Help with chemistry extension questions - due tomorrow.

[Mellyboo]

These are optional extension questions for my chem class. I would really appreciate some help! Although they are optional, it is imperative that I complete some or all of these questions, my chem teacher's respect for me went down the drain last year. I've attempted, but couldnt answer anything correctly. Even just some clues on how to answer some of these would be a lot of help!

http://imgur.com/Ahxcuj7
http://imgur.com/Bt2VBvc
http://imgur.com/LnBYlaK
http://imgur.com/1R6n1ij

The answers to the questions are(also found at the bottom of the last page):
1. N=2
2.X=10
3.1.8mol/L
4.73.9%
5.57.2%
6.66.3%
7.X=7

Thanks everyone.

[paper-back]

1)So the reaction is as said in the equation
We have the;
V(NaOH)=0.025L
C=0.1M
n=CV,
Therefore n=0.1x0.025=0.0025

0.0025/2=n(Succinic acid)
n(Succinic acid reacted)=0.00125
V(Succinic acid reacted)=0.0184L
V(Succinic acid in volumetric flask)=0.25L

So to get to the number of mol from the Succinic acid reacted to mol of Succinic acid in volumetric flask, we go;
V(Succinic acid in volumetric flask)/V(Succinic acid reacted)xn(Succinic acid reacted)
(0.25/0.0184)x0.00125
mol(Succinic acid in volumetric flask)=0.017mol

Using equation n=m/M
We now know that in the 2 grams of Succinic acid that was dissolved in the volumetric flask there was 0.017 mol of Succinic acid
Now we need molar mass, to do this we use equation n=m/M
Rearrange to make M subject
m/n=M
2/0.017=118
Therefore molar mass of Succinic acid is around 118

The formula of Succinic acid is (CH2)n(COOH)2
The molar mass of the (COOH)2 part gives  90
So 118-90=28
That leaves molar mass of 28 for the CH2 part
Molar mass of (CH2)=14
28/14= 2
Therefore there are 2 parts of CH2 for every (COOH)2 part
Hence, n =2


2) I'm not getting the right answer to this, sorry

3) CH3COOH+NaOH->CH3COONa+H2O
C(NaOH)=0.1
V(NaOH)=0.025
n=0.0025

That gives n(CH3COOH)=0.0025, in a volume of 0.0139L
To get to the number of mol in 0.25L of CH3COOH
Go 0.25/0.0139x0.0025= 0.045
So n(CH3COOH in 250ml)=0.045
Hence n(CH3COOH in 25ml)=0.045 (As they had just diluted it to 250ml, so added water, they didn't change number of mol through taking NaOH out or putting any in)

n=CV
0.045/0.025=1.8M

4) Similar to Question 3, just multiply number of mol of C2H2O4.2H2O by molar mass to get mass and find percentage purity

5) Similar to Question 4

6) Similar to Question 4,5

Note: Please correct me if I've made a mistake

[ras]

7. Firstly, I’d start by working out the mass of water which was evaporated from the sample.
13.2 g - 7.4 g = 5.8 g of H2O that was evaporated.

Next, figure out the moles of both the ZnSO₄ remaining and the H₂O evaporated, with our fave n=m/M
n(H₂O) = 5.8/18.0 = 0.322222 = 0.32 mol
n(ZnSO₄) = 7.4/161.5 = 0.045820433 = 0.0458 mol

Next, we find the mole ratio of H₂O to ZnSO₄ by dividing the moles of water evaporated by the moles of zinc sulfate which remained.
n(H₂O)/n(ZnSO₄) = 7.0322 = 7.0 to correct sig figs

Thus, the ratio of n(H₂O) : n(ZnSO₄) is 7:1, or, for every 1 ZnSO₄ molecule there are 7 H₂O molecules, so in the equation:
ZnSO₄.xH₂O --> ZnSO₄ + xH2O
x=7


I’ll have a crack with 2 as well.

2. Start with discovering the moles of HCl titrated with our trusty n=cV, given that c(HCl) = 0.1 M  and V(HCl) = 0.0245 L
n(HCl) = 0.1 x 0.0245 = 0.00245 mol

The equation tells us that the mole ratio of Na₂CO₃.xH₂O:HCl is 1:2, so therefore n(Na₂CO₃.xH₂O)/n(HCl) = 1/2.

Thus, n(Na₂CO₃.xH₂O) = n(HCl) x (1/2) = 0.00245 x (1/2) = 0.001225 mol

This is the amount of Na₂CO₃.xH₂O in the 25.0 mL aliquot, thus we can work out the concentration of this aliquot by using c=n/V
c = 0.001225/0.0250 = 0.0490 M

So knowing the concentration of the solution, we can discover the amount of Na₂CO₃.xH₂O in the 250 mL sample, again using n=cV.
n(Na₂CO₃.xH₂O) = 0.0490 x 0.250 = 0.01225 mol

Thus, in a 3.5 g sample there is 0.01225 mol of the hydrated sodium carbonate. We then need to use our classic n=m/M equation, which rearranged gives us M=m/n
M(Na₂CO₃.xH₂O) = 3.5/0.01225 = 285.7143, or approx. 286 g.mol^-1
M(Na₂CO₃) = 106 g.mol-1, thus M(xH₂O) = xM(H₂0) = 286 – 106 = 180 g.mol^-1

So, x = 180/M(H₂0) = 180/18 = 10

Sig figs aren’t correct in the working due to the 0.1 M HCl but work out in the end kinda Also, this question is a tad confusing for me too, so hopefully even if I’m wrong and there are small errors, some of it may help.

[Mellyboo]

These questions are absolutely brutal. Thanks so much guys, does anyone else have some ideas/ or like to confirm with the above posts?

[Mellyboo]

Tried to do 4,5,6 on my own to no avail. Help hahaha! So sorry !

[paper-back]

What part/parts are you having trouble with?

4) C(NaOH)=0.1
V(NaOH)=0.025
n(NaOH)=0.0025

C2H2O4.2H2O reacts with NaOH in a ratio 1:2
So n(NaOH)/2=n(C2H2O4.2H2O reacted)
0.0025/2=0.00125

n(C2H2O4.2H2O reacted)=0.00125
V(C2H2O4.2H2O reacted)=0.0213L
V(C2H2O4.2H2O volumetric flask)=0.25L
V(C2H2O4.2H2O volumetric flask)/V(C2H2O4.2H2O reacted)xn(C2H2O4.2H2O reacted)=n(C2H2O4.2H2O volumetric flask)
0.25L/0.0213L x 0.00125
Therefore n(C2H2O4.2H2O volumetric flask)=0.01467mol

0.01467mol x M(C2H2O4.2H2O)=1.85g
Mass of sample =2.5g
1.85/2.5=73.9%
Therefore Percentage purity = 73.9%

Try to do Question 5 again following the method used in the last question as it is practically the same question

5) C(NaOH)=0.1
V(NaOH)=0.025
n(NaOH)=0.0025

NaHSO4 reacts with NaOH in a ratio 1:1
So n(NaOH)=n(NaHSO4 reacted)
0.0025=0.0025

n(NaHSO4 reacted)=0.0025
V(NaHSO4O reacted)=0.0231L
V(NaHSO4 volumetric flask)=0.25L
V(NaHSO4 volumetric flask)/V(NaHSO4 reacted)xn(NaHSO4 reacted)=n(NaHSO4 volumetric flask)
0.25L/0.0231L x 0.0025
Therefore n(NaHSO4 volumetric flask)=0.0271mol

0.0271mol x M(NaHSO4)=3.24g
Mass of sample =5.678
3.24/5.678=57.18%
Therefore Percentage purity = 57.2%

6) So we have 2 equations
(1) SiCl4+2H2O->SiO2+4HCl

As the solution in the first flask is transferred to the one with NaOH, we can assume that an acid base reaction occurs between the products of the last equation and NaOH. SiO2 is a solid so it won't react which leaves us with HCl and the reaction;

(2) HCl+NaOH->NaCl+H2O

The V of Alkali, i.e. was 0.0187
V(NaOH)=0.0187
C(NaOH)=0.1
n(NaOH)=0.00187

Reacts in a 1:1 ratio so,
n(NaOH)=n(HCl)
n(HCl)=0.00187

To get from n(HCl reacted) to n(HCl in original flask) multiply by 10
0.00187x10=0.01878
n(HCl in original flask)=0.01878

Back to the equation;
(1) SiCl4+2H2O->SiO2+4HCl

n(HCl in original flask)/4=n(SiCl4 that reacted)
0.01878/4= 0.00468
n(SiCl4 that reacted)= 0.00468
0.00468 x M(SiCl4)=0.794g

0.794/1.2=66.23%

[Mellyboo]

Turns out I did begin question 4 correctly, so much doubt.... Thanks so much PB !!! 

PB, would you say this is easy, hard or about the same level as VCAA questions?

[paper-back]

I see, try not to doubt yourself when doing question outside of SAC's and Exams. It kind of makes you 'not want' to proceed with questions, from my experience

Sorry, I'm not too sure as I haven't started on VCAA questions yet
Maybe ras would know?

[ras]

Of the ones I did, something like Q7 would be a medium VCAA level questions, erring on the side of difficult. Q2 was definitely challenging, but VCAA could definitely put it on an exam if they were feeling mean.

The real difference is that VCAA is kind enough to split these types of questions up into part a) b) c) etc. The first couple are usually gimme marks that give you the first steps, eg for Q2 they may ask for part a) "Find the amount of HCl titrated, and thus the amount of sodium carbonate which reacted in this titration" The parts get harder as they go along, but there are usually clues as to what to do, so it splits up what would be an extremely challenging question to something a lot more manageable.

So, these are probably harder than what you'd get on VCAA exams, but only because the questions aren't split up into smaller parts which makes it easier to get your head around what to do