How would you find the molecular formula for a compound with molar mass 180 amu, that is composed of 40% carbon 6,67% hydrogen 53.3 oxygen?

1 Answer
Jan 3, 2016

Answer:

#C_6H_12O_6#

Explanation:

We assume 100 g of compound.

Elemental composition is divided thru by the atomic mass of each element:

#C: (40*cancelg)/(12.011*cancelg*mol^-1)# #=# #3.33*mol#.

#H: (6.67*cancelg)/(1.00794*cancelg*mol^-1)# #=# #6,62*mol#.

#O: (53.3*cancelg)/(15.99*cancelg*mol^-1)# #=# #3.33*mol#.

So the empirical formula is #CH_2O#, after we divide thru by the lowest quotient.

Now we know that the molecular formula is always a whole number mulitple of the empirical formula:

i.e. #(12.011+2xx1.0074+15.99)_n*g*mol^-1# #=# #180*g*mol^-1.#

Thus #n# #=# #6#, and the molecular formula is #C_6H_12O_6#.