Question

How to calculate these ?

1. A compound consists only C,H, and O. When 1.0876 g of this compound burn completely in oxygen, 2.078 g of `CO_2` and 1.2738 g of `H_2O` are produced.

(a)(i) Calculate the number of grams of C in the `CO_2` released.

(ii) Calculate the number of grams of H in the `H_2O` released.

(iii) Calculate the number of grams of O in the sample.

Answer 1

This is not the best question I have seen...

Explanation:

We gots `C_xH_yO_z` the which is combusted to give `CO_2` and water....clearly, `n_(CO_2)` is equivalent to the number of moles of carbon in the sample...

`"(i) Moles of carbon dioxide"=(2.078*g)/(44.01*g*mol^-1)=0.04722*mol`

...i.e. a mass of `0.5672*g` with respect to carbon..

`"(ii) Moles of water"=(1.2738*g)/(18.01*g*mol^-1)=0.07073*mol`

...i.e. a mass of `0.1426*g` with respect to hydrogen..

`"(iii) Moles of oxygen"=({1.0876-0.5672-0.1426}*g)/(16.00*g*mol^-1)=0.02361*mol`

...i.e. a mass of `0.3778*g` with respect to oxygen.. Clearly we had to access the mass of oxygen by the balance with respect to the original sample after we extracted the masses of hydrogen, and carbon...

And while the question did not ask for it, CLEARLY, we can access the `"empirical formula"` of the stuff....by dividing thru by the LOWEST molar quantity (that of oxygen) to get an `"empirical formula"` of....

`C_((0.04722*mol)/(0.02361*mol))H_((0.1415*mol)/(0.02361*mol))O_((0.02361*mol)/(0.02361*mol))=C_2H_6O`

`"Ughhh arithmetic...."` and I hope you can see the numbers on the screen, because I am having trouble and cannot find my spex...