Equilibria, Energetics and Elements (F325) - June 2011 Exam.

Right ok so lets look at the key facts here:

5.119 = CO2 1.575 = H2O

You know that CO2 Mr = 44 and in it C is 12

so you can work out a fraction of C in 5.119 of CO2 which is 12/44 x 5.119 = 1.396

Same with H2O = 18 and H is 2

2/18 x 1.575 = 0.175

Now you have the g of H2 and C, so you can subtract that from the total to find O2

4.362 - (1.396 + 0.175) = 2.791

Now you can work out the empirical firmula

C: H: O
1.396: 0.175: 2.791

divide through by their Mr

C 1.396/12 = 0.1163
H 0.175/1 = 0.175
O 2.791/16 = 0.1744

0.1163:0.175:0.1744
simplifying it

1: 1.5: 1.5
to get a whole number ratio you can multiply by 2 to get
2: 3: 3

This gives you C2H3O3 to get 75

150/75 = 2

2 x (C2H3O3) = C4H6O6

Hope this helps

Can anyone please help me with this question? It was on the Jan 09 UC paper.

"Compound E has the formula CxHyOz. 4.362g of E was burned in oxygen to form 5.119g of CO2 and 1.575g H2O. Mass spec showed a peak at 150.

Calculate the molecular formula of E."

I had a look at the mark scheme but I can't follow what it's doing :s Do you think we'll get something similar in our paper?

hi to everybody. i was just wondering how much revision everyone has done for this exam so far, as i have also got the biology exam which is like 2 days before the chemistry exam and i am finding it quite hard to fit everything in.

Ive been told that we wont be asked about explaining differences in ligands colour !

There is like 10 mark question in 2007 ..but apparently it was taken out from syllabus some time later

Hey, I'm finding it hard to get my head around buffers. It is composed of a weak acid, the conjugate base and the salt of the weak acid. How does the salt of the weak acid form? On adding acid, does the conjugate base react to form the weak acid? I revise over it again and again, I guess I just don't understand it properly. Could someone explain it to me please?

Ok, here goes. You seem a bit confused on how a buffer forms, so here it is:

You make a buffer solution by e.g. Adding NaOH to an EXCESS of CH3COOH. Therefore you get CH3COOH and some CH3COONa, which is the salt of the weak acid. Some CH3COOH remains.

The weak acid partially dissociates but the ionic salt fully dissociates. So from the salt we get large amounts of CH3COO- and Na+. For the purposes of the buffer solution we can ignore the Na+.

An equilibrium is set up: CH3COOH <-> CH3COO- + H+
(that should be an equilibrium arrow above)

There are large amounts of the undissociated acid, because it is a weak acid, and large amounts of the conjugate base, because the salt fully dissociates. There is only a small amount of H+ because it comes from the weak acid, which only partially dissociates.

Are we alright so far? Let me know if you need more help

Thanks That has helped me. So what happens when you add acid/alkali?
And do we need to know the equation to make a buffer solution, so like the equation to react NaOH with excess CH3COOH?

Lets see how much I remember..Hmmm

If u add OH- (alkali)

It combines to H+ ...>forms water

Equil. shifts to right ( cuz u removed H+) to minimise the change---> more H+ produced

so minimises ph changes

I let u guess the other one

So if you add acid, the conjugate base reacts with H+ to form the weak acid. The equilibrium shifts to the left to remove the extra H+.

Hope this comes up now. Haha

Thanks That has helped me. So what happens when you add acid/alkali?
And do we need to know the equation to make a buffer solution, so like the equation to react NaOH with excess CH3COOH?

You don't need to know any specific equations, but they may ask you to apply general principles. For example 'A buffer solution was made by reacting ammonia and HCOOH. Suggest the equation'. It's just acid base reactions really.

Thats right

Am slightly worried about colour of transition metals:

WOuld be epic if someone list them all here

1. Cro4 2- Yellow
2.cucl2- Pale blue
..