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Grade 12th PassAnalytical Geometry

A carbon compound contains 12.8% of carbon, 2.1% of hydrogen & 85.1% of bromine. The molecular weight of the compound is 187.9. Calculate the molecular formula of the compound? (Atomic weights:- H = 1.008, C = 12.0, B r= 79.9).

Profile image of Charitha Gorrela
9 Years agoGrade 12th Pass
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5 Answers

Profile image of Siri
ApprovedApproved Tutor Answer9 Years ago
C=12.8% at.wt. of C=12H=2.1%. at.wt. of H=1Be=85.1%. at.wt. of Br=80Now, 12.8/12 =1.067 2.1/1 = 2.1 85.1/80=1.067Dividing with 1.067 to get simple atomic ratio,1.067/1.067=12.1/1.067=21.067/1.067=1E.F= CH2BrE.F weight=12+(2×1)+80=94Molecular weight=187.9 (given)n=187.9/94= 2M.F= (E.F)×2 =(CH2Br)×2 =C2H4Br2
Profile image of Siri
9 Years ago
C=12.8% at.wt. of C=12H=2.1%. at.wt. of H=1Be=85.1%. at.wt. of Br=80Now, 12.8/12 =1.067 2.1/1 = 2.1 85.1/80=1.067Dividing with 1.067 to get simple atomic ratio,1.067/1.067=12.1/1.067=21.067/1.067=1E.F= CH2BrE.F weight=12+(2×1)+80=94Molecular weight=187.9 (given)n=187.9/94= 2M.F= (E.F)×2 =(CH2Br)×2 =(C2H4Br2)
Profile image of Siri
9 Years ago
C=12.8% at.wt. of C=12H=2.1%. at.wt. of H=1Be=85.1%. at.wt. of Br=80Now, 12.8/12 =1.067 2.1/1 = 2.1 85.1/80=1.067Dividing with 1.067 to get simple atomic ratio,1.067/1.067=12.1/1.067=21.067/1.067=1E.F= CH2BrE.F weight=12+(2×1)+80=94Molecular weight=187.9 (given)n=187.9/94= 2M.F= (E.F)×2 =(CH2Br)×2 =(C2H4Br2).
Profile image of Charitha Gorrela
9 Years ago
nice answer. thankyou.
                        but y did u send it 3 times????????????????????????????????????????????????????????????????????????????
Profile image of Vaibhav Selokar
9 Years ago
Mze carbon ki information cahie kam se kam 20 points batao please friends ..................................................................

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