Turn on thread page Beta

Chemistry question watch

Announcements
    • Thread Starter
    Offline

    2
    ReputationRep:
    a compound contains C 62.08%, H 10.34% and O 27.58% by mass. find its emperical formula given that its relative molecular mass is 58
    Offline

    3
    ReputationRep:
    (Original post by itsastudentyo)
    a compound contains C 62.08%, H 10.34% and O 27.58% by mass. find its emperical formula given that its relative molecular mass is 58
    You should post a question like this in chemistry section in future.

    62.08/12 : 10.34/1 : 27.58/16

    5.17 : 10.34 : 1.72

    3 : 6 : 1

    Empirical formula = C3H6O
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by TeachChemistry)
    You should post a question like this in chemistry section in future.

    62.08/12 : 10.34/1 : 27.58/16

    5.17 : 10.34 : 1.72

    3 : 6 : 1

    Empirical formula = C3H6O
    thanks i already got that answer but i thought it was wrong becauce i thought u were to divide somwthing with the molecular mass '58' are you not meant to do anytging with that ?
    Offline

    3
    ReputationRep:
    (Original post by itsastudentyo)
    thanks i already got that answer but i thought it was wrong becauce i thought u were to divide somwthing with the molecular mass '58' are you not meant to do anytging with that ?
    No. You would only need that to confirm a molecular formula.
    Offline

    7
    ReputationRep:
    I always did these with a formula i workedout on the toilet, take the molecular mass, multiply it by the percentage/100, then divide by the element with that percentage.

    Eg. Carbon; (58 x 0.6208)/12 = 3

    Repeat for other elements

    This helps because it gives you the exact number of atoms of each element in the compound
    Offline

    3
    ReputationRep:
    (Original post by Spudgunhimself)
    I always did these with a formula i workedout on the toilet, take the molecular mass, multiply it by the percentage/100, then divide by the element with that percentage.

    Eg. Carbon; (58 x 0.6208)/12 = 3

    Repeat for other elements

    This helps because it gives you the exact number of atoms of each element in the compound
    But that assumes you know the molecular mass which is not always the case.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: March 31, 2016

University open days

  • University of Buckingham
    Psychology Taster Tutorial Undergraduate
    Fri, 14 Dec '18
  • University of Lincoln
    Mini Open Day at the Brayford Campus Undergraduate
    Wed, 19 Dec '18
  • University of East Anglia
    UEA Mini Open Day Undergraduate
    Fri, 4 Jan '19
Poll
Were you ever put in isolation at school?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.