Q.no. :48,plz solve it step by step wih calculations,thanks

Asked by vishakhachandan026 | 28th Jun, 2019, 05:40: PM

Expert Answer:

Given:
C =70.6%, H=4.2%, N=11.8%, O=13.4%.
 
Consider, 100 gm of kevlar will have,
C =70.6 gm,
H=4.2 gm,
N=11.8 gm,
O=13.4 gm.
 
Lets find out the moles of each element, as we know the molar masses,
 
C=12, H=1, N=14, O=16
 
straight C equals fraction numerator 70.6 over denominator 12 end fraction space equals space 5.88 space mol
straight H equals fraction numerator 4.2 over denominator 1 end fraction space equals space 4.2 space mol
straight N equals fraction numerator 11.8 over denominator 14 end fraction space equals 0.84 space mol
straight O equals fraction numerator 13.4 over denominator 16 end fraction space equals space 0.84 space mol
Lets divide each with smallest moles i.e., 0.84 mol we get,
 
straight C equals fraction numerator 5.88 over denominator 0.84 end fraction space equals 7
straight H equals fraction numerator 4.2 over denominator 0.84 end fraction space equals 5
straight N equals fraction numerator 0.84 over denominator 0.84 end fraction space equals 1
straight O equals fraction numerator 0.84 over denominator 0.84 end fraction space equals 1
 
Hence the emperical formula will be C7H5NO
 
 
 

Answered by Ramandeep | 28th Jun, 2019, 06:27: PM

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