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,

it's c. elements are placed in rows based on how may valencies electrons have.

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