Empirical formula from Combustion Analysis

[FCMike11]

Not really sure where this should go so going to stick it here and hopefully someone bites.

A 12.983 g sample of warfarin (Warfarin only consists of C, H, & O) was subjected to combusion analysis and produced 35.12 g CO2 and 6.07 g of H2) as the only products

So I came up with this reaction:

CxHyOz (12.983 g) + excess O2 ----> CO2 (35.12 g) + H2O (6.07 g)

Calculated moles of C: (shortening to stop from being so messy) (35.125 * 12)/ 44(MW of CO2) = 9.6 g of C thus 9.6/12 = .8 mol C

Calculated moles of H: (6.07 * 2) / 18 (MW of H2O) = .674 g of H thus ..674 mol of H

Calculate grams & moles of O = 12.983 (sample) - 9.6 (g of C) - .674( g of H) = 2.709 g of O. 2.709/16 (MW of O = 16) = .169 mol of O

C.8 H .674 O .169

Divided this all by .169 and got C4.73 H 3.98 O, then multiplied by 4 and Subsequently got an empirical formula of C19 H16 O4.

This still just seems odd, worked this identically to the professors practice problem in class. Does this look okay?

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[AdrianVeidt]

Not really sure where this should go so going to stick it here and hopefully someone bites.

A 12.983 g sample of warfarin (Warfarin only consists of C, H, & O) was subjected to combusion analysis and produced 35.12 g CO2 and 6.07 g of H2) as the only products

So I came up with this reaction:

CxHyOz (12.983 g) + excess O2 ----> CO2 (35.12 g) + H2O (6.07 g)

Calculated moles of C: (shortening to stop from being so messy) (35.125 * 12)/ 44(MW of CO2) = 9.6 g of C thus 9.6/12 = .8 mol C

Calculated moles of H: (6.07 * 2) / 18 (MW of H2O) = .674 g of H thus ..674 mol of H

Calculate grams & moles of O = 12.983 (sample) - 9.6 (g of C) - .674( g of H) = 2.709 g of O. 2.709/16 (MW of O = 16) = .169 mol of O

C.8 H .674 O .169

Divided this all by .169 and got C4.73 H 3.98 O, then multiplied by 4 and Subsequently got an empirical formula of C19 H16 O4.

This still just seems odd, worked this identically to the professors practice problem in class. Does this look okay?

Didn't look at any of the numbers but the steps appear to be correct from a quick glance.

1. grams to moles of known
2. grams to moles of known
3. find grams of unknown
4. grams to moles of unknown
5. divide by lowest moles
6. multiply until all numbers close to even

Be careful when you do hydrogen though. It's easy to just throw it in as 1.01g/mol but remember 2 hydrogens per water molecule. You did it all in one step, I would suggest doing it in two to avoid messing up.

 

[chenzt]

Looks good to me, your math seems sound.

If you googled warfarin, on Wikipedia it says that the formula is C19H16O4. So you're right

 

[Aerus]

SDN isn't for homework help. Why not ask your TA or professor during office hours? Work through it with a friend?

 

[FCMike11]

I know and I normally wouldn't. Just in a hard place 2 hrs till test and can't get a hold of my teacher. Thanks for the help guys.

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