Question

Question #373c7

Answer 1

Empirical formula is `C_4H_9`

Molecular formula is `C_8H_18`

Explanation:

You've been given the composition in percent which we can write as grams and then divide by the mass of the element.

`CrArr(84.21cancel(g) cancel (C))/(12 cancel(g) cancel( C))= 7.0175 mol`

`HrArr(15.79 cancel(g)cancel(C))/(1 cancel(g) cancel (C))=15.79 mol`

Now divide each by the smallest mole. i.e: `7.0175`

`CrArr(7.0175cancel (mol))/(7.0175cancel (mol))=1.0`

`HrArr(15.79cancel(mol))/(7.0175cancel(mol))=2.25`

You realize that 2.3 is not a whole number as compared to 1.0. So now we multiply by a number that will give us a whole number for H.

If you multiply by `1,2, or 3`, you still don't get a whole number until you multiply by 4.

Rememember whatever you do to H, you do same to C.

`CrArr1.0*4=4`

`HrArr2.25*4=9`

`:.` The empirical formula is `C_4H_9`

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Molecular formula is found by:

`n`(empirical formula)

`n=` (molecular weight)/(molar mass)

Molar mass of `C_4H_9= (12*4) +(1*9)=57(g)/(mol)`

`n=(114 cancel(g)/cancel(mol))/(57 cancel(g)/cancel(mol))=2`

`n(C_4H_9)=2(C_4H_9)`

`:.` the molecular formula is `C_8H_18`