Question

A 12.95 g sample of Mo2O3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 13.82 g. Add subscripts below to correctly identify the empirical formula of the new oxide.

Answer 1

To find the empirical formula of the new molybdenum oxide, calculate the added mass of oxygen, determine the moles of elements, and convert them to the smallest whole number ratio.

Step-by-step explanation:

The student is asking how to determine the empirical formula of a new molybdenum oxide that forms when Mo₂O₃ gains oxygen and increases in mass.

To solve this, we need to calculate the moles of O added to the original compound by subtracting the mass of Mo₂O₃ from the mass of the new oxide to find the mass of O added.

Then, determine the moles of Mo and O in the product using their atomic masses, find their ratio, and adjust it to the smallest whole number ratio to get the empirical formula.

1. Calculate the mass of oxygen added: 13.82 g - 12.95 g = 0.87 g of O.
2. Determine the moles of Mo and O in Mo₂O₃ and the added O.
3. Divide the moles of each element by the smallest number of moles obtained.
4. If needed, multiply by an integer to get whole numbers for the subscripts.

Answer 2

(18.75 g Mo2O3) x [(1 mole Mo2O3)/(239.88 g)] x [(2 moles Mo)/(1 mole Mo2O3)] = 0.1563 moles

We know that Mox = 0.1563 moles
Mo = 15.00 g Mo, which in turn means that there must be (22.50 - 15.00) grams of O or 7.50 g O, which converts to:

(7.50 g O) x [(1 mole O) / (15.999 g O)] = 0.4688 moles O

Taking the ratio of Mo to O results in:

O/Mo = [(0.4688)/(0.1563)] = 3