my favourite pet is the mole!

Before you begin to refer to the suggested solutions, some reminders:

• Please get your file ready for “File Check”. Submit your files on 9 September, Tuesday.
• You may download the following from the portal.
  1. 2006 and 2007 Promo papers for revision practice
  2. Test 2 suggested solutions
  3. Kinetics Tutorial suggested solutions
  4. TYS list, based on the 2008 edition. (These are the questions to skip during your revision.)

Question 1

Part (a)

Comparing experiment 1 and 2, keeping the [HCl] constant, when the [sucrose] multiplied 1.5 times, initial rate also multiplied 1.5 times. Hence, reaction is first order with respect to sucrose. (Do NOT say “increase BY 1.5 times”. Also, it’s NOT “first order with respect to [SUCROSE]“.)

Comparing experiment 1 and 3, keeping [sucrose] constant, when [HCl] is doubled, initial rate also doubled. Hence, reaction is first order with respect to HCl.

k = 2.4 mol^-1 dm³ s^-1

Part (b)

(If the question is worth more than 2 marks, please draw the Mazwell-Boltzmann diagram. In the diagram, make sure:

• your curve does NOT resemble a normal distribution curve
• label the E_a,_cat and E_a,_uncat BOTH on the x-axis and in the LEGENDS
• label your axes WITH UNITS)

1. Catalyst is a compound which alters the reaction rate by providing an alternative reaction pathway with lower E_a.
2. With lower E_a, fraction of particles with energy ≥ E_a increases. (It’s NOT “greater than E_a“, it should be ”greater than OR EQUAL TO E_a“.)
3. Frequency of collisions increases.
4. Frequency of effective collisions increases. (It’s NOT “number of effective collisions”, it should be “number of effective collisions PER UNIT TIME“. Else just use “frequency of effective collisions”. Also, there’s NO such thing as “frequency of effective collisions per unit time”. It’s either one or the other.)
5. Hence, rate of reaction increases.

Question 2

Part (a)

Its important to remembe that the “order of reaction” is defined with respect to the reactants. Otherwise, the definition was ok.

Part (b)

Check these in your graph:

• A: Axes should be labelled WITH UNIT. The x-axis of “time” has the units of “minute”, NOT “s”!.
• S: Scale used should result in a big enough graph. Also, scale used should not be “odd”.
• P: Plotting of points.
• G: Graph drawn should be sharp (don’t use a blunt pencil), smooth, without double lines or discontinued line. All points should be passed through. Some uses flexible curves as drawing aid which results in the line not crossing one or two points. The flexible curve should be a help, not an hindrance. Don’t use it if you can.

Marking Points:

• Half-life is constant,
• at 40 minutes. (approximately there)
• Reaction is first order with respect to RCOOC₂H₅.
• (Dotted construction lines showing at least 2 half-lives in the graph. The values of the respective half-lives need to be mark out/ written on the graph, with the units.)

Part (c)

rate = k [RCOOC₂H₅] [HCl]

units of k = mol^-1 dm³ min^-1 (Time here is NOT “second”! You lose an easy mark just like that!)

(It is not necessary to calculate the half-life, although there are some who do so. If you do, ”t_½ = ln 2 / k” is NOT true. The overall order is second order, NOT first!)

Part (d)

The mechanism suggest that one molecule of RCOOC₂H₅ and one molecule of HCl is involved in the slow step of the mechanism.

(Note: This is actually a simplification of what may actually happen. Whichever the case, the rate certainly doesn’t tell us about the fast step, nor does it tells us how many steps there are.)

These are the explanations for the various MCQs in the recent Mid-year. Only questions with many answering incorrectly are presented as follow. For the rest of the questions not presented, please se Mr. Tong if you need help.

Question 1

The question asked for “electrons“, not “atoms of C”!

Number of mole of C = 4.0 / 12.0 = 0.3333 mol

One C atom has 6 electrons.

Number of electrons in 0.3333 mol of C = 0.3333 x 6 x 6.02 x 10²³ = 1.20 x 10²⁴ (Answer: C)

Question 3

Assume mass, in g, of SO₃ is x. Hence, the mass of NO₂ is 20-x.

Number of mole of SO₃ = x/80.1

Number of mole of NO₂ = (20-x)/46.0

1H₂SO₄ ≡ 2H⁺ and 1HNO₃ ≡ 1H⁺.

Number of mole of H⁺ = 0.450 = 2x/80.1 + 0.5(20-x)/46.0

Solving for x, x = 16.5 g

Hence, percentage of SO₃ = 82.5% (Answer: C)

Question 4

Write out balanced equation for all 4 compounds, the one which needs 4 mol of oxygen for every mole of organic compound burnt is CH₃COCH₃. (Answer: C)

Question 5

angle of deflection α |charge/mass|

|charge/mass| of Mn²⁺ = 2/54.9 = 0.03643

|charge/mass| of Se^2- = 2/79.0 = 0.02532

Let x be the angle of deflection of Se^2-.

0.03643/0.02532 = 15/x

x = 10.43 ≈ 10^o, to the nearest degree (Answer: B)

Question 6

Dissappointing for those who got it wrong. It’s a very simple question, as long as you remember these 2 things:

1. Electronic configuration for the neutral Cu atom must be drawn first, before that of Cu⁺ ion is obtained. You cannot calculate the number of electron first and then write the configuration!
2. Cu and Cr has slightly different manner of writing their electronic configuration. These two are the exceptions in the syllabus!

Cu atom: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹

Cu⁺ ion: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰

Hence, there is no half-filled orbital. (Answer: A)

Question 9

The shapes of the various compounds given are, stated in order of them appearing in the various options:

A: trigonal planar, trigonal pyramidal, trigonal pyramidal

B: linear, bent, bent

C: linear, bent, bent

D: tetrahedral, tetrahedral, tetrahedral (Answer: D, since all the species have the same shape.)

Question 10

|LE| α |(q₊ x q_-)/(r₊ + r_-)|

You have to compare the 4 parameters one by one, giving priority to the charge rather than the radius, if need be. You cannot substitute the actual values into the formula to get a number for comparison!

Mg²⁺ has a higher cationic charge than Na⁺. Hence, option C and D are eliminated.

O^2- has smaller radius than S^2-. (Answer: A, since MgO has the largest magnitude of lattice energy.)

Question 11

The phrase “poor conductor of electricity” eliminated option B and C.

Melting at 317 K (or 44 ^oC) and boiling at 553 K (280 ^oC) are relatively low temperatures! Note that these are temperature in K, not ^oC! Hence, answer is D: simple molecular.

Question 13 and 15 will be discussed during lessons, together with a few other questions that is specific only to the classes concerned. (Meaning to say, only one or two classes need to get them reviewed, and hence they are not posted here.)

Question 1 of s17/ s21

MgCO₃ → MgO + CO₂

CaCO₃ → CaO + CO₂

Mass of MgCO₃ = (47.3 / 100) x 78.3 = 37.04 g

Amount of MgCO₃ = 37.04 / 84.3 = 0.4394 mol = Amount of CO₂ produced from MgCO₃

Mass of CaCO₃ = (52.7 / 100) x 78.3 = 41.26 g

Amount of CaCO₃ = 41.26 / 100.1 = 0.4122 mol = Amount of CO₂ produced from CaCO₃

Total amount of CO₂ produced = 0.4122 + 0.4394 = 0.8516 mol

Total volume of CO₂ at rtp = 20.4 dm³

(this question is skipped, please see Mr. Tong if you need help)

MnO₄^- ≡ 5e^-

Number of mole of electron accepted by MnO₄^- = 1.16 x 10^-3 x 5 = 5.80 x 10^-3 mol = Number of mole of electron of electron released by Xⁿ⁺

Hence,  1.93 x 10^-3 Xⁿ⁺ ≡ 5.80 x 10^-3 electron

1 Xⁿ⁺ ≡ 3 electron

Oxidation number of X in XO₃^- = +5 (after each Xⁿ⁺ gives out 3 electron)

n = 5 – 3 = 2 

Number of mole of MnO₄^- in 250 cm³ of FA3 = (42.00 / 1000) x 0.100 = 4.20 x 10^-3 mol

Number of mole of MnO₄^- in 21.65 cm³ of FA3 = 4.20 x 10^-3 x (21.65 / 250) = 3.637 x 10^-4 mol

1 MNO₄^- ≡ 5 Fe²⁺

Amount of Fe²⁺ in 25.0 cm³ of FA2 = 3.637 x 10^-4 x 5 = 1.819 x 10^-3 mol

Amount of Fe²⁺ in 250 cm³ of FA2 = 1.819 x 10^-3 x (250 / 25.0) = 1.819 x 10^-2 mol = Number of mole of anhydrous FeSO₄ in 250 cm³ of FA2

Mass of anhydrous FeSO₄ in 250 cm³ of FA2 = 2.762 g

Amount of water of crystallisation in 250 cm³ of FA2 = (4.733 – 2.762) ÷ 18.0 = 0.1095 mol

x = 0.1095 / 1.819 x 10^-2 = 6

Please refer to the experiment A or B, whichever it is that concerns you.

Assume these values:

volume of FA1 used = 44.50 cm³

average titre value of FA3 used = 22.20 cm³

Part 5(i)

Concentration of MnO₄^- of FA3 = 2.85 / 158.0 = 0.01804 = 0.0180 mol dm^-3

Part 5(ii)

Amount of MnO₄^- in 22.20 cm³ of FA3 = 0.01804 x (22.20 / 1000) = 4.0049 x 10^-4 = 4.00 x 10^-4 mol

• Some ppl confuse the 22.20 cm³ here with 25.0 cm³. The 25.0 cm³ is for Fe²⁺ in FA2, not MnO₄^- in FA3!
• Many ppl wrote “Amount of MnO₄^- that react with ….”, please provide the volume of MnO₄^- and the FA_ in the statement.

Part 5(iii)

1MnO₄^- ≡ 5Fe²⁺

Amount of  Fe²⁺ in 25.0 cm³ of FA2 = 5 x 4.0049 x 10^-4 = 2.002 x 10^-3 mol

Amount of Fe²⁺ in 250 cm³ of FA2 =  2.002 x 10^-3 x (250 / 25.0) = 2.002 x 10^-2 = 2.00 x 10^-2 mol 

•  Don’t keep rounding off your intermediate values. If not, you’ll end up with a significant error eventually.
• Maximum sf for intermediates is 5.

Part 5(iv)

Amount of Fe²⁺ in 250 cm³ of FA2 = Amount of Fe²⁺ in 44.50 cm³ of FA1 = 2.002 x 10^-2 mol

Amount of Fe²⁺ in 1 dm³ of FA1 = 2.002 x 10^-2 x (1000 / 44.50) = 0.4499 mol

Mass of Fe²⁺ in 1 dm³ of FA1 = 0.4499 x 55.8 = 25.1 g

• Many ppl conveniently forget to convert for dilution involving from 44.50 cm³ of FA1 to 250 cm³ of FA2.

Assume these values:

Volume of FA1 = 44.50 cm³

Average titre value of FA4 used = 23.55 cm³

Part 9(i)

Concentration of Cr₂O₇^2- in FA4 = 3.24 / 216.0 = 0.0150 mol dm^-3

• Many ppl used 294.2 instead of 216.0, because they took the M_r for K₂Cr₂O₇ instead. The question gave 3.24 g mol^-1 for Cr₂O₇^2-, not for K₂Cr₂O₇!

Part 9(ii)

Amount of Cr₂O₇^2- in 23.55 cm³ of FA4 = 0.0150 x (23.55 / 1000) = 3.533 x 10^-4 = 3.53 x 10^-4 mol

• Some ppl confuse the 23.55 cm³ here with 25.0 cm³. The 25.0 cm³ is for Fe²⁺ in FA2, not Cr₂O₇^2- in FA3!
• Many ppl wrote “Amount of Cr₂O₇^2- that react with ….”, please provide the volume of Cr₂O₇^2- and the FA_ in the statement.

Part 9(iii)

1Cr₂O₇^2- ≡ 6Fe²⁺

Amount of  Fe²⁺ in 25.0 cm³ of FA2 = 6 x 3.533 x 10^-4 = 2.12 x 10^-3 mol

Amount of Fe²⁺ in 250 cm³ of FA2 =  2.12 x 10^-3 x (250 / 25.0) = 2.12 x 10-2 mol 

•  Don’t keep rounding off your intermediate values. If not, you’ll end up with a significant error eventually.
• Maximum sf for intermediates is 5.

Part 5(iv)

Amount of Fe²⁺ in 250 cm³ of FA2 = Amount of Fe²⁺ in 44.50 cm³ of FA1 = 2.12 x 10^-2 mol

Amount of Fe²⁺ in 1 dm³ of FA1 = 2.12 x 10^-2 x (1000 / 44.50) = 0.4764 mol

Mass of Fe²⁺ in 1 dm³ of FA1 = 0.4764 x 55.8 = 26.6 g

• Many ppl conveniently forget to convert for dilution involving from 44.50 cm³ of FA1 to 250 cm³ of FA2.

Have to give credits to many people. The report and presentation is definitely improving. Well done!

General

1.  Some do not know how to use the weighing balance. If you use the whole volume of the burette column of FA2 during titration and still did not get an end-point, you’re having this problem. It’s because you only took 2+ g of FA1 when you thought you took 6 g. Look for me please, MAJOR error!
2. Please present your answer to standard form from now on.
3. There’s no need to copy the question when doing calculation. On the other hand, there’s a need to follow the format of the report.

Calculation

Some of the error during calculation was not being clear with whether a certain mass is in 250 cm³ or 1 dm³. E.g. the mass of 6 g is in 250 cm³, not 1 dm³. 

• Part 6: Present answer to 3 sf, and to standard form.
• Part 8(ii): The opposite of “anhydrous” is not “hydrous”! It’s “hydrated”.
• Part 8(iii): “Mass of water in 1 dm³ of FA2 =” is a wrong statement. It should be something like, “Mass of water of crystallisation in 1 dm³ of FA2 =”. The first is wrong because “water” includes all the water in the solution and not just that of the water of crystallisation, and it’s an awfully a lot of water! This is not what we want, we only want that associated with Na₂S₂O₃.
• Part 10: After obtaining the value of x, use the new formula of Na₂S₂O₃.5H₂O to calculate the percentage by mass of water of crystallisation. Do not use the other values calculated previously in Part 8. Meaning to say, if you obtain x =5, the percentage would be 36.3%, regardless of your calculated values in previous parts.

Some common mistakes were still recurring. Refer to the previous post on “Experiment 1: Common Mistake”. Some more notable reminders again are: 

1. Writing clear and complete statements. Examples:
   • For number of mol: “Amount of NaOH in 22.45 cm³ of FA3 =”
   • For concentration: “Concentration of NaOH in FA3 =”
   • For mass: “Mass of NaOH in 22.45 cm³ of FA3 =”
2. Using the correct dp and sf, whether in recording, calculation workings or final answers.
   • All burette readings to 2 dp.
   • All pipette readings to 1 dp.
   • All A_r, M_r or molar mass to 1 dp.
   • All mass (in this experiment) to 3 dp, although 2 dp is also acceptable. You need to be consistent, whether 2 or 3.
   • All final answers to 3 sf, except for the above.
   • All sf during calculation workings not to exceed 5 sf, except for the above.

One new comment to be highlighted:

• Do not use short form of words. E.g. “amt”, “vol” and “conc” should be spelt out fully as “amount”, “volume” and “concentration”.

•  The question stated the mass to be “about 1.4 g”. The suggested range is then 1.35 to 1.45 g, do not exceed this range.

Experiment A (H₂C₂O₄.2H₂O and NaOH)

1. In calculation Part (8), the mass of 1.4 g is dissolved in 250 cm³ of solution, not in 25.o cm³ which is the volume taken using pipette (or 10% of the amount).
2. Also in Part (8), the formula of ethanedioic acid given is H₂C₂O₄.2H₂O. Do not ignore the water of crystalisation in the formula when calculating M_r. Hence, the M_r is 126.0 and not 90.0.

Experiment B (H₂C₂O₄.2H₂O and KMnO₄)

1. The end-point colour change for Experiment B is colourless to “pale pink”, not “pale purple”. The pale pink is due to the production of Mn²⁺ ion, which is pale pink. [In actual fact, the pale pink is hardly noticeable. That's why too dark a shade pink means that you've definitely over-titrated, so is purple.]
2. Manganate(VII) ion is MnO₄^- and manganese(II) ion is Mn²⁺. Do not confuse the two.
3. In calculation Part (15), the question asked for “number of mol” and not “concentration”. Also, question asked for “KMnO₄” and not “MnO₄^-“. Please make sure you’re answering the question. Although the differences between them doesn’t seem to be much, they’re quite very different.

1. Pipette readings are recorded to 1 dp and burette readings are recorded to 2 dp, not just during tabulation but also during workings. In other words, it is the case for the WHOLE paper.
2. ONLY titre values that are consistent ≤ 0.10 cm³ are chosen. Please demonstrate common sense when selecting consistent titre values. If you have 22.50 cm³, 22.50 cm³ and 22.60 cm³ , you only take both of the 22.50 cm³, not all three values! Calculation of average titre values must be shown, even when the values used are the same.
3. Clear and specific statements MUST be written.
   • For amount: e.g. “No. of mole of HCl in 27.85 cm³ of FA2 = “
   • For concentration: e,g, “Concentration of HCl in FA2 = “
   • There’s no such thing as “mol = “
4. Units must always accompany all numerical answer, unless there’s no units. The word or unit “mols” doesn’t exist.
5. Usage of “→” is only reserve for chemical equation. It is not interchangable with “=”. Hence:
   • “Amount of HCl in 27.85 cm³ of FA2 → 0.00250 mol” is wrong.
   • The presentation “27.85 cm³ → 0.00250 mol” is improper. Nor is “27.85 cm³ = 0.00250 mol”, saying so is like saying “2 pens = 7 egg yolks”. Doesn’t make sense to you, eh? Me too.
6. All answers should be presented to 3 sf. You should use the calculator values in subsequent calculations, but please do not copy out the whole string of numbers onto the paper. Record/ write a maximum of 4 or 5 sf in workings, and 3 sf in answers (except for A_r or M_r, to 1 dp).
7. In this experiment, there’s no dilution between Q8 ad Q9 (both are FA2 of the same concentration). Dilution is only between Q9 (FA2, diluted) and Q10 (FA1, concentrated). A very clear mind and understanding of these details will get you the right calculations and answer.

08s17 and 08s30:

• final calculated value for concentration of HCl in FA1: 1.019 ≤ [HCl] ≤ 1.035 mol dm^-3; except that your final value should be presented, as stressed numerous times by now, to 3 sf.

08s21:

• your solution used on Monday was more diluted than intended. It’s slightly more than 0.800 mol dm^-3, rather than the suggested value of about 1.00 mol dm^-3. Exact value not available.

Hi there! By the time you successfully see this, you’d have already done the following:

1. did the ice-cream stick thing.
2. select your buddy and buddy-pair.
3. got the handouts (one white, one pink, one yellow). Pls file the pink and yellow. The white is to be submitted to me first, and will be returned to you shortly.
4. have your periodic table, since a couple of weeks ago!
5. done your stoichiometry and redox assignment. (it’s meant to be done during the holidays!)

Now, you may or need to do these.

1. If you’ve provided information for the setting up the ”edublogs.org”,  please check the email you provided. The information to activate your blog should have been sent to you. Please activate your blog.
2. If a blog account has been created for you and activated by you, you may want to change the password of that account. Go to: Users > Your Profile. If not created yet, pls be patient.
3. You may want to change that blog’s background/ wallpaper/ whatever-you-call-it. Go to: Presentation > Themes. Again, if not created yet, pls be patient. (Do not create one yourself. We’re trying to link everyone in the class.)
4. You need to email me at tong_yong_kiang@moe.edu.sg with ALL of these info:

• your handphone number
• your birthdate
• your buddy (one name here)
• your buddy-pair (2 names here, 3 with approval)
• one of your parents’ handphone number (state father or mother; or if guardian, state relationship)

Your email subject should be: “[08sxx] name”, where “xx” is your ct and “name” is your name. Please make sure that this email account is the one that can reach you, not your junkmail account.

Announcement:

1. You can find the SPA generic mark scheme in the college online portal. You should get your portal username and password within these 2 days. The generic mark scheme will be available at every SPA assessment.
2. You have to download and print your own notes and tutorials for Chemistry from the online portal from now onwards. Reminder: You should have your Atomic Structure notes by now!
3. Please remember to bring your practical worksheets EVERY WEEK for lab sessions from this week onwards, unless otherwise stated.

After you’ve finished doing and read all the above, exclaimed ”that silly fox just jumped over the lazy cat!” Your friends around will wonder whether you’re crazy, but it’s ok. They’ll get it sooner or later. Do it, so that I know you are done!

Lastly, if the “edublogs.org” account has been created for you, please proceed to the discussion forum page and brainstorm your suggestion to the forum question there. You should post something!

 Let’s make Chemistry fun! You’ll either hate it or like it, so why not make the best of it and love it? And all of us need to do our part, you and me, no exception…