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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: lacoste on February 02, 2009, 05:22:48 pm

Title: Empirical Formula qst
Post by: lacoste on February 02, 2009, 05:22:48 pm: How do we know what factor to multiply the emprical numbers to so that it is very close to a whole number?

Eg. C : H

1 : 2.3

>C₇H₁₆

(i figured out that it was 7 after a long waste of time with the calc, i thought that maybe there is a way that is not trial and error?)

thanks, mates
Title: Re: Empirical Formula qst
Post by: dekoyl on February 02, 2009, 05:40:50 pm: I thought it might be 3?

C : H
1 : 2.3
3 : 6.9

$\approx C_3H_7$

As for how I did it (if it's right).. I can't give you a direct answer, sorry (because I don't have one).
Title: Re: Empirical Formula qst
Post by: lacoste on February 02, 2009, 05:47:15 pm: The solution said it was C₇H₁₆.

But your answer seems more accurate because its the lowest figure and both are out by 0.1.

hmmm, but does anyone know a way to get the 'factor' to be multiplied with ease?
Title: Re: Empirical Formula qst
Post by: lacoste on February 02, 2009, 05:48:08 pm: Quote from: dekoyl on February 02, 2009, 05:40:50 pm

As for how I did it (if it's right).. I can't give you a direct answer, sorry (because I don't have one).

trial and error?
Title: Re: Empirical Formula qst
Post by: Edmund on February 02, 2009, 05:57:51 pm: I would say C3H7

But does the question ask about having alkane or alkene, that sort of stuff?
Title: Re: Empirical Formula qst
Post by: lacoste on February 02, 2009, 06:00:39 pm: Quote from: edmund372 on February 02, 2009, 05:57:51 pm
I would say C3H7

But does the question ask about having alkane or alkene, that sort of stuff?

Nope, I don't know why they got that. If it was alkane or alkene what would it be?
Title: Re: Empirical Formula qst
Post by: /0 on February 02, 2009, 06:09:45 pm: $C_3H_7$ is neither alkane nor alkene. I don't think it can exist.

$C_7H_{16}$ follows the rule $C_{n}H_{2n+2}$ for alkanes.
Title: Re: Empirical Formula qst
Post by: lacoste on February 02, 2009, 06:58:45 pm: Ahh thanks, i remember that now. thanks.
Title: Re: Empirical Formula qst
Post by: lacoste on February 02, 2009, 07:05:07 pm: Another q.

If 3.72 g of element X reacts with exactly 4.80 g of oxygen to form a compound whose molecular formula is shown, from other experiments, to be X₄O₁₀, what is the relative atomic mass of X?

I don't get the steps in the solution.
Title: Re: Empirical Formula qst
Post by: Mao on February 02, 2009, 08:20:05 pm: $n(X)=4\times \frac{n(O)}{10}=\frac{4}{10}\times (2\times n(O_2)) = \frac{4}{5}\times \frac{4.80}{32} = 0.12\; mol$

$A_r(X)=\frac{m}{n} = \frac{3.72}{0.12}= 31.0\; \mbox{g mol}^{-1}$
Title: Re: Empirical Formula qst
Post by: lacoste on February 02, 2009, 08:33:53 pm: Quote from: Mao on February 02, 2009, 08:20:05 pm
$n(X)=4\times \frac{n(O)}{10}=\frac{4}{10}\times (2\times n(O)) = \frac{4}{5}\times \frac{4.80}{32} = 0.12\; mol$

$A_r(X)=\frac{m}{n} = \frac{3.72}{0.12}= 31.0\; \mbox{g mol}^{-1}$

Don't get that? O__O

Why is the 10 there for, what does it mean? .. $4\times \frac{n(O)}{10}$

And also, I don't get the 2 x n(O)?
Title: Re: Empirical Formula qst
Post by: Mao on February 02, 2009, 08:53:11 pm: oops, typo. I meant 2 x n(O₂)

the number of moles of oxygen consumed has a ratio to the number of moles of X by the chemical formula X4O10. that is, for every 10 oxygen atoms, $\frac{O}{10}$ , there are 4 X, $4\times \frac{n(O)}{10}$