
One of the hydrides of boron has molar mass $ 122gmo{l^{ - 1}} $ and is $ 11.55\% $ by mass hydrogen. Determine its molecular formula?
Answer
483.6k+ views
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.
Recently Updated Pages
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

Trending doubts
What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

An example of chemosynthetic bacteria is A E coli B class 11 biology CBSE

Name the metals and nonmetals in the first twenty class 11 chemistry CBSE

Which one of the following is not a method of soil class 11 biology CBSE

