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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?

Answer 1

Owen Ambrose

#"Empirical formula "=CH_2O#

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

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.

It is now a proven fact that the empirical formula is always multiplied by the molecular 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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Answer 2

Jose Ayala

The compound's molecular formula is C6H12O6, while its empirical formula is CH2O.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.