What would be the empirical formula of a hypothetical compound that contains 60% by mass of element A (#A_(r1) = 10#) and 40% by mass of element B (#A_(r2) =20#)?

a. #"A"_3"B"_2#
b. #"AB"_3#
c. #"AB"_2#
d. #"A"_3"B"#

The answer is d. #"A"_3"B"#, why?

1 Answer
Stefan V.
Feb 17, 2016

#"A"_3"B"#, indeed.

Explanation:

Assuming that #A_1# and #A_2# represent the molar masses of the two elements, here's what's going on here.

A compound's percent concentration by mass essentially tells you how many grams of each element that makes up said compound you get per #"100 g"# of compound.

In this case, your hypothetical compound contains two elements, #"A"# and #"B"#, each having the following percent concentration by mass

#"For A: " 60%#

#"For B: "40%#

So, what this tells you is that in #"100 g"# of this compound, you will have #"60 g"# of #"A"# and #"40 g"# of #"B"#.

Now, in order to find the compound's empirical formula, you need to know the smallest whole number ratio that exists between its constituent elements.

Use the two molar masses to determine how many moles of each element are present in this #"100-g"# sample

#"For A: " 60 color(red)(cancel(color(black)("g"))) * "1 mole A"/(10color(red)(cancel(color(black)("g")))) = "6 moles A"#

#"For B: " 40color(red)(cancel(color(black)("g"))) * "1 mole B"/(20color(red)(cancel(color(black)("g")))) = "2 moles B"#

To get the smallest whole number ratio, divide both values by the smallest one

#"For A: " (6color(red)(cancel(color(black)("moles"))))/(2color(red)(cancel(color(black)("moles")))) = 3#

#"For B: " (2color(red)(cancel(color(black)("moles"))))/(2color(red)(cancel(color(black)("moles")))) = 1#

This means that the empirical formula of the compound is

#"A"_3"B"_1 implies color(green)("A"_3"B")#

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