Emprical Formula Question

Hi,

I'm really stuck with this question - I've had a look at the markscheme as well but I still don't know how to approach it.

Crocetin consists of the elements carbon, hydrogen and oxygen. Determine the empirical formula of crocetin, if 1.00 g of crocetin forms 2.68 g of carbon dioxide and 0.657 g of water when it undergoes complete combustion.

Thanks in advance :smile:

Edit: I've just realised I've spelt empirical wrong in the title - oops! :p:

Reply 1

Laura373

16

hmmm.
let me think about it and i'll get back to you. xD

Reply 2

Laura373

16

I got CH6O2? Is that right? If it's not, I haven't a clue how to do it.

Reply 3

OP

8

Well in the markscheme it says the answer is C5H6O..I have no idea how to do it..

Reply 4

Laura373

16

In that case, I haven't the foggiest. Sorry!

Reply 5

OP

8

Laura373

In that case, I haven't the foggiest. Sorry!

No worries! Thanks anyway =D

Reply 6

Marcian0

somehow
someones made a thread similar to this before
they getting weird answers aswell

http://www.thestudentroom.co.uk/showthread.php?t=388924&page=5

Reply 7

Eggs-over_easy

No worries! Thanks anyway =D

yup have you tried this page?
http://dbhs.wvusd.k12.ca.us/webdocs/Mole/EmpiricalFormula.html
merry christmas
x

Reply 8

b2Zd1wKAmT

What paper/year is this from?

Reply 9

OP

8

Eggs-over_easy

yup have you tried this page?
http://dbhs.wvusd.k12.ca.us/webdocs/Mole/EmpiricalFormula.html
merry christmas
x

Thanks I'll try that =)

Zak XCI

What paper/year is this from?

IB May 2007 HL Paper 2..

Reply 10

Achilles99

5

Dont Know where the o came from - probably explained by filling bonds

I did this to get the C :h:

ratio

2.68/44= 0.0609 moles of C in 1g Cortin

0.657/18= 0.0365

therefore 0.0365x2 moles H in Cortin= 0.0730

0.730/0.0609= 1.2

therefore 5:6 C :h:

so C6H5

For the O- who knows :biggrin:

hope this works?

was : H btw lol

OP

Thanks I understood that! Now we need to know how to get the O..I had a go at it and here's what i got..hope I'm going along the right lines..

Crocetin + O2 --> CO2 + H20
1.00g + x --> 2.68g + 0.657g

So to find the mass of O2 (2.68 + 0.657) - 1.00g = 2.337?

I hope thats getting somewhere :confused:

Reply 13

michaelyus

10

Thanks I understood that! Now we need to know how to get the O..I had a go at it and here's what i got..hope I'm going along the right lines..

Crocetin + O2 --> CO2 + H20
1.00g + x --> 2.68g + 0.657g

So to find the mass of O2 (2.68 + 0.657) - 1.00g = 2.337?

That's what I used as well and I got the 1 × O they wanted.

Spoiler

Reply 14

Achilles99

5

ah that makes sense :P good to know- otherwise I would probably have ignored the O there if in an exam :biggrin:

Reply 15

OP

8

michaelyus

That's what I used as well and I got the 1 × O they wanted.

Spoiler

Thanks a lot mate! I understand it now :biggrin:

but i think i need some more practice on similar questions like that - anybody have some good sites?

Thanks everybody for your help :smile:

Reply 16

NLCeducation

Try this method

We are able to work
the mass of oxygen = (12/44 )X 2.68 ( oxygen in CO2_
Mass of H = 2/18 X 0.657 ( H in H2O)
Mass of O = 1.00g of crocetin subtracting the above two values. ( subtraction)
Continue this with how you calculate for empirical formula. ( divide by Ar , divided by smallest no. of mole , then find the ratio.
Answer should be C5H6O.

This question is indeed very unique.

I am trained and qualified IB private chemistry tutor in Singapore.
I am contactable at [email protected]
if you have questions pertaining to chemistry , i am most welling to help you guys.
Best wishes
Matthew

Reply 17

kirstensolevad

Original post by Achilles99

Dont Know where the o came from - probably explained by filling bonds

I did this to get the C :h:

ratio

2.68/44= 0.0609 moles of C in 1g Cortin

0.657/18= 0.0365

therefore 0.0365x2 moles H in Cortin= 0.0730

0.730/0.0609= 1.2

therefore 5:6 C :h:

so C6H5

For the O- who knows :biggrin:

hope this works?
[how do you end up getting 5 to 6?? from the 1.2?? ]

Original post by <liverpoolfc>

Hi,

I'm really stuck with this question - I've had a look at the markscheme as well but I still don't know how to approach it.

Crocetin consists of the elements carbon, hydrogen and oxygen. Determine the empirical formula of crocetin, if 1.00 g of crocetin forms 2.68 g of carbon dioxide and 0.657 g of water when it undergoes complete combustion.

Thanks in advance :smile:

Edit: I've just realised I've spelt empirical wrong in the title - oops! :p:

This is a typical IB question.

You are told that it contains C, H and O, so your first task is to determine how many grams and hence moles of each there are in the 1.00 g you start with.

2.68 g of carbon dioxide (the proportion of carbon in CO2 is 12/44 parts)

All of the carbon in the CO2 has come from the original compound. = 2.68 x 12/44 g = 0.73 g

0.657 g of water (proportion hydrogen = 2/18 parts) = 0.657 x 2/18 = 0.073 g

The remainder MUST be oxygen = 1.00 - (0.73 + 0.073) = 0.197 g

Now convert all of the masses into moles:

Carbon = 0.73/12 = 0.0608
Hydrogen = 0.073/1 = 0.073
Oxygen = 0.197/16 = 0.0123

Make into integers by dividing through by the smallest number

Carbon = 0.0608/0.0123 = 4.94
Hydrogen = 0.073/0.0123 = 5.93
Oxygen = 0.0123/0.0123 = 1

Empirical formula = C₅H₆O

Reply 19

Invicibella

5

Original post by charco

This is a typical IB question.

You are told that it contains C, H and O, so your first task is to determine how many grams and hence moles of each there are in the 1.00 g you start with.

2.68 g of carbon dioxide (the proportion of carbon in CO2 is 12/44 parts)

All of the carbon in the CO2 has come from the original compound. = 2.68 x 12/44 g = 0.73 g

0.657 g of water (proportion hydrogen = 2/18 parts) = 0.657 x 2/18 = 0.073 g

The remainder MUST be oxygen = 1.00 - (0.73 + 0.073) = 0.197 g

Now convert all of the masses into moles:

Carbon = 0.73/12 = 0.0608
Hydrogen = 0.073/1 = 0.073
Oxygen = 0.197/16 = 0.0123

Make into integers by dividing through by the smallest number

Carbon = 0.0608/0.0123 = 4.94
Hydrogen = 0.073/0.0123 = 5.93
Oxygen = 0.0123/0.0123 = 1

Empirical formula = C₅H₆O

I know that this from many, many, many years ago and you probably won't see this but THANK YOU SO MUCH

Emprical Formula Question

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