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March 30, 2017

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6.10g of boron reacted completely with 22.9g of chlorine to give 29.0g of the metallic chloride. Calculate the empirical formula of the chloride.

  • chemistry - ,

    Use the following method:
    Mass
    Formula Mass
    Moles
    Divide by smallest ratio
    Ratio

  • chemistry - ,

    I've gone through this time and time again and something isn't clicking. Here is the way to do it BUT the answers are not reasonable.
    %B = (6.1/29)*100 = 21.03
    %Cl = (22.9/29)*100 = 78.97
    Take 100 g sample which then gives you 21.03 g B and 78.97 g Cl.

    mols B = 21.03/10.81 = 1.95
    mols Cl = 78.97/35.45 = 2.23
    B = 1.95/1.95 = 1
    Cl = 2.23/1.95 = 1.14

    Then you multiply both values by a whole number; i.e. by 2, 3, 4, 5, 6, etc until you come up with numbers that can be rounded to whole numbers. Multiplying by 7 gives 1*7 = 7 for B
    1.14*7 = 7.98 which rounds to 8.0 and that gives B7Cl8. As far as I know there is no such compound. The person making up the problem either made a typo, didn't calculate it right, OR just made up numbers that resulted in no such compound. Some profs do that but I never did. I think it's unfair to make up fake numbers to give fake compounds.

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