# Chemistry

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
0-16.00

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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

SO2

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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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3. IT IS STIL GOING TO BE THE SAME

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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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