What is molecular formula of compound that weighs 46 g/mol and analyzes to 52.2% C, 34.8% O, and 13% H?

1 Answer
Dec 3, 2016

Answer:

#"Molecular formula"# #-=# #C_2H_6O#

Explanation:

As with all these problems, we assume that there are #100*g# unknown material:

First we calculate an empirical formula...........

#"Moles of carbon:"# #=# #(52.2*g)/(12.011*g*mol^-1)=4.35*mol#.

#"Moles of hydrogen:"# #=# #(13.0*g)/(1.079*g*mol^-1)=12.04*mol#.

#"Moles of oxygen:"# #=# #(34.8*g)/(15.999*g*mol^-1)=2.18*mol#.

Note (i) that here we have simply divided the atomic masses thru by the #"atomic mass"#, and (ii), normally #"% oxygen content"# would not be measured. You would be given #"% carbon content"#, and #"% hydrogen content"#, and #"% nitrogen content"#, and #"% oxygen content"# would be assessed by the balance.

And now we normalize the formula by dividing thru by the lowest molar ratio, that of oxygen to get the empirical formula:

#O=(2.18*mol)/(2.18*mol)# #=# #1#

#C=(4.35*mol)/(2.18*mol)# #=# #2#

#H=(12.04*mol)/(2.18*mol)# #=# #6#

And thus our empirical formula is #C_2H_6O#.

But the molecular formula is always a mulitple of the empirical formula:

i.e. #"(empirical formula)"xxn="(molecular formula)"#.

We KNOW the molecular mass because it has been measured in a separate experiment, and provided for us:

So #nxx(2xx12.011+6xx1.00794+15.999)*g*mol^-1=46.0*g*mol^-1#,

and we solve for #n#.

Clearly #n=1#, and here

#"molecular formula "-=" empirical formula"=C_2H_6O#.