An unknown sample with a molecular mass of 180.0g is analyzed to yield 40% C, 6.7% H, and 53.3% O. What is the empirical formula and the molecular formula of this compound?

1 Answer
anor277
Apr 19, 2016

#"Empirical formula "=CH_2O#

#"Molecular formula"# #=# #C_6H_12O_6#

Explanation:

We assume #100*g# of compound and work out the atomic proportions:

#C: (40.0*g)/(12.011*g*mol^-1)# #=# #3.33# #mol#.

#H: (6.7*g)/(1.0794*g*mol^-1)# #=# #6.65# #mol#.

#O: (53.3.0*g)/(15.999*g*mol^-1)# #=# #3.31# #mol#.

We divide thru by the lowest molar quantity to give us the empirical formula, #CH_2O#, which is the simplest whole number ration defining constituent atoms in a species.

Now it is a fact that the molecular formula is always a multiple of the empirical formula.

Thus #"Molecular formula " =" (Empirical formula")xxn#

#(12.011+2xx1.00794+15.999)*g*mol^-1xxn = 180.0*g*mol^-1#

Clearly #n# #=# #6#, and #"molecular formula"# #=# #C_6H_12O_6#.

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