Question

An organic compound has a percentage composition of `40.1%` `C`, and `6.6%` `H`. What is its probable empirical formula?

Answer 1

I makes it `CH_2O`

Explanation:

The empirical formula is the simplest while number ratio defining constituent atoms in a chemical species.

We assumes that we got `100*g` of some compound....

And thus `"moles of carbon"-=(40.1*g)/(12.011*g*mol^-1)=3.34*mol`.

And `"moles of hydrogen"-=(6.6*g)/(1.00794*g*mol^-1)=6.55*mol`.

But you have undoubtedly already noted that the quoted percentages do not add up to 100%. In this scenario IT IS ALWAYS ASSUMED that the BALANCE, the missing percentage, is DUE TO OXYGEN.....

And `"moles of oxygen"-=(100*g-40.1*g-6.6*g)/(15.999*g*mol^-1)=3.33*mol`.

And please note that here WE CAN MAKE NO OTHER ASSUMPTION. `%O` is usually UNREPORTED because there are few analytical methods for measurement of oxygen in microanalysis, and it is assumed to be the missing percentage. This is a standard practice in analysis.

And so we divide each molar quantity thru by the SMALLEST molar quantity to get an empirical formula of.....

..........................`CH_2O`.

We could quote a molecular formula PROVIDED that we get a measurement of molecular mass.