Serotonin (molar mass 176 g/mol) is a compound that conducts nerve impulses in brain and muscle. It contains 68.2% C, 6.86% H, 15.9% N and 9.08% O. What is the molecular formula?

1 Answer
anor277
Feb 16, 2017

#"Serotonin"-=C_10H_12N_2O#

Explanation:

As with all these problems, we ASSUME #100*g# of compound, and we work out the percentages of each element on a molar basis:

#"Moles of carbon"# #=# #(68.2*g)/(12.011*g*mol^-1)=5.68*mol#.

#"Moles of hydrogen"# #=# #(6.86*g)/(1.00794*g*mol^-1)=6.81*mol#.

#"Moles of nitrogen"# #=# #(15.9*g)/(14.01*g*mol^-1)=1.14*mol#.

#"Moles of oxygen"# #=# #(9.08*g)/(15.999*g*mol^-1)=0.568*mol#.

Note that normally the analyst would not give you #%O#; you would be expected to work it out by difference.

And now we divide thru by the SMALLEST molar quantity, that of oxygen, to give an empirical formula of:

#C_10H_12N_2O#.

And this is the empirical formula, the simplest whole number ratio defining constituent atoms in a species.

Now the molecular formula is ALWAYS a multiple of the empirical formula. That is #"Molecular formula"-="(empirical formula)"_n#.

Thus the molecular mass must show this proportionality:

#(C_10H_12N_2O)_n=176*g*mol^-1#

So #nxx(10xx12.011+12xx1.00794+2xx14.01+15.999)*g*mol^-1=176*g*mol^-1#

Clearly, #n=1#, and the #"empirical formula"# is the same as the #"molecular formula"#.

This approach is standard, however, most of the time you would not be quoted a percentage oxygen composition. You would be expected to calculate this yourself by difference.

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