A10.63 g sample of mo2o3(s) is converted completely to another molybdenum oxide by adding oxygen. the new oxide has a mass of 11.340 g. add subscripts below to correctly identify the empirical formula of the new oxide.

MoO2 First, lookup the atomic weights of the elements involved Atomic weight molybdenum = 95.94 Atomic weight oxygen = 15.999 Now calculate the molar mass of Mo2O3 2 * 95.94 + 3 * 15.999 = 239.877 g/mol Now determine how many moles of the original Mo2O3 you had 10.63 g / 239.877 g/mol = 0.044314378 mol Determine how much oxygen was added 11.340 g - 10.63 g = 0.71 g How many moles of oxygen was added 0.71 g / 15.999 g/mol = 0.044377774 mol Looking at the number of moles of oxygen added and the number of moles of the original compound, they're the same. So 1 oxygen atom was added to each molecule. Since the formula was Mo2O3, the new formula becomes Mo2O4. But since you're looking for the empirical formula, you need to reduce it. Both 2 and 4 are evenly divisible by 2, so the empirical formula becomes MoO2

Wyyeurudydudyeye2y3y3y3ueuudisusuisus7w7uwh2u