Some Empirical/Molecular formula questions...

[tl]

Recently a teacher gave out a hand-out showing us what some "harder" chemistry questions look like.

Because I don't fully understand the wording, I find these questions almost "impossible".

So if anyone could figure and also explain it to me, that would be ace.

Question 1:
"A compound contains carbon, hydrogen, oxygen and nitrogen. 0.3006g of the compound was burnt and gave 0.1804g of water, and 0.3530g of carbon dioxide. A further 0.2500g of the compound evolved 38.76ml of nitrogen at 13 degrees Celsius, 788mmHg on decomposition. The relative molecular mass was found to be 75.00. Find the molecular formula of the compound."

What I have trouble in that question is the nitrogen part, and the ratio between hydrogen, carbon and oxygen. If that could be explained that would be great.

Question 2:
"0.220g of a compound contains carbon, hydrogen and chlorine. On combustion in oxygen it yields 0.195g of carbon dioxide, and 0.0804g of water.
0.132g of the compound yielded after suitable treatment, 0.38822g of silver chloride.
i) Find the empirical formula of the compound.
0.1089g of the compound occupied a volume of 37.15ml at 135 degrees Celsius and 767mmHg.
ii)Find the molecular formula of the compound.

Again, simply not understanding.. what I take is that this Compound X is combusted CO2 and H20 was left over. The second part with the silver chloride makes no sense to me.. 

Thanks friends 

[fady_22]

1) I won't work it all out here, but will just explain what I did:

You have the masses of carbon dioxide and water yielded from .3006 g of the compound, so you can figure out the number of mole of both. From this, you can figure out the number of mole of hydrogen and carbon in the compound.

You can also figure out the number of mole of nitrogen gas yielded from .2500 g of the compound. Take note that this value cannot be used to figure out the ratio: you must find the number of mole of nitrogen in .3006 g.
i.e. and cross multiply to find the n(nitrogen in 0.3006g).

Once you know the number of mole of carbon, hydrogen and nitrogen, you can multiply these values by the molar masses of each element, and subtract the total of the masses from the original mass of the compound to find the mass of oxygen. This value can be used to find the mole of oxygen.

Now you have the number of mole of oxygen, nitrogen, hydrogen, and carbon. Use the same technique as always to find the ratio.
The molecular formula should be

[fady_22]

Whoops, forgot number 2!

2) The same sorta question as 1).
i) After you find the number of mole of carbon and hydrogen, you can find the mole of AgCl, and therefore Cl in the compound (however this is from 0.132 g of the compound). Use the cross-multiplication method again to find the mole of Cl in .220g, and use this value in conjunction with the mole of carbon and hydrogen, to find the ratio of the elements in the compound. This will give you the empirical formula.
The rest is as usual.

[tl]

i.e.  and cross multiply to find the n(nitrogen in 0.3006g).

With that part.. I do not fully understand.. Is it meant to be n(Nitrogen from 0.3006) = 0.3006*n(nitrogen from 0.2500)/0.2500?? With that i don't know the mols of nitrogen to begin with.. sorry I'm rather slow  

EDIT: OHH, I get it, yea, pV=nRT formula, ok ok, thanks heaps!

EDIT AGAIN: I get the ratio of H:C:N:O as 10:4:1:4.5  .. I don't know what I'm doing incorrectly