Question

​When a 1.50 g sample of aluminum metal is burned in an oxygen atmosphere, 2.83 g of aluminum oxide are produced. However, the combustion of 1.50 g ultrafine aluminum in air results in 2.70 g of a product, which is a mixture of 80% aluminum oxide and 20% aluminum nitride (% by mass). Use this information to determine the empirical formulas of aluminum oxide and luminum nitride.

Answer 1

Answer – Given, mass of Al metal sample = 1.50 g , mass of aluminum oxide = 2.83 g

Mass of ultrafine aluminum = 1.50 , mass of product = 2.70 and in that 80% aluminum oxide and 20% aluminum nitride (% by mass)

So we can calculate the mass of produced the aluminum oxide and aluminum nitride

First the empirical formula for the of aluminum oxide

From the first reaction –

Mass of sample of Al = 1.50 and product = 2.83 g ,

So the mass of O = 2.83 g – 1.50 g

                            = 1.33 g

So, moles of Al = 1.50 g / 26.982 g.mol^-1

                          = 0.0556 moles

Moles of O = 1.33 g / 15.998 g.mol^-1

                   = 0.0831 moles

So, the Al = 0.0556 / 0.0556 = 1

For O = 0.0831 / 0.0556 = 1.5

We need the whole number, so we need to multiply these both by 2

Al = 1*2 = 2

O = 1.5 *2 = 3

So empirical formula for the of aluminum oxide is Al₂O₃

Now we need to calculate mass of aluminum oxide and aluminum nitride for second reaction

2.70 g of product contain 80% aluminum oxide and 20% aluminum nitride (% by mass)

So, mass of aluminum oxide = 2.70*80 % /100 % = 2.16 g

Mass of aluminum nitride = 2.70*20 % / 100 % = 0.54 g

So from the first reaction we know

1.50 g of Al produced 2.83 g of Al₂O₃

So, 2.16 g of Al₂O₃ = ?

= 1.14 g of Al

Now need to calculate the how much Al used for the formed aluminum nitride is equal

= total Al - mass of Al used in the Al₂O₃

= 1.50-1.14

= 0.36 g of Al

Mass of N is from the mass of aluminum nitride and we already calculated

So, mass of N = 0.54 – 0.36 = 0.18 g

So moles of Al = 0.36 g / 26.982 g.mol^-1

                      = 0.0133 moles

Moles of N = 0.18 g / 14.007 g.mol^-1

                  = 0.0128 moles

So, the Al = 0.0133 / 0.0128 = 1

For N = 0.0128 / 0.0128 = 1

So the empirical formula for the aluminum nitride is AlN