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An oxide of aluminum contains 0.545 g of Al and 0.485 g of O. Find the empirical formula
for the oxide.

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yersanabria

Answer:

The answer is: Al2O3

Explanation:

The data they give us is:

  • 0.545 gr Al
  • 0.485 gr O.

To find the empirical formula without knowing the grams of the compound, we find it per mole:

  • 0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al
  • 0.485 g O * 1 mol O / 16 g O = 0.03 mol O

Then we must divide the results obtained by the lowest result, which in this case is 0.02:

  • 0.02 mol Al / 0.02 = 1  Al
  • 0.03 mol O / 0.02 = 1.5  O

Since both numbers have to give an integer, multiply by 2 until both remain integers:

  • 1Al * 2 = 2Al
  • 1.5O * 2 = 3O

Now the answer is given correctly:

  • Al2O3

The empirical formula for the oxide is:  Al₂O₃

Given:

Mass of Al= 0.545g

Mass of O=0.485 g

Empirical formula

It is the simplest whole number ratio of atoms present in a compound.

In order to find empirical formula:

0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al

0.485 g O * 1 mol O / 16 g O = 0.03 mol O

Then we must divide the results obtained by the lowest result, which in this case is 0.02:

0.02 mol Al / 0.02 = 1  Al

0.03 mol O / 0.02 = 1.5  O

In order to get a integer value we will multiply it by 2:

1 Al * 2 = 2Al

1.5 O * 2 = 3O

Thus, the empirical formula so formed will be: Al₂O₃.

Find more information about "Empirical formula" here:

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