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

1 Answer
Sep 6, 2017

Answer:

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.