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 your calculations.

Answer 1

First, we determine the mass of each element from the data collected. We can get the mass of molybdenum Mo from the difference between the mass of crucible and molybdenum and the mass of crucible:
     Mass of molybdenum = 39.52 – 38.26 = 1.26 g Mo

We can calculate for the mass of molybdenum oxide from the difference between the mass of crucible and molybdenum oxide and the mass of crucible:
     Mass of molybdenum oxide = 39.84 – 38.26 = 1.58g 

We can now compute for the mass of oxygen O by subtracting the mass of molybdenum from the mass of molybdenum oxide:
     Mass of oxygen in molybdenum oxide = 1.58 – 1.26 = 0.32g O

To convert mass to moles, we use the molar mass of each element.
     1.26 g Mo * 1 mol Mo / 95.94 g Mo = 0.0131 mol Mo
     0.32 g O * 1 mol O / 15.999 g O = 0.0200 mol O

0.0131 mol is the smallest number of moles. We divide each mole value by this number:
     0.0131 mol Mo / 0.0131 = 1
     0.0200 mol O / 0.0131 = 1.53

Multiplying these results by 2 to get the lowest whole number ratio,
     0.0131 mol Mo / 0.0131 = 1 * 2 = 2
     0.0200 mol O / 0.0131 = 1.5 * 2 = 3
Thus, we can write the empirical formula as Mo2O3.

Answer 2

Given,
Mass of Molybdenum = Mass of crucible and Molybdenum - Mass of crucible
                                    = 39.52gm - 38.26gm
                                    = 1.26 gm
Mass of Molybdenumoxide= Mass of crucible & Molybdenum oxide- mass of crucible
                                          = 39.84gm -38.26gm
                                          =1.58 gm

Mass of oxide = Mass of molybdenum oxide- Mass of molybdenum
                       = 1.58-1.26
                       = 0.32 gm
Now,

No. of moles of Molybdenum =  mass of molybdenum/ molecular mass of molybdenum
                                                = 1.26/96
                                                = 0.01 moles

Simillarly,
No. of moles of oxide =  0.32/16
                                   = 0.02 moles

Here, it is evident that moles of molybdenum and oxide is in 1:2 ratio.
Hence, the empirical formula would be : MoO2