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Chemistry+Empirical formula

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This is kind of tricky for me.

What os the empirical formula (lowest whole number ratio) of the compounds below?

A.

25.3% Cu, 12.9% S, 25.7% O, 36.1% water

How would i set this up? The water at the end is confusing me. I think I need to find the percentage of water inside the whole thing first, but I would be wrong. Can you help?

  • Chemistry+Empirical formula - ,

    Do it just as you did the MgBr2.
    25.3/atomic mass Cu =
    12.9/atomc mass S =
    25.7/atomc mass O =
    36.1/molar mass H2O =
    You already have the percent water in the entire molecule. The formula will come out
    as CuxSyOz*WH2O
    I don't know that this is the formula but it would look somthing like this.
    CuSO4*5H2O

  • Chemistry+Empirical formula - ,

    I'm still confused. Okay I solved everything but H20

    25.3g Cu 1 mole/63.54 = .40
    12.9g S 1 mole/32.064 = .40
    25.7g O 1 mole/15.999 = 1.61

    But what do i do with the H20? Can you work this part for me? I want to see how you did it

  • Chemistry+Empirical formula - ,

    Cu = 25.3/63.54 = 0.398 (You have 3 places in 25.3 so don't throw anything away, at least not yet).
    S = 12.9/32.064 = 0.402
    O = 25.7/15.999 = 1.606
    H2O = 36.1/18.015 = 2.00
    Now we divide everything by the smallest, which is 0.398

    Cu = 0.398/0.398 = 1.000
    S = 0.402/0.398 = 1.01
    O = 1.606/0.398 = 4.035
    H2O = 2.00/0.398 = 5.025

    So we round off to whole numbers now.
    Cu 1.000 becomes 1.00
    S 1.01 becomes 1.00
    O 4.035 becomes 4.00
    H2O 5.025 becomes 5.00
    The formula is
    Cu1S1O4*5H2O or since the 1s are not needed, the formula is CuSO4*5H2O

  • Chemistry+Empirical formula - ,

    My teacher always told us to round to the hundreths place. How did you get 18.015 ??

  • Chemistry+Empirical formula - ,

    aah wait i see how now.

  • Chemistry+Empirical formula - ,

    I use a web site to do all my molar mass stuff. Here is a link if you want to use it. http://environmentalchemistry.com/yogi/reference/molar.html

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