What is the percentage of oxygen in the sulphuric acid?
Answer
553.2k+ views
Hint: Mass percentage composition is usually expressed as the mass or mass percentage of each element which is present in a particular compound. Mass percentage is defined as the ratio of mass of a particular element to the total mass of the compound.
Complete answer:
As we know, the chemical formula of sulphuric acid is $ {H_2}S{O_4} $ which contains two hydrogen atoms, one sulphur atom and four oxygen atoms. In order to determine the percentage of oxygen in the sulphuric acid we should know the mass of each of its constituents. Some basic steps to calculate the percentage of oxygen is-
$ \to $ Calculate the mass of the compound by adding mass of its constituent including oxygen also.
$ {M_{\left( {{H_2}S{O_4}} \right)}} = {M_{\left( H \right)}} + {M_{\left( S \right)}} + {M_{\left( O \right)}} $
Where $ {M_{\left( {{H_2}S{O_4}} \right)}} $ $ = $ mass of sulphuric acid
$ {M_{\left( H \right)}} $ $ = $ mass of hydrogen atom which is equal to $ 1 $
$ {M_{\left( S \right)}} $ $ = $ mass of sulphur atom which is equal to $ 32 $
$ {M_{\left( O \right)}} $ $ = $ mass of oxygen which is equal to $ 16 $
Multiply the numerical digit which is equal to the total number of particular atoms present in the chemical formula.
$ {M_{\left( {{H_2}S{O_4}} \right)}} = 2 \times {M_{\left( H \right)}} + {M_{\left( S \right)}} + 4 \times {M_{\left( O \right)}} $
Put all the values in the above equation
$ {M_{\left( {{H_2}S{O_4}} \right)}} = 2 \times 1 + 32 + 4 \times 16 $
On simplifying the above equation, we get
$ {M_{\left( {{H_2}S{O_4}} \right)}} = 2 + 32 + 64 $
$ {M_{\left( {{H_2}S{O_4}} \right)}} = 98g/mol $
Hence, the molecular formula of sulphuric acid is $ 98 $ $ g/mol $ .
Calculate the total mass of oxygen present in the sulphuric acid
$ {M_{\left( O \right)}} $ $ = 2 \times 16 $
$ {M_{\left( O \right)}} $ $ = 64 $ $ g/mol $
Hence, $ 64 $ $ g/mol $ oxygen is present in $ 98 $ $ g/mol $ of the sulphuric acid.
Take the ratio of mass of oxygen to the total mass of sulphuric acid multiplied by $ 100 $ to calculate in terms of percentage.
$ \% {M_{\left( O \right)}} = \dfrac{{{M_{\left( O \right)}}}}{{{M_{\left( {{H_2}S{O_4}} \right)}}}} \times 100 $
Now put all the values in the above equation-
$ \% {M_{\left( O \right)}} = \dfrac{{64}}{{98}} \times 100 $
After calculating the above equation, we get
$ \% {M_{\left( O \right)}} = 65.30\% $
$ \Rightarrow $ The percentage of oxygen in the sulphuric acid is $ 65.30\% $ .
Note:
Mass percentage composition is calculated for $ 1 $ mole of the compound. Remember to convert the number of particles of an element into its molar mass before putting the value in the above formula. Mass percentage composition is dimensionless parameter. Mass percentage composition is used to determine the relative abundance of a particular element in a compound.
Complete answer:
As we know, the chemical formula of sulphuric acid is $ {H_2}S{O_4} $ which contains two hydrogen atoms, one sulphur atom and four oxygen atoms. In order to determine the percentage of oxygen in the sulphuric acid we should know the mass of each of its constituents. Some basic steps to calculate the percentage of oxygen is-
$ \to $ Calculate the mass of the compound by adding mass of its constituent including oxygen also.
$ {M_{\left( {{H_2}S{O_4}} \right)}} = {M_{\left( H \right)}} + {M_{\left( S \right)}} + {M_{\left( O \right)}} $
Where $ {M_{\left( {{H_2}S{O_4}} \right)}} $ $ = $ mass of sulphuric acid
$ {M_{\left( H \right)}} $ $ = $ mass of hydrogen atom which is equal to $ 1 $
$ {M_{\left( S \right)}} $ $ = $ mass of sulphur atom which is equal to $ 32 $
$ {M_{\left( O \right)}} $ $ = $ mass of oxygen which is equal to $ 16 $
Multiply the numerical digit which is equal to the total number of particular atoms present in the chemical formula.
$ {M_{\left( {{H_2}S{O_4}} \right)}} = 2 \times {M_{\left( H \right)}} + {M_{\left( S \right)}} + 4 \times {M_{\left( O \right)}} $
Put all the values in the above equation
$ {M_{\left( {{H_2}S{O_4}} \right)}} = 2 \times 1 + 32 + 4 \times 16 $
On simplifying the above equation, we get
$ {M_{\left( {{H_2}S{O_4}} \right)}} = 2 + 32 + 64 $
$ {M_{\left( {{H_2}S{O_4}} \right)}} = 98g/mol $
Hence, the molecular formula of sulphuric acid is $ 98 $ $ g/mol $ .
Calculate the total mass of oxygen present in the sulphuric acid
$ {M_{\left( O \right)}} $ $ = 2 \times 16 $
$ {M_{\left( O \right)}} $ $ = 64 $ $ g/mol $
Hence, $ 64 $ $ g/mol $ oxygen is present in $ 98 $ $ g/mol $ of the sulphuric acid.
Take the ratio of mass of oxygen to the total mass of sulphuric acid multiplied by $ 100 $ to calculate in terms of percentage.
$ \% {M_{\left( O \right)}} = \dfrac{{{M_{\left( O \right)}}}}{{{M_{\left( {{H_2}S{O_4}} \right)}}}} \times 100 $
Now put all the values in the above equation-
$ \% {M_{\left( O \right)}} = \dfrac{{64}}{{98}} \times 100 $
After calculating the above equation, we get
$ \% {M_{\left( O \right)}} = 65.30\% $
$ \Rightarrow $ The percentage of oxygen in the sulphuric acid is $ 65.30\% $ .
Note:
Mass percentage composition is calculated for $ 1 $ mole of the compound. Remember to convert the number of particles of an element into its molar mass before putting the value in the above formula. Mass percentage composition is dimensionless parameter. Mass percentage composition is used to determine the relative abundance of a particular element in a compound.
Recently Updated Pages
Which will be the least stable resonating structure class 11 chemistry CBSE

How many 5 digit telephone numbers can be construc-class-11-maths-CBSE

How do you find the angle of the resultant vector class 11 physics CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

What is the need and importance of classification class 11 biology CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

The teeth used for biting and cutting food are called class 11 biology CBSE

