Question

1. The experimental empirical formula of a magnesium oxide compound was determined. To do this a magnesium ribbon weighing 0.252 g was heated in a crucible. From the data listed, determine the empirical formula for the magnesium oxide product. Please show all work! I'm trying to really understand

          Mass of empty crucible and cover: 20.74 g

          Mass of crucible, cover, and final product: 21.17 g

Answer 1

We know that,

Molar mass of Mg = 24.30g/mol

Molar mass of O (oxygen) = 16.00g/mol

And

Number of of x =( mass of x)/(molar mass of x) .....(1)

Given data ,

Mass of Mg ribbon = 0.252g

Mass of empty crucible and cover = 20.74g

Mass of empty crucible and cover and final product = 21.17g

Step(1) Calculation For mass of Oxygen.

mass of Oxygen will be given as follows,

mass of Oxygen = (mass of empty crucible and cover and final product) - [( mass of empty crucible and cover) + (mass of magnesium)]

Mass of O = (21.17g) - (20.74g + 0.252g)

Mass of O = (21.17g) -(20.992g)

Mass of O = 0.168g

step(2) Calculation For number of moles of Mg and O.

Using the equation number (1) and above data

Number of moles of Mg = (0.252g)/(24.30g/mol)

Number of moles of Mg = 0.010mol

And

Number of mole of O = (0.168g)/(16.00g/mol)

Number of mole of O = 0.010mol

Step(3)

Converting the number of mole of Mg and O in whole number by multiplying 100

So,

Number of mole of Mg = (0.010mol)(100)

Number of mole of Mg = 1mol

amd

Number of mole of O = (0.010mol)(100)

Number of mole of O = 1mol

step(4) Chemical formula of product will be given as follows.

chemical Formula

= Mg(number of mol of Mg)O(number of mole of O)

Chemical Formula = Mg​​​​​1​​​​O​​​1

Or

Chemical Formula = MgO