[MakoEyes]

An organic compound containing only C, H, and possibly O was subjected to combustion analysis. A
sample weighing 0.8881 g yielded 1.650 g CO2 and 0.338 g H2O. What is the empirical formula of the
compound?

[Borek]

Please read forum rules.

You have to show your attempts to receive help. This is a forum policy.

Even searching forums for elemental analysis will help.

[MakoEyes]

I did some additional work:

1.650g CO2 * (1 mole CO2 / 44.01g) = 0.03749 mole

0.338g H2O * (1 mole H2O / 18.016g) = 0.0187 mole

Then I divided these by 0.0187 and it gave me the formula CO2H2O....Which I reduced to COHO... but this does not make sense I feel I'm doing something wrong here and not using what I'm given properly.

[sjb]

I did some additional work:

1.650g CO2 * (1 mole CO2 / 44.01g) = 0.03749 mole

0.338g H2O * (1 mole H2O / 18.016g) = 0.0187 mole

You can see that C  CO₂, so what is the mass of 0.03749 mole of carbon? Similarly for hydrogen?

So how much (if any) is left?

Then I divided these by 0.0187 and it gave me the formula CO2H2O....Which I reduced to COHO... but this does not make sense I feel I'm doing something wrong here and not using what I'm given properly.

Well, given that you haven't divided the carbon by half here that also doesn't really make sense anyway, but you're on the right lines in dividing by the hcf to start at getting an empirical formula.

[MakoEyes]

I did some additional work:

1.650g CO2 * (1 mole CO2 / 44.01g) = 0.03749 mole

0.338g H2O * (1 mole H2O / 18.016g) = 0.0187 mole

You can see that C  CO₂, so what is the mass of 0.03749 mole of carbon? Similarly for hydrogen?

So how much (if any) is left?

Then I divided these by 0.0187 and it gave me the formula CO2H2O....Which I reduced to COHO... but this does not make sense I feel I'm doing something wrong here and not using what I'm given properly.

Well, given that you haven't divided the carbon by half here that also doesn't really make sense anyway, but you're on the right lines in dividing by the hcf to start at getting an empirical formula.

Oh I didn't see that ratio before. So I can write:

Mass of C: 0.03749 mole * (12.01g / 1 mole C) = 0.4502549g
Mass of H: 0.0187 mole * (1.008g / 1 mole H) = 0.0188496g

How can I calculate how much is left? And what can I do as the next step?
(Sorry this question really confused me.)

[DrCMS]

Mass of C: 0.03749 mole * (12.01g / 1 mole C) = 0.4502549g

Yes

Mass of H: 0.0187 mole * (1.008g / 1 mole H) = 0.0188496g

No, how many H in water 1 or 2?

Once you have the weight of C and H in the original sample you can calculate how much O there was if any from the total weight of the sample.

Turn those weights into moles of C, H and O.

Divide by the smallest figure to get whole numbers and that the empirical formula.

[MakoEyes]

Whopps.

Thus:

Mass of H: 0.0187 mole * (1.008 / 2) =  0.0094248g

Once you have the weight of C and H in the original sample you can calculate how much O there was if any from the total weight of the sample.

How do I do this step?? Thanks again for your help.

[sjb]

Whopps.

Thus:

Mass of H: 0.0187 mole * (1.008 / 2) =  0.0094248g

Not quite. Any hydrogen in your sample will be turned into water by your analysis, so C_wH_x(other elements) + O₂  w CO₂ + ^x/₂ H₂O + other products (possibly).

So if you had 0.0187 mole of water formed, you had 0.0187 mol x 2 of "H" in your original sample, with what mass?

Once you have the weight of C and H in the original sample you can calculate how much O there was if any from the total weight of the sample.

How do I do this step?? Thanks again for your help.

Consider a total of 500 grams of fruit, containing bananas, apples and possibly oranges. If 135 grams has been found to be apples, 278 grams found to be bananas, what mass of oranges do I have?