Acompound 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 : The empirical formula is,

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 =

Moles of O =

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

For Mo =

For O =

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,

the answer to your question is c.

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