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Chemistry

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A 12 g sample of an unknown compound contains 3.489 g of sodium and 4.870 g of sulfur. The rest is oxygen. Name the compound

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  • Chemistry - ,

    3.489 g Na
    4.870 g S
    12.000-3.489-4.870 = 3.541 g oxygen.

    I have rounded the atomic numbers I used below; you should use correct values from the periodic table and that may change the values below slightly.
    Convert to moles.
    3.487/23 = 0.152
    4.87/32 = 0.152
    3.541/16 = 0.221

    Now we find the ratio of these elements to each other in small whole numbers with the lowest number being 1.000. The easy way to do this is to divide everything by the smallest number.
    Na = 0.152/0.152 = 1.00
    S = 0.152/0.152 = 1.00
    O = 0.221/0.152 = 1.46 and these are not small whole numbers. (It is ok to round SOME but we can't round 1.46 to 1 or 2.) So what can we do. We can multiply everything by a sequence of numbers beginning at 2,3,4,5,6 etc until we get a set of whole numbers. We try the first one (2) and
    Na = 1.00 x 2 = 2.00
    S = 1.00 x 2 = 2.00
    O = 1.46 x 2 = 2.92 which we can round to 3 so the empirical formula is
    Na2S2O3. Do you know what that is called? Na2SO4 is sodium sulfate. When an oxygen atom is replaced with a S atom, we call it a thio compound so this one be called sodium ....sulfate. You supply the ....

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