# 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#)?

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a. #"A"_3"B"_2#

b. #"AB"_3#

c. #"AB"_2#

d. #"A"_3"B"#

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

a.

b.

c.

d.

The answer is d.

##### 1 Answer

#### Explanation:

Assuming that **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**

In this case, your hypothetical compound contains **two elements**,

#"For A: " 60%#

#"For B: "40%#

So, what this tells you is that in

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

#"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")#