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
