Edexcel AS Chemistry Unit 1 22/5/2015 Unofficial Mark Scheme

Watch
C0balt
Badges: 16
Rep:
?
#1
Report Thread starter 5 years ago
#1
Completed Unit 2 mark scheme here: http://www.thestudentroom.co.uk/show....php?t=3375301
Thanks to sat_freak I can now attempt to make a complete unofficial mark scheme for unit 1. However, my unit 2 performance was better than unit 1, so the quality of this mark scheme may be lower than unit 2.


Multiple Choice - 20 marks


Question 17 - Mass spectrometry - 16 marks

Part (a): Bromine has two isotopes, Br-79 and Br-81. Explain the term isotopes, by reference to sub-atomic particles. (1)
- Atoms of the same element with the same number of protons but different number of neutrons.

Part (b): The presence and abundance of these isotopes can be determnied by using a mass spectrometer such as that shown in the diagram below.
(i) Explain how ions are produced in the area labelled A. (2)
- High energy electrons are bombarded on the gaseous sample, and an electron is knocked off.

(ii) State what is used to deflect the ions moving through the mass spectrometer in the area labelled B. (1)
- By an electromagnet.

(iii) Explain why there is a vacuum in a mass spectrometer. (1)
- To ensure that the ions do not collide with air molecules.

Part (c): Complete the mass spectrum below for a sample of bromine gas that contains approximately half Br-79 isotope and half Br-81 isotope.
- Peak at 79 and 81
- of the same height
- Peak at 158, 160 and 162
- 1 : 2 : 1 height for the molecular ion peaks

Part (d): Calculate the relative atomic mass of bromine for a sample which was found to contain 47.0% Br-79 and 53.0% Br-81. Give your answer to thee significant figures. (2)
- 80.1

Part (e): What would be the effect, if any, on the m/e value of the peak if the ion detected had lost two electrons rather than one electron? (1)
- m/e value will be halved.

(f) One of the uses of mass spectrometers is for the detection of banned substances, such as anabolic steroids, in a blood or urine sample taken from competitors in sports events.
(i) Suggest two precautions that are necesary to ensure that the result of any analysis would be valid. (2)
This is the only question I have no idea how the mark schemes will be like (sat_freak's suggestions below)
- Samples are fresh / not contaminated / collected properly with witnesses and all legal requirements.
- Proper ID / tagging / labelling of the samples is very crucial as the stakes are too high.
- Transport & storage of samples.
- Equipment calibration / Use of control.
- Accuracy, precision, reliability, resolution of equipment

(ii) These substances can give competitors an unfair advantage. Sugges twhy the use of these substances may be of concern to the user. (1)
- Side effects on health.

Part (g): Suggest one other use for mass spectrometers. (1)
- Determining the relative atomics mass.
- Carbon dating
- Space research etc



Question 18 - Organic chemistry - 20 marks

Alkenes and cycloalkanes have the same general formula, but react very differently with halogens.
Part (a) Give the general formula that applies to both alkene and cycloalkanes. (1)
- CnH2n

(b) Using structural formulae, complete the overall equation for the reaction of an alkene of your own choice, containing fewer than four carbon atoms, with liquid bromine. Name the product. (3)
- CH2=Ch2 + Br2 --> CH2BrCH2Br
- 1,2-dibromoethane

(c) An example of an alkene with six carbon atoms is 2-methylpent-1-ene. It reacts with chlorine by means of an electrophilic addition reaction. The diagram below shows a student's attempt at drawing the mechanism for this reaction.
(i) Identify the three errors in this student's drawing of the mechanism. (3)
- The partial charges on the chlorine molecule are other way around.
- The double headed arrow must be from the double bond to the top chlorine atom.
- A - charge is lacking

(ii) The structure of the carbocation intermediate is correctly drawn. Explain why the positive charge is on the carbon atoms how. (1)
- The carbon atom has lost an electron from the double bond to the chlorine. (not sure about the exact wording)
- Tertiary carbocation is more stable than primary carbocation.
Honestly the wording of this question is insanely vague. Primary carbocation CAN form but it just more unlikely, so my interpretation to this question was why that particular carbon has a positive charge (rather than let's say, the entire molecule).

Part (d) There are five possible cycloalkanes, each containing five carbon atoms. Three of the isomers are given below. Complete the other two boxes, by adding the skeletal formulae of the other two structural isomers. (2)
http://imgur.com/VyMI6Gz

Part (e) Define the them structural isomerism. (1)
- Structural isomers have the same molecular formula but different arrangement of atoms.

Part (f) Another example of a cycloalkane is cyclobutane. This compound, like or the cycloalkanes, can also react with chlorine.
The overall reaction of cyclobutane with chlorine is as follows:
C4H8 + Cl2 --> C4H7Cl +HCl
(i) This reaction can occur at room temperature an pressure. What further condition is needed for this reaction to take place? (1)
- UV light (sunlight is usually allowed)

(ii) Using the appropriate arrows, complete the equation for the initiation step of the reaction mechanism for the reaction of chlorine with cyclobutane. (2)
- Single headed arrows from the bond to each of the chlorine atoms.
http://imgur.com/j2ElrD2

(iii) Using molecular formulae, write equations fore the two propagation step for this mechanism. (2)
- C4H8 + Cl. --> C4H7. + HCl
- C4H7. + Cl2 --> C4H7Cl +Cl.

(iv) Name the type of bond fission which occurs in these propagation steps. (1)
- Homolytic fission

(v) There are also termination steps in this mechanism. Explain how these differ from the other steps in the mechanism and why these result in the reaction ending. (2)
- Two free radicals come together
- No free radical is generated

Part (g) If the reaction with cyclobutane is carried out with an excess of chlorine how are the products of the reaction affected? (1)
- More dichloro, trichloro...formed


Question 19 - Sodium chloride - 10 marks

Part (a) Draw the electron density map for a chlorine molecule to show covalent bonding. (1)
http://imgur.com/1NI97hE

Part (b) Sodium chloride is ionically bonded. What is meant by the term ionic bond? (1)
- The electrostatic attraction between a cation and an anion

Part (c) Electrolysis is an experiment which you could carry out in a school or college laboratory on an aqueous solution of sodium chloride, to provide evidence for the presence of ionic bonding.
Draw a labelled diagram of the apparatus that you would use for this experiment, indicating how your results would show that the bonding was ionic. (3)
- An electrolysis setup
- by means of showing current (a bulb, an ammeter etc)
- What you would see if the bonding is ionic

Part (d) Chlorine gains an electron when it reacts with soidium to form sodium chloride.
(i) Draw the dot and cross diagram of a chloride ion showing outer electrons only. (1)
http://imgur.com/5HViVPo

(ii) Give the formula of an ion from Period 3 that is isoelectronic with the chloride ion. (1)
- S2-

Part (e) Sodium and sodium chloride can both be good conductors of electricity. Under what conditions do these substances conduct electricity? Compare the method of conductivity in each case. (3)
- Sodium in solid and liquid states, sodium chloride in aqueous and liquid state.
- In sodium delocalised electrons which are free to move carry charge
- In aqueous and liquid sodium chloride the ions are mobile and they carry charge


Question 20 - Enthalpy change - 14 marks

The reaction of calcium oxide with hydrochloric acid is an exothermic reaction.
CaO(s) + 2HCl(aq) --> CaCl2(aq) + H2O(l)
In an experiment to investigate this reaction, the following procedure was carried out.
1. 50.0cm^3 of hydrochloric acid, concentration 2.0 mol/dm^3 (an excess) was pipetted into a polystyrene cup and the initial temperature measured using thermometer with 0.5C graduations.
2. 1.46g of calcium oxide powder was weighed out and added to the acid. The mixture was stirred and the maximum temperature measured.
Max: 35.0C
Initial: 19.5C

Part (a): Calculate the enthalpy change, in joules, for the quantities in this experiment. Assume that the specific heat capacity of the solution is 4.18J/g/C. (1)
Q = 50.0 x 4.18 x (35.0-19.5) = 3239.5J

Part (b): Using your answer from (a), calculate the molar enthalpy change for the reaction between calcium oxide and hydrochloric acid. Include a sign and units in your answer. (2)
n(CaO) = 1.46/(40.1+16.0) = 0.026mol
ΔH = -3239.5/0.026mol ÷ 1000 = -124.5kJ/mol

Part (c) The standard molar enthalpy change for the reaction between calcium oxide and hydrochloric acid is -196.8kJ/mol
(i) Suggest three reasons why the calculated value in part (b) is different from this value (3)
- Density of the solution was assumed to be the same as water
- Heat loss from the container
- Impurity in calcium oxide

(ii) Using the standard enthalpy change of -196.8Kj/mol, calculate the minimum mass of calcium oxide that would be needed to raiseh the temperature of 250cm^3 of hydrochloric acid (an excess) by 25.0C. (3)
- 7.45g
http://imgur.com/yd6ygz5

(d) The reaction of calcium carbonate with hydrochloric acid has the following standard molar enthalpy change.
ΔH = -18.8kJ/mol
This value can be used, with the enthalpy change for the reaction of calcium oxide with hydrochloric acid, to determine the enthalpy change for the thermal decomposition of calcium carbonate. This cannot be measured directly.
(i) Complete the Hess energy cycle below by adding the missing arrow and entities.
Use the cycle, and the standard enthalpy change for the reaction of calcium oxide and hydrochloric acid (-196.8kJ/mol), to determine the standard enthalpy change for the decomposition of calcium carbonate. (4)
- +178kJ/mol
http://imgur.com/x29XZLs


(iii) Complete and label the enthalpy level diagram below, for the series of reactions in (d)(i).
Your diagram does not have to be to scale. (1)
http://imgur.com/d938P5W

1
reply
Feraligatr
Badges: 20
Rep:
?
#2
Report 5 years ago
#2
:woo:
0
reply
GrandMasti
Badges: 11
Rep:
?
#3
Report 5 years ago
#3
Cheers C0balt
0
reply
sat_freak
Badges: 2
Rep:
?
#4
Report 5 years ago
#4
For Q 17 (c), as they specify that the sample is bromine gas, the peaks for 79 & 81 are probably not needed, as bromine is diatomic.
0
reply
sat_freak
Badges: 2
Rep:
?
#5
Report 5 years ago
#5
Q 18 (c) (ii) As the secondary carbocation is more stable than the primary (will form the major product).
0
reply
GrandMasti
Badges: 11
Rep:
?
#6
Report 5 years ago
#6
(Original post by sat_freak)
For Q 17 (c), as they specify that the sample is bromine gas, the peaks for 79 & 81 are probably not needed, as bromine is diatomic.
Yes but fragmentation? I dunno
0
reply
C0balt
Badges: 16
Rep:
?
#7
Report Thread starter 5 years ago
#7
(Original post by sat_freak)
For Q 17 (c), as they specify that the sample is bromine gas, the peaks for 79 & 81 are probably not needed, as bromine is diatomic.
No, you do need it.
By your argument, just because propane is a gas, there would only be molecular ion peak, which obviously isn't the case.
State of the sample does not affect the final spectrum. Even if the bromine liquid was put into the mass spectrometer, the first step is vaporisation so it would become a gas anyways. Which is then bombarded by electrons which knocks electrons off and this can break the bond between the atoms. Putting gaseous state bromine will not prevent this from happening, and end result is same as putting bromine liquid or even solid.

Posted from TSR Mobile
0
reply
GrandMasti
Badges: 11
Rep:
?
#8
Report 5 years ago
#8
(Original post by sat_freak)
Q 18 (c) (ii) As the secondary carbocation is more stable than the primary (will form the major product).
Tbh that question was worded very weirdly, and so it could be interpreted as literally WHY is there a positive charge on that carbon, since it has 'lost an electron' or why that carbon and nothing else i.e. what you said
0
reply
sat_freak
Badges: 2
Rep:
?
#9
Report 5 years ago
#9
Q17 (f) (i) Suggest two precautions that are necesary to ensure that the result of any analysis would be valid.

- Samples are fresh / not contaminated / collected properly with witnesses and all legal requirements.
- Proper ID / tagging / labelling of the samples is very crucial as the stakes are too high.
- Transport & storage iof samples.
- Equipment calibration / Use of control.
- Accuracy, precision, reliability, resolution of equipment
0
reply
C0balt
Badges: 16
Rep:
?
#10
Report Thread starter 5 years ago
#10
(Original post by sat_freak)
Q 18 (c) (ii) As the secondary carbocation is more stable than the primary (will form the major product).
It is tertiary carbocation

Posted from TSR Mobile
0
reply
C0balt
Badges: 16
Rep:
?
#11
Report Thread starter 5 years ago
#11
(Original post by GrandMasti)
Tbh that question was worded very weirdly, and so it could be interpreted as literally WHY is there a positive charge on that carbon, since it has 'lost an electron' or why that carbon and nothing else i.e. what you said
I said both but it's tertiary I believe lol

Posted from TSR Mobile
0
reply
GrandMasti
Badges: 11
Rep:
?
#12
Report 5 years ago
#12
(Original post by C0balt)
I said both but it's tertiary I believe lol

Posted from TSR Mobile
Wait, the wording of the question was: The structure of the carbocation is correctly drawn. Explain why the positive charge on the carbon atom shown.

So, I guess they may accept that the carbon atom has lost an electron in the formation of the C-Cl bond (since it is dative covalent). Although the tertiary argument is probably more correct. Is there any chance they may be lenient given the ambiguous wording of the question?
0
reply
Dinaa
Badges: 17
Rep:
?
#13
Report 5 years ago
#13
Thanks, have forgotten my answers though.


Didn't like that exam tbh :sad:
0
reply
C0balt
Badges: 16
Rep:
?
#14
Report Thread starter 5 years ago
#14
(Original post by GrandMasti)
Wait, the wording of the question was: The structure of the carbocation is correctly drawn. Explain why the positive charge on the carbon atom shown.

So, I guess they may accept that the carbon atom has lost an electron in the formation of the C-Cl bond (since it is dative covalent). Although the tertiary argument is probably more correct. Is there any chance they may be lenient given the ambiguous wording of the question?
Yeah it was so vague. I took it as why that particular carbon rather than, say, the entire molecule, has a positive charge
Honestly, primary can still form, just less likely, that's why I took the question as I said above. Because even if the student drew a primary carbonation it wouldn't have technically been incorrect.


Posted from TSR Mobile
0
reply
GrandMasti
Badges: 11
Rep:
?
#15
Report 5 years ago
#15
(Original post by C0balt)
Yeah it was so vague. I took it as why that particular carbon rather than, say, the entire molecule, has a positive charge
Honestly, primary can still form, just less likely, that's why I took the question as I said above. Because even if the student drew a primary carbonation it wouldn't have technically been incorrect.


Posted from TSR Mobile
Yeah. Hopefully I have got the mark since it does make sense!
0
reply
sat_freak
Badges: 2
Rep:
?
#16
Report 5 years ago
#16
(Original post by C0balt)
It is tertiary carbocation

Posted from TSR Mobile
Yeah sorry, it is tertiary. And tertiary is more stable than primary.
0
reply
sat_freak
Badges: 2
Rep:
?
#17
Report 5 years ago
#17
(Original post by GrandMasti)
Yeah. Hopefully I have got the mark since it does make sense!
Maybe they wanted to hear the word(s) : "inductive effect of the alkyl groups"

Or Markovnikov's rule.
0
reply
sat_freak
Badges: 2
Rep:
?
#18
Report 5 years ago
#18
(Original post by C0balt)
No, you do need it.
By your argument, just because propane is a gas, there would only be molecular ion peak, which obviously isn't the case.
State of the sample does not affect the final spectrum. Even if the bromine liquid was put into the mass spectrometer, the first step is vaporisation so it would become a gas anyways. Which is then bombarded by electrons which knocks electrons off and this can break the bond between the atoms. Putting gaseous state bromine will not prevent this from happening, and end result is same as putting bromine liquid or even solid.

Posted from TSR Mobile
Yeah that's correct but how would one figure out the size of the peak (relative intensity) of the atomic ions. For the molecular ions, as you said, 1:2:1 is evident. Gaseous bromine will still be mostly molecular.

As the question has four marks, I believe 3 marks for three peaks and one for the correct proportion between them.
0
reply
GrandMasti
Badges: 11
Rep:
?
#19
Report 5 years ago
#19
(Original post by sat_freak)
Maybe they wanted to hear the word(s) : "inductive effect of the alkyl groups"

Or Markovnikov's rule.
Ah well, it is only 1 mark. Also, can you please upload the unit 1 and 2 biology papers onto the biology thread. That would be great! Thanks
0
reply
C0balt
Badges: 16
Rep:
?
#20
Report Thread starter 5 years ago
#20
(Original post by sat_freak)
Yeah that's correct but how would one figure out the size of the peak (relative intensity) of the atomic ions. For the molecular ions, as you said, 1:2:1 is evident. Gaseous bromine will still be mostly molecular.

As the question has four marks, I believe 3 marks for three peaks and one for the correct proportion between them.
I don't think relative abundance between the atomic peaks and molecular peaks would be a marking point. Two atomic peak having the same height can be a marking point because it said half 81 and half 79
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Have you experienced financial difficulties as a student due to Covid-19?

Yes, I have really struggled financially (42)
18.26%
I have experienced some financial difficulties (66)
28.7%
I haven't experienced any financial difficulties and things have stayed the same (87)
37.83%
I have had better financial opportunities as a result of the pandemic (29)
12.61%
I've had another experience (let us know in the thread!) (6)
2.61%

Watched Threads

View All