Question

A compound has a percent composition of 40.0% carbon, 6.72% hydrogen and 53.28% oxygen.if its molar mass is 180 g/mol, what is its molecular formula?

Answer 1

First act as if each percentage is mass, then divide by the element's molar mass.


`C = \frac {40}{12.01} = 3.331`

`H= \frac {6.72}{1.01}=6.653`

`O= \frac {53.28}{16} = 3.330`

Then, out all the solutions we must figure which one is the least. Continued by dividing each solution by the smallest solution. We can see that 3.330 is the least solution. Each answer should be rounded to a whole number, so they can become subscripts for the empirical formula.


`C=(3.331)/(3.330) = 1.0003 = 1`

`H= (6.653)/(3.330) = 1.998=2`

`O= (3.330)/(3.330)= 1`

The empirical formula is
`CH_2O`

Now we find the molar mass of the empirical formula and divide the molecular mass by the empirical formula mass. Finally round to a whole number.

Empirical Formula Mass:
`30.04g/mol`


`\frac {180g/mol}{30.04g/mol} = 5.992 = 6`

Finally we multiply 6 by the Empirical Formula. That will give the molecular formula.


`6*CH_2O=C_6H_(12)O_6`

The answer is
`C_6H_(12)O_6`.