Question

Question #df0c6

Answer 1

If you're dealing with a hydrocarbon, which is a compound that only contains carbon and hydrogen, then it should actually have `"14.29%"` hydrogen, rather than `"13.29%"`.

Because a hydrocarbon only contains carbon and hydrogen, the percentages must add to 100%. I'll show you how to solve this using `"14.29%"` for hydrogen.

So, you know from the percentages given that a 100-g sample of this compound will contain 85.71 g carbon and 14.29 hydrogen. The first step in determining a compound's empirical formula is to divide each element's sample mass by its atomic mass. This wil ltell you how many moles of each element the sample contains

`"For C": "85,71 g"/"12.0 g/mol" = "7.14 moles carbon"`

`"For H": "14.29 g"/"1.01 g/mol" = "14.1 moles hydrogen"`

Now divide both these numbers by the smallest one to get the mole ratio of the two elements in the compound

`"For C": "7.14"/"7.14" = 1`

`"For H": "14.1"/"7.14" = 1.97 ~= 2`

Therefore, the empirical formula for your compound is `C_1H_2`, or `CH_2`.

SIDE NOTE If you use `"13.29%"` for hydrogen, you'll get the ratio for hydrogen equal to `13.29/7.14 = 1.87`, which could be considered equal to `2`.