An organic compound containing 78.5% carbon, 13.1% nitrogen, 8.4% hydrogen. If the molar mass of the compound is 107 g/mol, what is the empirical and molecular formulas?

1 Answer
Jan 10, 2017

Answer:

#"Empirical formula"# #=# #"Molecular formula"# #=# #C_7H_9N#.

Explanation:

With all these problems, we assume a mass of #100*g# and divide the individual elemental masses thru by the atomic mass, i.e.

#C:# #(78.5*g)/(12.011*g*mol^-1)=6.54*mol.#

#H:# #(8.4*g)/(1.00794*g*mol^-1)=8.33*mol.#

#N:# #(13.1*g)/(14.01*g*mol^-1)=0.935*mol.#

We divide thru by the smallest molar quantity, that of nitrogen, to get an empirical formula of #C_7H_9N#. The hydrogen ratio took a bit of rounding up.

Now the #"molecular formula"# is always a multiple of the #"empirical formula"#. So we solve for #n# in the equation #nxx(7xx12.011+9xx1.00794+1xx14.01)*g*mol^-1=107*g*mol^-1.#

Clearly, #n=1#, and the #"molecular formula"# #-=# #C_7H_9N#.