# chemistry

A 15.27 g sample of Mo2O3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 18.33 g. Add subscripts below to correctly identify the empirical formula of the new oxide.

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1. Mo2O3 has mass 15.27g
% Mo in Mo2O3 = (2*95.94/239.88)*100 = about 80% but you should do that more accurately.
Out of a 15.27g sample, 15.27 x 80% = about 12.2 g Mo and the rest is oxygen.

If you started with 12.2 g Mo there must be 12.2 g in the new oxide which means oxygen in the new oxide is 18.33-12.21 = about 6g.

mols Mo = 12.2/95.94 = about 0.13
mols O = 6/16 = about 0.38
Find the ratio of Mo to O. The easy way to do that is to divide the smaller number and the other number by itself; ie, 0.13/0.13 = 1.00 Mo and 6/0.13 = 2.9 O which rounds to Mo1O3 or MoO3.
You can go through and clean up the numbers which will make the ratio come out a little better.

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2. m k

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