass propanedioic acid in

10.0 cm3 first then x 25

Allow TE from c(i) eg 0.00738 gives

19.188 (g)

Question Acceptable Answer Additional Guidance Mark

Number

8(c)(iii) example of calculation (2)

theoretical yield (1) theoretical yield

15.2 x104/76 = 20.8 g

% yield (1) % yield

answer to c(ii) x 100/20.8

9.594 x 100/20.8 =

46/46.1/46.12/46.13/46.125 (%)

use of 9.6 gives 46.15385

allow any number of sig figs except

one

Correct answer with or without

working scores 2 marks

TE on incorrect theoretical yield and

answer to c(ii)

Both marks will be lost for use of

15.2 as theoretical yield (gives

63.1%)

Question Acceptable Answer Additional Guidance Mark

Number

8(c)(iv) an answer that makes reference to one of the following (1)

points:

transfer losses Ignore spillage/impure

reactants/incompetence/references

to uncertainties

incomplete reaction/oxidation/ formation of Ignore other products formed/loss

aldehyde by evaporation

side reaction(s) Penalise additional incorrect reasons

ie +1 -1 = zero

(Total for Question 8 = 11 marks)

TOTAL FOR PAPER = 80 MARKS

How to answer it

Chemistry of Propane-1,3-diol and Propanedioic Acid

What this question tests

This multi-step question assesses core organic chemistry and quantitative skills: drawing alcohol structural isomers, understanding reflux conditions and oxidation reagents for primary alcohols, interpreting chromium colour changes, carrying out multi-stage titration calculations (stoichiometry, concentration, and volume scaling), working out percentage yield, and evaluating practical losses.

Part (a)

Structures of Isomeric Diols

✅ Correct Answer (2 marks)

Propane-1,3-diol: HOCH₂CH₂CH₂OH or full displayed formula showing all bonds correctly.
Isomeric diol: CH₃CH(OH)CH₂OH (propane-1,2-diol) or CH₃C(OH)₂CH₃ / CH₃CH₂CH(OH)₂ (gem-diols, though propane-1,2-diol is the standard structural isomer).

❌ Common Errors

Omitting hydrogen atoms on carbon chains or drawing incomplete bonds to hydroxyl groups ( -OH ).
Listing non-existent isomers or repeating the same structure with different rotations.

Mark scheme note: 1 mark per correct structure. Full displayed or skeletal formulae accepted. C-H-O bonds must connect correctly to oxygen atoms.

Part (b)(i)

Oxidation Reagents and Conditions

✅ Correct Answer (1 mark)

A: sodium dichromate(VI) + sulfuric acid, heating under reflux.

💡 Key Knowledge

Primary alcohols oxidise to aldehydes, then further to carboxylic acids under vigorous conditions.
Reflux prevents volatile organic vapours from escaping by condensing them back into the reaction vessel, ensuring complete oxidation to the di-acid.
Hydrochloric acid is avoided as a matrix because chloride ions can be oxidised instead, or side-reactions occur. Sulfuric acid provides the necessary acidic medium ( H⁺ ).

Mark scheme note: Option A only. B is incorrect due to HCl ; C and D are incorrect due to room temperature lacking reflux.

Part (b)(ii)

Colour Change in Oxidation

✅ Correct Answer (1 mark)

B: orange to green

🧠 Exam Technique

Remember the oxidation states of chromium: Dichromate(VI) ions ( Cr₂O₇²⁻ ) are orange, and when reduced during the oxidation of the alcohol, they form chromium(III) ions ( Cr³⁺ ), which are green.

Mark scheme note: Option B only. Ensure you state initial to final colour in the exact order requested.

Part (c)(i)

Moles of Propanedioic Acid in Titration Portion

📐 Calculation Steps (2 marks)

Moles of NaOH used:
18.45 × 0.400 / 1000 = 7.38 × 10⁻³ mol
Moles of acid in the 10.0 cm³ portion:
Using stoichiometry ( H₂X + 2NaOH → Na₂X + 2H₂O ):
7.38 × 10⁻³ / 2 = 3.69 × 10⁻³ mol

❌ Common Errors

Forgetting to divide by 2 for the 1:2 reacting ratio given in the equation is the most frequent way students lose the second mark.

Mark scheme note: 1 mark for correct moles of NaOH; 1 mark for applying the 1:2 ratio to find moles of acid.

Part (c)(ii)

Mass of Propanedioic Acid in 250 cm³ Solution

📐 Calculation Steps (2 marks)

Scale up to 250 cm³ flask:
250 / 10.0 = 25 scaling factor.
Moles in 250 cm³ = 3.69 × 10⁻³ × 25 = 0.09225 mol
Convert moles to mass:
Mass = moles × molar mass ( M = 104.0 g mol⁻¹ )
0.09225 × 104.0 = 9.594 g (accept 9.6 g )

🧠 Exam Technique

Always carry unrounded numbers through to final calculation steps to avoid intermediate rounding errors.

Mark scheme note: 1 mark for scaling moles to 250 cm³; 1 mark for multiplying by Mr (104.0) . Allow error carried forward (TE) from part (c)(i).

Part (c)(iii)

Percentage Yield Calculation

📐 Calculation Steps (2 marks)

Calculate theoretical yield:
Moles of reactant (propane-1,3-diol) = 15.2 / 76.0 = 0.200 mol
Theoretical mass of product = 0.200 × 104.0 = 20.8 g
Calculate percentage yield:
(Actual mass / Theoretical mass) × 100
(9.594 / 20.8) × 100 = 46.125% (accept 46.1% or 46% )

❌ Common Errors

Using the mass of the starting material ( 15.2 g ) directly as the theoretical yield instead of calculating moles and scaling by molar mass.

Mark scheme note: 1 mark for theoretical yield ( 20.8 g ); 1 mark for percentage yield calculation. TE applies from part (c)(ii).

Part (c)(iv)

Reason for Yield Less Than 100%

✅ Acceptable Answers (1 mark)

Transfer losses during pouring / filtration / separation.
Incomplete reaction or oxidation (formation of intermediate aldehyde).
Side reactions occurring simultaneously.

❌ Avoid Vague Terms

Examiners penalize vague answers like "human error" or generic "experimental error". Be specific (e.g., "loss of crystals when transferring from beaker").

Mark scheme note: Award 1 mark for any valid chemical or practical reason explaining loss of product. Ignore references to "apparatus uncertainties".

Topics

Organic Chemistry · Physical Chemistry · Topic 6: Organic Chemistry I · Topic 5: Formulae, Equations and Amounts of Substance

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.