What is the molecular formula of a compound that is 62.58% carbon, 9.63% hydrogen, 27.79% oxygen?

1 Answer
Mar 27, 2016

Answer:

#C_6H_11O_2# is the simplest whole number ration that defines constituent atoms in a species, and is, therefore, the empirical formula.

Explanation:

As is the standard with these problems, we assume #100# #g# of compound, and break the percentages up into atoms.

In this mass, there are #(62.58*g)/(12.011*g*mol^-1)# #C# #=# #5.21# #mol# #C#.

And, #(9.63*g)/(1.00794*g*mol^-1)# #H# #=# #9.55# #mol# #H#.

And, #(27.79*g)/(15.999*g*mol^-1)# #O# #=# #1.740# #mol# #O#.

We divide thru, by the smallest molar quantity (#O#), and get (almost) the empirical formula: #C_3H_5.5O#.

Because the empirical formula is by definition the smallest WHOLE number ratio that describes constituent atoms in a species, we DOUBLE this formula.

#C_6H_11O_2#