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Topic: Calculating Empirical and Molecular Formula  (Read 7986 times)

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mattg2006

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Calculating Empirical and Molecular Formula
« on: December 03, 2005, 08:18:15 PM »
I have this question:

An alcohol has a relative molecular mass of 74.0 and has the following composition by mass: C, 64.9%; H, 13.5%; O, 21.6%

Calculate the empirical formula of the alcohol and show that its molecular formula is the same as the empirical formula.

This is what I've done so far:

C 74 x 0.649 = 48.026
H 74 x 0.135 = 9.99
0 74 x 0.216 = 15.984

This is roughly C48 H10 and 016 (all add up to 74)

I have then divided each one by its mass.

C 48/12 = 4
H 10/1 = 10
O 16/16 = 1

I know the structure of an alcohol is CnH2n+1OH

There the empirical formula is: C4H9OH

First of all, is this correct?

How do I show its molecular formula is the same? (How do I calculate molecular formula?)

Thanks,

Unco

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Re:Calculating Empirical and Molecular Formula
« Reply #1 on: December 04, 2005, 01:41:10 AM »
G'day, mattg2006.

Recheck with your textbook about how this is done.

Say we have 100g of this substance. So we have 64.9g of C, 13.5g of H and 21.6g of O.

We need to convert those masses to amounts (in moles), by n = m/M. The molar ratio determines the empirical formula.

Once you have C, H and O in moles, determine which has the least number of moles. Divide the other two's number of moles by this to get two molar ratios of each of  the two higher-amount elements to the lowest-amount one, and hence the empirical formula.

The final step is to check this empirical formula gives a molecular mass equal to that stated in the question. If it is half the actual molecular mass, the ratio must be doubled, for example.

mattg2006

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Re:Calculating Empirical and Molecular Formula
« Reply #2 on: December 04, 2005, 11:16:29 AM »
I've had another go and came up with the same answer! Where am I going wrong?

First of all, with that question, should I presume 100g of substance?

If so,

C - 64.9g, H - 13.5g, 0 - 21.6g

Moles = Mass/Molar mass

C - 64.9/12 = 5.408333333
H - 13.5g/1 = 13.5
O - 21.6g/16 = 1.35

Oxygen has the least number amount of moles.

So,

5.40833333/1.35 = 4.00617, roughly 4
13.5/1.35 = 10
1.35/1.35 = 1

This is exactly the same numbers I had before.

Where am I going wrong?

Thanks for your help Unco.

Offline Borek

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Re:Calculating Empirical and Molecular Formula
« Reply #3 on: December 04, 2005, 11:54:19 AM »
There the empirical formula is: C4H9OH

First of all, is this correct?

Seems OK with me. Both methods give the same answr.

Quote
How do I show its molecular formula is the same? (How do I calculate molecular formula?)

Molar mass is given so there is no problem with checking how many times C4H9OH must be repeated to obtain molceluar formula.
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mattg2006

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Re:Calculating Empirical and Molecular Formula
« Reply #4 on: December 04, 2005, 11:58:42 AM »
Doesnt it need to only be repeated once to obtain the molecular formula?

I'm slightly confused about how to even calculate the molecular formula from that.

Thanks.

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Re:Calculating Empirical and Molecular Formula
« Reply #5 on: December 04, 2005, 12:02:36 PM »
Doesnt it need to only be repeated once to obtain the molecular formula?

Yes

Quote
I'm slightly confused about how to even calculate the molecular formula from that.

Multiply all coefficients by 1 :)
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mattg2006

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Re:Calculating Empirical and Molecular Formula
« Reply #6 on: December 04, 2005, 12:09:20 PM »
Is it because:

4 Carbons = 4 x 12 = 48
10 Hydrogens = 10 x 1 = 10
1 Oxygen = 1 x 16 = 16

48+10+16 = 74.

Is this why?

Offline Borek

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Re:Calculating Empirical and Molecular Formula
« Reply #7 on: December 04, 2005, 12:39:19 PM »
Yes. You have calculated molar mass of the empirical formula - which can be repeated several times in reality. In this case it is repeated once.
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mattg2006

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Re:Calculating Empirical and Molecular Formula
« Reply #8 on: December 04, 2005, 12:44:03 PM »
Could you give an example when I would need to repeat the empirical formula?

Thanks.

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