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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: Wizard on December 08, 2008, 07:18:09 pm

Title: Stoichiometry Question
Post by: Wizard on December 08, 2008, 07:18:09 pm: This is a question I got for Holiday Homework. I suspect that it is from Neap, so if anyone has the book already, providing an answer to this question would be much appreciated :) It is question 1.1.21 in the Stoichiometry and Gravimetric Analysis Section (1.1)

Here is the question:

A 1.60g sample of an organic compound was burnt in excess Oxygen. 0.96g of water and 2.40g of Carbon Dioxide were produced. In a separate experiment, a 2.58g of the organic compound was vapourised. The vapour occupied 1.15L at 130 degrees Celcius (403 Kelvin) and 125 KPa.

Determine the Molecular formula of the compound. I got ethene (C₂H₄), but probably did something wrong.

Thanks
Title: Re: Stoichiometry Question
Post by: unknown id on December 08, 2008, 07:31:52 pm: Hmm, I got C₂H₄O₂ after a fair bit of calculation

I can see how you derived C₂H₄ but remember that the molar mass of the compound has to be ~60 g/mol through the equation PV = nRT. The molar mass of ethane is 28 g/mol. Therefore, that leaves 32 g/mol (60 - 28) to be taken by another element present in the compound (most likely, oxygen). This means that the molecular formula of the compund should have 2 oxygen atoms, thus leading to the formula C₂H₄O₂.
Title: Re: Stoichiometry Question
Post by: Collin Li on December 08, 2008, 07:40:00 pm: Organic compounds burnt in excess oxygen: combustion

If only water and carbon dioxide is produced, then the organic compound is restricted to only containing C, H and O atoms.

$\mbox{C}_x\mbox{H}_y\mbox{O}_z + \mbox{O}_2 \rightarrow \mbox{H}_2\mbox{O} + \mbox{CO}_2$

(You must have assumed $z=0$ , i.e.: there is no oxygen)

* 2.40g of carbon dioxide implies there is $2.40 \times \frac{12}{44} = 0.65\mbox{ g}$ of carbon that must have come from the organic compound (couldn't have come from the oxygen).

* 0.96g of water implies there is $0.96 \times \frac{2}{18} = 0.11\mbox{ g}$ of hydrogen that must have come from the organic compound (also couldn't have come from the oxygen).

* This leaves us with $1.6 - 0.65 - 0.11 = 0.84\mbox{ g}$ of oxygen in the organic compound.

Now: $n(C) = 0.055\mbox{ mol}$ , $n(H) = 0.11\mbox{ mol}$ , and $n(O) = 0.052\mbox{ mol}$

So: $\mbox{n(C) : n(H) : n(O)} = 1:2:1$

Hence, the empirical formula is $\mbox{CH}_2\mbox{O}$

The second piece of information tells us that: $n = \frac{pV}{RT} = \frac{125\times 1.15}{8.31\times 403} = 0.043\mbox{ mol}$

And since $m = nM \implies M = \frac{m}{n} = \frac{2.58}{0.043} = 60.1\mbox{ gmol}^{-1}$

Since $\mbox{M(CH}_2\mbox{O)} = 30$ , then the molecular formula must be two times the empirical formula, as 60 is twice as large as 30.

Therefore, the organic compound is $\mbox{C}_2\mbox{H}_4\mbox{O}_2$ as unknown id calculated.
Title: Re: Stoichiometry Question
Post by: Wizard on December 08, 2008, 08:17:22 pm: lol, I thought it was a hydrocarbon, so I left out the O from the formula. Thanks! :)
Title: Re: Stoichiometry Question
Post by: Collin Li on December 08, 2008, 08:23:04 pm: Ah.

Hydrocarbons: C, H
Carbohydrates: C, H, O
Proteins: C, H, N, O (and possibly others)
Organic compounds: anything, but must contain C

In VCE, you only need to know the results of molecular combustion involving molecules with C, H, O, though. (For curiosity, all the N atoms will become nitrogen dioxide, if I remember correctly).
Title: Re: Stoichiometry Question
Post by: Wizard on December 08, 2008, 09:44:19 pm: Another Stoich Question :)

The mass of Iron obtained by the complete reduction of 1kg of haematite(Fe₂O₃) if the ore is 90% pure will be:
a) 350 g
b) 700 g
c) 777 g
d) 389 g

I got 629.5 g. But the answers say C. Could someone tell me if I am right or wrong? Thanks
Title: Re: Stoichiometry Question
Post by: Collin Li on December 08, 2008, 09:47:20 pm: $0.9 \times 1000 = 900\mbox{ g}$ of pure $\mbox{Fe}_2\mbox{O}_3$

This means there is: $900 \times \frac{2(55.9)}{2(55.9) + 3(16.0)} = 629.7\mbox{ g}$

Yeah, you're right.