A 0.596 g sample of a gaseous compound containing only boron and hydrogen occupied 484 cm^3 at STP. When ignited in excess oxygen all the hydrogen was recovered as 1.17 g of water, and all the boron was present as B2O3. What is the empirical formula?

1 Answer
Oct 29, 2015

Answer:

Empirical formula: #"BH"_3#
Molecular formula: #"B"_2"H"_6#

Explanation:

!! LONG ANSWER !!

The trick here is to recognize the fact that you can use the mass of water produced by the reaction to determine how much hydrogen the original compound contained, then use the original sample's mass to determine how much boron it contained.

Basically, you don't need to worry about the compound that contains the boron, #"B"_2"O"_3#, since that has no relevance to finding the mass of boron.

So, start by figuring out how much hydrogen you get per mole of water. You know that every mole of water, #"H"_2"O"#, contains

  • *2 moles of hydrongen*
  • *1 mole of oxygen*

This means that if you use the molar masses of the two elements, you can determine how much hydrogen you get per mole of water, i.e. the percent composition of water

Don't forget about the fact that you have #2# moles of hydrogen!

#(2 xx 1.00794color(red)(cancel(color(black)("g/mol"))))/(18.015color(red)(cancel(color(black)("g/mol")))) xx 100 = 11.19% "H"#

This means that you get #"11.19 g"# of hydrogen for every #"100 g"# of water. This means that the reaction produced

#1.17color(red)(cancel(color(black)("g water"))) * "11.19 g H"/(100color(red)(cancel(color(black)("g water")))) = "0.1309 g H"#

SInce all the hydrogen that was originally present in the compound is now a part of the water, it follows that your compound contained #"0.1309 g"# of hydrogen.

This means that it also contained

#m_"compound" = m_"hydrogen" + m_"boron"#

#m_"boron" = "0.596 g" - "0.1309 g" = "0.465 g B"#

To get the compound's empirical formula, you need to figure out what the mole ratio betyween boron and hydrogen is in the compound.

#0.1309color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "0.12987 moles H"#

and

#0.465color(red)(cancel(color(black)("g"))) * "1 mole B"/(10.811color(red)(cancel(color(black)("g")))) = "0.043012 moles B"#

Now divide both these values by the smallest one to get

#"For H: " (0.12987 color(red)(cancel(color(black)("moles"))))/(0.043012color(red)(cancel(color(black)("moles")))) = 3.0194 ~~ 3#

#"For B: " (0.043012 color(red)(cancel(color(black)("moles"))))/(0.043012color(red)(cancel(color(black)("moles")))) = 1#

The empirical formula of the compound wil thus be

#"B"_1"H"_3 implies "BH"_3#

Notice that they tell you that the compound occupies a volume of #"484 cm"^3# at STP, which means that you can use the molar volume of a gas at STP to find how many moles you ahve in that sample.

At STP conditions, a pressure of #"100 kPa"# and a temperature of #0^@"C"#, one mole of any ideal gas occupies exactly #"22.7 L"#. This means that your sample will contain

#484color(red)(cancel(color(black)("cm"^3))) * (1color(red)(cancel(color(black)("L"))))/(1000color(red)(cancel(color(black)("cm"^3)))) * "1 mole"/(22.7color(red)(cancel(color(black)("L")))) = "0.02132 moles"#

Now you can find the compound's molar mass

#M_"M" = m/n = "0.596 g"/"0.02132 moles" = "27.95 g/mol"#

You can use this value to find the compound's molecular formula

#("10.811 g/mol" + 3 xx "1.00794 g/mol") xx color(blue)(n) = "27.95 g/mol"#

This means that you have

#color(blue)(n) = (27.95color(red)(cancel(color(black)("g/mol"))))/((10.811 + 3 * 1.00794)color(red)(cancel(color(black)("g/mol")))) = 2.0203 ~~ 2#

The molecular formula will thus be

#("BH"_3)_color(blue)(2) = "B"_2"H"_6 -># diborane