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1 A is an alcohol that has the following percentage composition by mass: carbon 60.0%, hydrogen 13.3% and oxygen 26.7%.

[Ar values: C = 12.0, H = 1.0, O = 16.0]

a i Calculate the empirical formula of A. [2]
ii The molecular mass of A is 60. Calculate its molecular formula. [1]

b Give the structural formulae of the isomers of A. [2]

c If A is heated with a mixture of sulfuric acid and potassium dichromate(VI), there is
a colour change and propan­2­one is formed. Identify A and explain your answer. [2]

d Give the structural formulae for the organic products from the reaction of A with the following reagents:
i heated pumice or concentrated sulfuric acid [1]

ii ethanoic acid in the presence of an acid catalyst. [1]
2 B and C are isomeric alcohols with the molecular formula C4H10O.

a Draw the four skeletal formulae for the isomeric alcohols with this molecular formula. [4]
b When B is heated with a mixture of sulfuric acid and potassium dichromate(VI), there
is no change in colour.
i Identify B and explain your answer. [2]
ii Give the name, and the structural and skeletal formulae, of the alkene formed
when B is dehydrated. [3]
c When C is passed over heated pumice, it gives the same alkene as B does.
i Give the structural formula and name of C. [2]
ii What is formed when C is refluxed with an excess of acidified potassium
dichromate(VI) mixture? [1]
iii Explain the term reflux. [1]
3 D has the following percentage composition by mass: 53.3% carbon, 11.1% hydrogen and 35.6% oxygen.
[Ar values: C = 12.0, H = 1.0, O = 16.0]
a i Calculate the empirical formula of D. [2] ii The relative molecular mass of D is 90. What is its molecular formula?
Show your working.
b D is a diol. What does the term diol mean?
[2] [1]

1 Answer

3 votes

1. a)
i) To calculate the empirical formula of A, we assume 100 g of the compound. Therefore, the mass of each element in the compound is:
Carbon = 60.0 g
Hydrogen = 13.3 g
Oxygen = 26.7 g

We can then convert these masses to moles by dividing by their respective atomic masses:
Carbon: 60.0 g / 12.0 g/mol = 5.0 mol
Hydrogen: 13.3 g / 1.0 g/mol = 13.3 mol
Oxygen: 26.7 g / 16.0 g/mol = 1.67 mol

We divide each of these moles by the smallest number of moles, which is 1.67, to get the simplest whole-number ratio:
Carbon: 5.0 mol / 1.67 mol = 2.99 ≈ 3
Hydrogen: 13.3 mol / 1.67 mol = 7.96 ≈ 8
Oxygen: 1.67 mol / 1.67 mol = 1

Therefore, the empirical formula of A is C3H8O.

ii) The molecular mass of A is 60, which is equal to the empirical formula mass. Therefore, the molecular formula is the same as the empirical formula: C3H8O.

b) There are two isomers of A: propanol-1 (1-propanol) and propanol-2 (2-propanol).

c) A is propanol-2 (2-propanol). When heated with a mixture of sulfuric acid and potassium dichromate(VI), propan-2-one is formed via oxidation of the alcohol functional group.

d)
i) When heated with pumice or concentrated sulfuric acid, A undergoes dehydration to form propene (propylene):

CH3CH(OH)CH2OH → CH3CH=CH2 + H2O

ii) When reacted with ethanoic acid in the presence of an acid catalyst, A undergoes esterification to form propyl ethanoate:

CH3CH(OH)CH2OH + CH3COOH → CH3COOCH2CH2CH3 + H2O

2. a)
There are four isomers of C4H10O:
- Butan-1-ol (1-butanol)
- Butan-2-ol (2-butanol)
- 2-methylpropan-1-ol (isobutanol)
- 2-methylpropan-2-ol (tert-butanol)

b)
i) B is butan-2-ol (2-butanol). When heated with a mixture of sulfuric acid and potassium dichromate(VI), there is no change in colour because butan-2-ol cannot be oxidized by the dichromate ion.
ii) The alkene formed when butan-2-ol is dehydrated is but-1-ene (1-butene):

CH3CH(OH)CH2CH3 → CH2=CHCH2CH3 + H2O

c)
i) C is butan-1-ol (1-butanol).
ii) When refluxed with an excess of acidified potassium dichromate(VI) mixture, C is oxidized to butanone (methyl ethyl ketone):

CH3CH2CH2CH2OH + [O] → CH3C(O)CH2CH3 + H2O

iii) Reflux is a process in which a reaction mixture is heated to boiling and the vapours are condensed and returned to the reaction vessel. This ensures that the reaction proceeds to completion and that no volatile components are lost.

3. a)
i) To calculate the empirical formula of D, we assume 100 g of the compound. Therefore, the mass of each element in the compound is:
Carbon = 53.3 g
Hydrogen = 11.1 g
Oxygen = 35.6 g

We can then convert these masses to moles by dividing by their respective atomic masses:
Carbon: 53.3 g / 12.0 g/mol = 4.44 mol
Hydrogen: 11.1 g / 1.0 g/mol = 11.1 mol
Oxygen: 35.6 g / 16.0 g/mol = 2.23 mol

We divide each of these moles by the smallest number of moles, which is 2.23, to get the simplest whole-number ratio:
Carbon: 4.44 mol / 2.23 mol = 1.99 ≈ 2
Hydrogen: 11.1 mol / 2.23 mol = 4.98 ≈ 5
Oxygen: 2.23 mol / 2.23 mol = 1

Therefore, the empirical formula of D is C2H5O.

ii) The relative molecular mass of D is 90. The empirical formula mass of C2H5O is 45 (2 x 12.0 + 5 x 1.0 + 16.0 = 45). Therefore, the molecular formula of D is twice the empirical formula: C4H10O2.

b) A diol is a molecule that contains two hydroxyl (-OH) functional groups. Therefore, D is a diol.
User Samuel Toh
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