Question #373c7

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Answer 1 · 2017-03-27T16:35:16

Empirical formula is #C_4H_9#

Molecular formula is #C_8H_18#

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#

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#