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Physical Chemistry
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ashwin menon
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Joined: 29 Jun 2007
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3 Mar 2008 20:38:10 IST
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CH3CHO
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3 Mar 2008 21:47:51 IST
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it'll be CH4O.
you have 24g Carbon in 64g of the compound ie. 2-C. n rest is O and H. O will be minimum 16g. remaining will be H (64-24-16= 20). 20 H with 2C is not possible. so there has to be one more O. thus remaining H will be 64-24-32=8.
thus it'll be C2H80. n the empirical formula will be CH4O.
3 Mar 2008 22:25:33 IST
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See, consider a compound containing C,H,O with molecular wt. 64. if the weight of C in it is 24, then we can say 2-C are present. n rest 40g is O and H. If 1-O is present weight of H remaining will be 24. which is not possible for a compound containing 2C. so we include 2O. thus remaining H will be 40-32=8.
its formula will be C2H8O2 n simplest will be CH4O.











