VCE mid-year exams
Chris Dwyer helps VCE Chemistry students prepare for their mid-year exam on June 10
What you achieve in the 90 minutes after you start writing in Examination 1 will depend on how effectively you “interpret each question’’.
Given the “application of chemical principles” is the focus of the course, the performance on some questions on last year’s examinationmay provide some insight.
Section A Question 2.
Serotonin (C10H12N2O; molar mass = 176 g mol-1) is a compound that conducts nerve impulses in the brain and muscles. A sample of spinal fluid from a volunteer in a study was found to contain a serotonin concentration of 1.5 ng L-1 (1.5 nanograms per litre).
How many molecules of serotonin are there in one millilitre of the spinal fluid?
A. 5.13x109
B. 9.03x1011
C. 5.13x102
D. 9.03x1029
The main challenge in that question was to quickly recognise the steps:- convert ng to g; divide by molar mass; multiply by NA; divide by 1000. Only 43 % of students followed that pathway successfully.
Section A Question 3
Xylose is a compound that has five carbon atoms in each molecule and contains 40% carbon by mass.
What is the molar mass of xylose?
A. 30
B. 67
C. 150
D. It cannot be determined without further information.
This question can be done quite quickly if the information is effectively ‘interpreted’. 5 C atoms in 1 molecule indicates there is 5 mol C atoms in 1 mol xylose. So the mass of 5 mol C (5 x 12.0 g) is 40 % of the mass of 1 mol xylose. Only 42 % identified the correct answer, and 49 % selected alternative D.
Section B Question 1
This question described the method used in a gravimetric analysis and provided a series of data associated with that analysis. You needed to identify the data from which to start.
Section B Question 1
This question described an analysis using UV-visible spectroscopy and also provided experimental data, which had to be effectively interpreted.
I strongly recommend you attempt those questions soon and use them as a benchmark as to your current effectiveness in these areas.
Though you may feel that you have been totally immersed in spectroscopy, organic reaction pathways, biochemical fuels, biomolecules and DNA, do not forget that data interpretation and chemical calculations are fundamental skills that should still be practised.
Being able to write a balanced equation is a fundamental skill in VCE Chemistry. Because you study hydrocarbons and biochemical fuels it is expected that you can write balanced equations for the combustion of an alkane, or ethanol or a biodiesel such as methyl linoleate.
The formula for methyl linoleate – deduced from the formula of linoleic acid (C17H31COOH) provided in the data book is C17H31COOCH3.
The equation for its combustion is 2C17H31COOCH3(l) + 53O2(g) → 38CO2(g) + 34H2O(g).
The requirement to ‘Explain why ethanol produced by fermentation is referred to as a biochemical fuel’ was not well handled in last year’s exam.
Consider what you would write before checking the discussion in the Exam Assessment Report.
There was a strong emphasis on ‘drawing structural’ formulae, showing all bonds, on the 2008 examination.
Though this was well handled when alkanols, acids and esters were involved, Question 5 relating to amino acids, dipeptides and protein structure proved more challenging.
Since you are given the structure of the 20 amino acids in your data book, you are expected to be able to interpret and use those structures.
When required to write the molecular formula of phenylalanine most students faltered. It was apparent that students did not know how to interpret the benzene ring.
Ensure that you are able to show that the molecular formula is C9H11NO2, and that the molecular formula of tyrosine is C9H11NO3.
Because amino acids have a “basic” amino functional group and an “acidic” carboxyl functional group their structures are different at low and high pH.
Data provided by IR, NMR and Mass spectroscopy enable us to identify and distinguish between compounds.
There is a sample question and answer relating to this on this website.
One aspect of 1H NMR spectroscopy that can be confusing is the “splitting” of signals by H atoms on adjacent C atoms.
In the n+1 rule the n is the “number of equivalent neighbouring H atoms”.
In CH3CH2OH, the signal for the H atoms on CH3 is split into a triplet (2+1) by the 2 equivalent H atoms on the adjacent CH2.
The signal for the H atoms on CH2 is split into a quartet (3+1) by the 3 equivalent H atoms on the adjacent CH3.
1-propanol, CH3CH2CH2OH has four different H environments.
The signal for H on OH is a singlet, since splitting only applies to H atoms bonded to C. The signal for H on CH3 is split into a triplet (2+1) by the 2 equivalent H on the adjacent CH2. The signal for H on CH2 adjacent to the hydroxyl group is split into a triplet (2+1) by the 2 equivalent H on its adjacent CH2.
But when there are neighbouring hydrogen atoms in two or more ‘different’ environments the situation is more complex.
CH3CH2CH2OH
These 2 H atoms, above in bold, have 3 equivalent neighbouring H atoms on the adjacent CH3 and 2 different equivalent neighbouring H atoms on the adjacent CH2. Hence the signal will be split into a quartet (3+1) by the H atoms on CH3 and into a triplet (2+1) by the H atoms on CH2.
The effect of this complex “dual splitting” shows up on 1H NMR spectrum as a multiplet.
Often the best way to identify the “number of different C or H environments” is to draw the structural formula of the molecule. Can you show that:
(i)pentan-1-ol has 5 different C environments and 6 different H environments?
(ii)methyl propanoic acid has 3 different C environments and 3 different H environments?
Good luck in the exam.
Chris Dwyer teaches at Vermont Secondary College