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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: Scarlett on April 24, 2008, 09:22:21 pm

Title: Please help - Empirical Formula question
Post by: Scarlett on April 24, 2008, 09:22:21 pm: Hi

Could someone please help me with this question?

Methyl salicylate, a compound of hydrogen, carbon and oxygen, is commonly known as oil of wintergreen. It is often used in medical rubs. If a 20.00g sample of methyl salicylate was burned in air and it produced 9.46g of water and 46.30g of carbon dioxide, show that the empirical formula of this compound must be C8H8O3.

Thanks
Title: Re: Please help - Empirical Formula question
Post by: Collin Li on April 24, 2008, 10:24:51 pm: Recall that the equation of the combustion for a carbohydrate (any compound made of carbon, hydrogen and oxygen) is:

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

Balancing by elements (atoms), yields:

$\mbox{n(C)} = \mbox{n(CO}_2\mbox{)} = \frac{46.30}{44.0} = 1.05\mbox{ mol}$

$\mbox{n(H)} = 2\cdot\mbox{n(H}_2\mbox{O)} = \frac{2\times 9.46}{18.0} = 1.05\mbox{ mol}$

Note that it is not possible to deduce the number of moles of oxygen formed in the same way as carbon and hydrogen, because there is an unknown input of oxygen in the reaction.

However, if $\mbox{n(C)} = \mbox{n(H)} = 1.05\mbox{ mol}$ , then there are $1.05\times 12.0 + 1.05\times 1.0 = 13.68\mbox{ g}$ of carbon and hydrogen originating from the carbohydrate. This means that $20.00 - 13.68 = 6.32\mbox{ g}$ must be the oxygen in the carbohydrate!

So, $\mbox{n(O)} = \frac{6.32}{16.0} = 0.395\mbox{ mol}$

Therefore, $\mbox{n(C)}\,:\,\mbox{n(H)}\,:\,\mbox{n(O)} = 1.05\,:\,1.05\,:\,0.395$

$= 2.66\,:\,2.66\,:\,1.00$

$= 8\,:\,8\,:\,3$

Hence, the empirical formula is $\mbox{C}_8\mbox{H}_8\mbox{O}3$
Title: Re: Please help - Empirical Formula question
Post by: Scarlett on April 24, 2008, 10:31:03 pm: You really are a genius. Thank you so much!