What is the molecular formula of a material with percentage composition #C:40.5%#, #H:6.67%#, #O:53.33%#, and molecular mass of #180.0*g*mol^-1#?

1 Answer
May 7, 2017

Answer:

This is a standard question, and as a routine we ASSUME a mass of #100*g# of unknown compound.......and get a molecular formula of #C_6H_12O_6#.

Explanation:

So if there are #100*g# of unknown compound, we can access the empirical formula given the percentage composition:

#"Moles of carbon"=(40.0*g)/(12.011*g*mol^-1)=3.33*mol#

#"Moles of hydrogen"=(6.67*g)/(1.00794*g*mol^-1)=6.62*mol#

#"Moles of oxygen"=(53.33*g)/(16.00*g*mol^-1)=3.33*mol#

We divide thru by the LOWEST molar quantity, that of oxygen/carbon to give an empirical formula of:

#CH_2O#, which represents the simplest whole number ratio defining constituent atoms in a species, i.e. the #"empirical formula"#.

But we also have the molecular mass, #180*g*mol^-1#.

Now it is a fact that the molecular formula is always a whole number multiple of the empirical formula, and thus..............

#180*g*mol^-1=nxx(12.011+2xx1.00794+15.999)*g*mol^-1#,....................... and so we solve for #n#. Clearly, #n=6#, and the molecular formula is #C_6H_12O_6#.

Just on a practical issue, an analyst would normally NEVER give you the percentage oxygen by mass. Most of the time you get #%C,H,N#, and if their sum is not #100%#, then the BALANCE is assumed to be due to #"oxygen"#.