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A compound contains 53% Al and 47% O by mass. What
is the empirical formula of this compound?

A Compound Contains 53 Al And 47 O By Mass What Is The Empirical Formula Of This Compound class=

Sagot :

ChemhelperAZidn

Answer:

Al2O3

Explanation:

Empirical formula is found by converting percent composition into a mole ratio.

First, assume you have a 100g sample of the substance. This allows you to make your percent composition into grams.

100% = 100g

53% = 53g Al

47% = 47g O

Next, divide each substance's mass by their molar mass. This converts your number to moles.

[tex]53g Al / 27g Al = 1.96 mol\\\\47g O/16g O = 2.93mol[/tex]

Next, divide both mol answers by the smallest number of moles.

[tex]1.96mol/1.96mol = 1\\2.93mol/1.96mol = 1.5[/tex]

Usually, these answers would be your subscripts for your empirical formula. However, we cannot have a decimal subscript, so we will multiply our answers by a factor of 2 to get to a whole number. This gives us 2 moles of Al and 3 moles of oxygen, or a formula of Al2O3.

CintiaSalazaridn

Considering the definition of empirical formula, the correct answer is the second option: the empirical formula is Al₂O₃.

Definition of empirical formula

The empirical formula is the simplest expression to represent a chemical compound, which indicates the elements that are present and the minimum proportion in whole numbers that exist between its atoms, that is, the subscripts of chemical formulas are reduced to the most integers. small as possible.

Empirical formula in this case

Assuming a 100 grams sample, the percentages match the grams in the sample. So you have 53 grams of Al and 47 grams of O.

Then it is possible to calculate the number of moles of each atom in the molecule, taking into account the corresponding molar mass:

Al: [tex]\frac{53 grams}{27 \frac{grams}{mole} }[/tex]= 1.9629 moles

O: [tex]\frac{47 grams}{16 \frac{grams}{mole} }[/tex]= 2.9375 moles

The empirical formula must be expressed using whole number relationships, for this the numbers of moles are divided by the smallest result of those obtained. In this case:

Al: [tex]\frac{1.9629 moles}{1.9629 moles}[/tex]= 1

O: [tex]\frac{2.9375 moles}{1.9629 moles}[/tex]≅ 1.5

You cannot have a decimal subscript, so you will multiply our answers by a factor of 2 to get to a whole number:

Al: 1× 2= 2

O: 1.5× 2= 3

Therefore the Al:O mole ratio is 2: 3

Finally, the correct answer is the second option: the empirical formula is Al₂O₃.

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