How do you solve this question?

Question

A compound that is composed of molybdenum (Mo) and oxygen (O) was produced in a lab by heating molybdenum over a Bunsen burner. The following data was collected:

Mass of crucible: 38.26 g
Mass of crucible and molybdenum: 39.52 g
Mass of crucible and molybdenum oxide: 39.84 g

Solve for the empirical formula of the compound, showing (or explaining in complete sentences) your calculations.

Answer 1

First subtract the weight of the crucible from the mass of the molybdenum, then subtract the mass of both from the 3rd mass, so that you find the individual masses of the elements

 

 

Mass of Mo = 1.26g (from 39.52 - 38.26)

Mass of O = 0.32g   (from 39.84 - 39.52)

 

Divide these by their molar masses

 

Moles of Mo = 0.01313  (from 1.26/95.95)

Moles of O = 0.02         (from 0.32/16.00)

 

So for every 0.02 moles of oxygen, we have 0.01313 moles of Mo. We can simplify this ratio by dividing each number by 0.01313

 

0.01313/0.01313 =1

0.02/0.01313 = 1.52

 

So for every mole of Mo we have 1.5 moles of O. We need a whole number ratio, so double these numbers.

For every 2 moles of Mo, we have 3 moles of O.

 

Empirical formula: Mo₂O₃