How can I determine the empirical formula from percent composition?

I see this has been asked before, and I understand the method to solve it. My source question is this:

"A large family of boron-hydrogen compounds has the general formula BxHy.
One member of this family contains 88.5% B; the remainder is hydrogen. What
is its empirical formula? "

I have determined that there are 8.186 moles of B per 100g of BxHy and 11.409 moles of H. The next step as I understand is usually to divide the larger number of moles by the smaller to try and get a whole integer to us in your formula. But the quotient of these two numbers is 1.394. Did I make a mistake? Is there something else I should do in instances like this?

2 Answers
Sep 2, 2016

Answer:

You did nothing wrong. You just didn’t go far enough.
The empirical formula is #"B"_5"H"_7#.

Explanation:

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles. So our job is to calculate the molar ratio of #"B"# to #"H"#.

Your compound contains 88.5 % B and 11.5 % H.

Assume that we have 100 g of sample.

Then it contains 88.5 g B and 11.5 g H.

#"Moles of B" = 88.5 color(red)(cancel(color(black)("g B"))) × "1 mol B"/(10.81 color(red)(cancel(color(black)( "g B")))) = "8.187 mol B"#

#"Moles of H" = 11.5 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "11.41 mol H"#

From this point on, I like to summarize the calculations in a table.

#"Element"color(white)(m) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio"color(white)(m)"×5"color(white)(m)"Integers"#
#stackrel(—————————————————————)(color(white)(m)"B" color(white)(XXXml)88.5 color(white)(Xml)8.187 color(white)(mll)1color(white)(mmml)5color(white)(mmml)5)#
#color(white)(m)"H" color(white)(XXXXl)11.5 color(white)(mll)11.41 color(white)(Xml)1.394 color(white)(ml)6.97color(white)(mm)7#

The empirical formula is #"B"_5"H"_7#.

This result may seem strange, but boron forms a series of compounds called nido-boranes with general formula #"B"_"n""H"_text(n+4)#.

The compound is probably decaborane-[14], #"B"_10"H"_14#.

Sep 2, 2016

Answer:

Here's how you can do that here.

Explanation:

No, you did not make a mistake, your calculations are correct.

A #"100-g"# sample of this compound will indeed contain

  • #"8.186 moles B"#
  • #"11.409 moles H"#

In order to find the empirical formula of the compound, you must find the smallest whole number ratio that exists between boron and hydrogen.

After you divide both values by the smallest one, you get

#"For B: " (8.186 color(red)(cancel(color(black)("moles"))))/(8.186color(red)(cancel(color(black)("moles")))) = 1#

#"For H: " (11.409 color(red)(cancel(color(black)("moles"))))/(8.186color(red)(cancel(color(black)("moles")))) = 1.394#

The trick now is to realize that in order to get the smallest whole number ratio, you must find the smallest number that when multiplied with #1# and with #1.394# gets you two whole numbers.

In this case, that number will be #5#, since

# 5 xx 1 = 5#

#5 xx 1.394 = 6.96 ~~ 7#

The empirical formula for your boron-hydrogen compound will thus be

#color(green)(bar(ul(|color(white)(a/a)color(black)("B"_5"H"_7)color(white)(a/a)|)))#

As a final note, when you have quotients that are not whole numbers, focus on finding the smallest whole number that you can multiply your quotient by.

For example, you'll have instances where one of the quotients will be #1.33#. In such cases, multiplying by #3# will get you

#3 xx 1.33 = 4#

You can also have things like #2.25#, which can be multiplied by #4# to get

#4 xx 2.25 = 9#

or #1.6#, which can be multiplied by #5# to get

#5 xx 1.6 = 8#