Cr2O3 is the empirical formula of a compound that is 68.42% Cr and 31.58% O by mass.
- The masses in grams are pretend percentages: 68.4 g Cr 31.6 g O
- Divide both by the periodic table mass to get the moles: 68.4gCr/51.996=1.316 mol Cr 31.6gO/16.00=1.975 mol O
- The lesser quantity (1.316 is less than 1.975) should be used to divide the two mole values: Cr 1.316/1.316=1 O 1.975/1.316=1.5
- Since 1.5 is not a whole number, and multiplying 0.5 by 2 gets rid of the decimal, multiply both values by 2
The subscripts are Cr2O3.
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