A 0.596 gm sample of a gaseous compound containing only boron and hydrogen occupies 484 cc at 273°K and 1-atm pressure. When the compound was ignited in excess oxygen all its hydrogen was recovered as 1.17 gm of H2O, and all the boron was present as B2O3. What is the empirical formula, the molecular formula, and the molecular weight of the boron-hydrogen compound? What weight of B2O3 was produced by the combustion?
Answers
Answer:
start by figuring out how much hydrogen you get per mole of water. You know that every mole of water, H2O, 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!
2×1.00794g/mol18.015g/mol×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
1.17g water⋅11.19 g H100g 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
mcompound=mhydrogen+mboron
mboron=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.
0.1309g⋅1 mole H1.00794g=0.12987 moles H
and
0.465g⋅1 mole B10.811g=0.043012 moles B
Now divide both these values by the smallest one to get
For H: 0.12987moles0.043012moles=3.0194≈3
For B: 0.043012moles0.043012