Question

An oxide of aluminum is formed by the reaction of 4.151 g of aluminium and 3.692 g of oxygen. Calculate the empirical formula.​

Answer 1

Explanation:

FIrst, you calculate the number of moles that reacted for both elements - it's a little easier to use monoatomic oxygen, not diatomic oxygen

4.151 g

⋅

1 mole aluminium

26.98 g

=

0.1539 moles

A

l

, and

3.692 g

⋅

1 mole monoatomic oxygen

16.00 g

=

0.2308 moles

O

Now just divide these two numbers to get the mole ratio of aluminium to oxygen in the oxide

aluminium

oxygen

=

0.1539

0.2308

=

0.667

=

2

3

This means that the empirical formula for your oxide will be

A

l

2

O

3

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