Gen Chem Question

[Metswin677]

I have a question.....

A compound contains 27.3% carbon, and 72.7% oxygen by mass. What is the expirical formula....When I try this, I divide 72.7/27.3 and I get C:O ration of 1:2.66

The answer from Collins is CO2.....why is it not CO3?

 

[ASUGRL10]

Hi,

You need to divide both C and O by their atomic weight, so for C, 27/2= ~2

and for O, 72/16= ~4, and you get the molecular formula C2O4, and the empirical formula would be CO2. Hope that makes sense.

 

[Metswin677]

Hi,

You need to divide both C and O by their atomic weight, so for C, 27/2= ~2

and for O, 72/16= ~4, and you get the molecular formula C2O4, and the empirical formula would be CO2. Hope that makes sense.

I thought you stay consistent and divide both sides by the lowest molecular weight. Guess not....

 

[UyenRita]

First, get the moles for both C and O.
Divide 27.3 by the Molarity of C : 27.3/12 = 2.275
Divide 72.12 by the molarity of O: 72.7/16 = 4.54

Then, get all the numbers of moles divided by the smallest number of mole.

For C, 2.275/2.275=1
For O, 4.54/2.275 =~ 2

So you can get the empirical formula for this: CO2.
Hope this helps.

 

[Metswin677]

First, get the moles for both C and O.
Divide 27.3 by the Molarity of C : 27.3/12 = 2.275
Divide 72.12 by the molarity of O: 72.7/16 = 4.54

Then, get all the numbers of moles divided by the smallest number of mole.

For C, 2.275/2.275=1
For O, 4.54/2.275 =~ 2

So you can get the empirical formula for this: CO2.
Hope this helps.

Thanks Alot...Looks good!

 

[1ive2die]

Thanks Alot...Looks good!

Looks good except the part about molarity. I think you mean molar mass.

 

[UyenRita]

Looks good except the part about molarity. I think you mean molar mass.

Oh yeah, I think I'd better call it "molar mass". Sorry, I'm always mess up with those words. 🙄