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

39.84-38.26=Mo+O=1.58 grams
39.52-38.26=Mo=1.26 grams
1.58-1.26=O=.59 grams
.59/(16)=O=.036875 mol O
1.26/95.94=Mo=.0131332083 mol Mo
mol O/mol Mo=3mol O for every Mo, or MoO3

Answer 2

Answer : The empirical formula is,
`Mo_2O_3`

Solution :

First we have to calculate the mass of molybdenum and mass of oxygen.

Mass of molybdenum = Mass of crucible and molybdenum - Mass of crucible

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

Mass of oxygen = Mass of crucible and molybdenum oxide - Mass of crucible and molybdenum

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

Molar mass of Mo = 96 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of Mo =
`\frac{\text{ given mass of Mo}}{\text{ molar mass of Mo}}= (1.26g)/(96g/mole)=0.013moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (0.32g)/(16g/mole)=0.02moles`

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Mo =
`(0.013)/(0.013)=1`

For O =
`(0.02)/(0.013)=1.5`

The ratio of Mo : O = 1 : 1.5

In the whole number the ratio of Mo : O = 2 : 3

The mole ratio of the element is represented by subscripts in empirical formula.

Therefore, the empirical formula is,
`Mo_2O_3`