Question:

Determine the empirical formula of a compound containing 72.22% of C, 7.07% of H, 4.68% of N ,16.03% of O?

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Determine the empirical formula of a compound containing 72.22% of C, 7.07% of H, 4.68% of N ,16.03% of O?

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  1. C: 72.22% / 12  = C6

    H: 7.07% / 1 = H7

    N: 4.68% / 14 = 0.33N

    O: 16.03% / 16 = O1

    With N = 0.33 then, we need to multiply everything by 3 to give...

    C18H21NO3 which is Codeine (A pain killer).


  2. Empirical Formula is just basically the ratio of the molecular formula. For example H202 is the molecular formula of hydrogen perox. and the ratio of H:0 is 2:2 or 1:1. so the empirical formula is HO.

    to solve this problem:

    C= 72.22 ( 1M / 12)=6.02

    H= 7.07 (1m/ 1) = 7.07

    N = 4.68 (1/14) = 0.334

    O = 16.03 (1/ 16)= 1.001

    until this point, divide by by the smallest number which in this case is 0.334.

    C= 6.02/ .334 = 18 C

    H= 7.07/ .334  = 21 H

    N 0.334/ 0.334 = 1 N

    O= 1.01/ 0.334 =3 0

    conclusion: the empirical formula is C18H21N03 (codeine) alleviates pain

    hope this helped!

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