Question

Question #a339b

Answer 1

I got `C_2H_4`.

Explanation:

Since the compound consists of `85.7%` of carbon and `14.3%` of hydrogen, then in `100 \ "g"` of the compound, there exist `85.7 \ "g"` of carbon and `14.3 \ "g"` of hydrogen.

Now, we need to find the amount of moles that exist in `100 \ "g"` of the compound.

Carbon has a molar mass of `12 \ "g/mol"`, while hydrogen has a molar mass of `1 \ "g/mol"`. So here, there are

`(85.7color(red)cancelcolor(black)"g")/(12color(red)cancelcolor(black)"g""/mol")~~7.14 \ "mol"`

`(14.3color(red)cancelcolor(black)"g")/(1color(red)cancelcolor(black)"g""/mol")=14.3 \ "mol"`

Since there is less carbon than hydrogen, we must divide by the number of moles of carbon.

`C=(color(red)cancelcolor(black)(7.14"mol"))/(color(red)cancelcolor(black)(7.14"mol"))=1`

`H=(14.3color(red)cancelcolor(black)"mol")/(7.14color(red)cancelcolor(black)"mol")~~2`

So, the empirical formula of the compound will be `CH_2`.

But, we are not finished yet. We know that the compound's molar mass is `32.1 \ "g/mol"`.

Since `CH_2` only has a molar mass of `14 \ "g/mol"`, then we must figure the multiplication factor to multiply `CH_2` by to get the full formula. To do that, we have to divide the molar mass of the compound by `CH_2`'s molar mass.

We got:

`(32.1color(red)cancelcolor(black)"g/mol")/(14color(red)cancelcolor(black)"g/mol")=2.29~~2`

So, we must multiply the empirical formula by `2`, and we get `C_2H_4`, which might be an isomer of ethylene.

Answer 2

`CH_2` (empirical) `approx` `C_2H_4` (molecular)

Explanation:

If we have `100g` of this compound then we can calculate the molar ratio of the two elements of the compound.

`(85.7/12.01) approx 7` mol of `Carbon`

and
`(14.3/1.01) approx 14` mol of `"Hydrogen"`

Now we divide by the smallest integer (7) to get the empirical formula:
`(7/7)=1 and (14/7)=2`

So the empirical formula becomes `CH_2`.
You mention `32.1 gram`, and I assume this is the weight of the actual molecule without empirical formula, in other words the molecular formula. However, I was unable to find a suitable molecule with the correct mass.

To find the molecular we must find the masses of the molecules in grams per mole.
Mass of `CH_2` (In grams per mole) `= 14.03`
Mass of actual molecule(in grams per mole) `= 32.1`

Their ratio is `0.43`, which, if we approximate, means you must multiply every molecule in `CH_2` with `2` to get the molecular formula , `C_2H_4`. However, this does not exactly correspond to your desired mass but it is as close we can get, from my calculations.