A 11.79 g sample of Mo2O3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 14.151 g . Add subscripts to correctly identify the empirical formula of the new oxide.

MoO₃

Explanation:

To solve this question we must find the moles of molybdenum in Mo2O3. The moles of Mo remain constant in the new oxide. With the differences in masses we can find the mass of oxygen and its moles obtaining the empirical formula as follows:

Moles Mo2O3 -Molar mass: 239,878g/mol-

11.79g * (1mol / 239.878g) = 0.04915 moles Mo2O3 * (2mol Mo / 1mol Mo2O3) = 0.09830 moles Mo

Mass Mo in the oxides:

0.09830 moles Mo * (95.95g/mol) = 9.432g Mo

Mass oxygen in the new oxide:

14.151g - 9.432g = 4.719g oxygen

Moles Oxygen:

4.719g oxygen * (1mol/16g) = 0.2949 moles O

The ratio of moles of O/Mo:

0.2949molO / 0.09830mol Mo = 3

That means there are 3 moles of oxygen per mole of Molybdenum and the empirical formula is:

11 players on a field that has a maximum of 11 players