
One of the hydrides of boron has molar mass $ 122gmo{l^{ - 1}} $ and is $ 11.55\% $ by mass hydrogen. Determine its molecular formula?
Answer
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Hint: Given that the percentage of hydrogen in borane compound. From this percentage of hydrogen and molar mass the number of moles will be calculated. From the number of moles of boron, hydrogen mole ratio will be calculated and, mole ratio gives the empirical formula. From the empirical mass and molecular mass, the molecular formula can be determined.
Complete answer:
Given that hydride of boron has a molar mass of $ 122gmo{l^{ - 1}} $
The percentage of hydrogen by mass is $ 11.55\% $
Let us consider hydride of boron has $ 100\% $
Moles of boron will be $ \dfrac{{\left( {100 - 11.55} \right)}}{{10.8}} = 8.18moles $
Moles of hydrogen will be $ \dfrac{{11.55}}{{1.008}} = 11.46 $
Now, divide the moles of boron and hydrogen with the moles of boron, as moles of boron were less.
Mole ratio of $ B:H $ is $ \dfrac{{8.18}}{{8.18}}:\dfrac{{11.46}}{{8.18}} = 1:1.4 $
As the mole ratio must be simple whole number, multiply the mole ratio with $ 5 $
Then the empirical formula will be $ {B_5}{H_7} $ as empirical formula is the simplest whole number
The empirical mass will be $ 5\left( {10.8} \right) + 7\left( 1 \right) = 61gmo{l^{ - 1}} $
Given molecular mass is $ 122gmo{l^{ - 1}} $
The empirical mass is not equal to molecular mass. To determine the molecular formula, the n-factor has to be determined which relates the empirical mass and molecular mass.
$ n = \dfrac{{61}}{{122}} = 2 $
Thus, the molecular formula is $ {B_{10}}{H_{14}} $ which can be known as deca borane.
Note:
Boranes are hydrides of boron, only boron and hydrogen atoms will be present in boron, Thus, the whole mass was taken as $ 100\% $ to determine the mass of boron. Moles will be obtained by dividing the mass of an atom with the molar mass of that atom.
Complete answer:
Given that hydride of boron has a molar mass of $ 122gmo{l^{ - 1}} $
The percentage of hydrogen by mass is $ 11.55\% $
Let us consider hydride of boron has $ 100\% $
Moles of boron will be $ \dfrac{{\left( {100 - 11.55} \right)}}{{10.8}} = 8.18moles $
Moles of hydrogen will be $ \dfrac{{11.55}}{{1.008}} = 11.46 $
Now, divide the moles of boron and hydrogen with the moles of boron, as moles of boron were less.
Mole ratio of $ B:H $ is $ \dfrac{{8.18}}{{8.18}}:\dfrac{{11.46}}{{8.18}} = 1:1.4 $
As the mole ratio must be simple whole number, multiply the mole ratio with $ 5 $
Then the empirical formula will be $ {B_5}{H_7} $ as empirical formula is the simplest whole number
The empirical mass will be $ 5\left( {10.8} \right) + 7\left( 1 \right) = 61gmo{l^{ - 1}} $
Given molecular mass is $ 122gmo{l^{ - 1}} $
The empirical mass is not equal to molecular mass. To determine the molecular formula, the n-factor has to be determined which relates the empirical mass and molecular mass.
$ n = \dfrac{{61}}{{122}} = 2 $
Thus, the molecular formula is $ {B_{10}}{H_{14}} $ which can be known as deca borane.
Note:
Boranes are hydrides of boron, only boron and hydrogen atoms will be present in boron, Thus, the whole mass was taken as $ 100\% $ to determine the mass of boron. Moles will be obtained by dividing the mass of an atom with the molar mass of that atom.
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