A compound is found to contain 50.05 % sulfur and 49.95 % oxygen by mass. What is the simplest (empirical) formula for this compound?

They said to assume that it is 100 grams. I found the molar mass.
S- 32.06
But now I don't know what to do. Please help. Thank you!

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  1. The total masses of S and O are approximately equal, so since O has half the mass/mole of S, that means the empirical formula is probably


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  2. There is nothing wrong with Steve's answer but here is a systematic way to solve the problem in case the percentages are not the same.
    Take 100 g sample, that gives you
    50.05 g S and
    49.95 g O.

    Convert to mols
    mols S = 50.05/32.06 = 1.561
    mols O = 49.95/16 = 3.121

    Now find the ratio of the two elements to each other with the smaller being no less than 1.00. The easy way to do that is to divide the smaller number by itself (thus making sure it is 1.000); then divide the other number by the smaller number.
    1.561/1.561 = 1.000
    3.121/1.561 = 1.999 which rounds to 2 so
    empirical formula is SO2.
    By the way, I would like to point out that the molar mass of S and O are not what you listed; those numbers are the atomic mass and not molar masses.

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  4. Ok I got O1S1 but I don't think that it is right and I am really confused.

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  5. S02

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