Question

How many grams of Hydrogen are there in $23.5\,g$ of ${H_2}O$ ?

Answer 1

We will use the method of percentage composition in which you have to divide molar mass of that component by molar mass of compound. It will help you by giving an idea about the percentage of a component in a $100\,g$ of compound, then you can easily calculate the amount of hydrogen in whatever amount you have given to find.

Complete solution:
We know that a water molecule that is ${H_2}O$ contains $2$ hydrogens and $1\,$ oxygen. If we consider this fact in terms of moles then, we can write that each mole of water will have $2mole$of hydrogens and $1\,mole$ of oxygen. Now why we are using this mole term because by this we can find out the percentage of each component in water.
For percentage of hydrogen means how much percentage of hydrogen is present in $100\,g$ of water. So, let’s start with finding percentage composition for hydrogen, it has molar mass of nearly $1\,g\,mo{l^{ - 1}}$ but to be more specific we take $1.0079\,g\,mo{l^{ - 1}}$ . Now as water has $2mole$ of hydrogen, molar mass of hydrogen divided by total mass of water gives us its percentage composition.
$\dfrac{{2 \times 1.0079\,g\,mo{l^{ - 1}}}}{{18.015\,g\,mo{l^{ - 1}}}} \times 100$ Here, we write molar mass of water as $18.015\,g\,mo{l^{ - 1}}$
Now on solving further that is by dividing the terms, we get $11.19\,\%$
The above percentage gives us the information that $100\,g$ of water contains $11.19\,g$ of hydrogen. So as we used to do our basic mathematical problems we attempt the same path.
$100\,g\,of\,water\,contains \to \,11.19\,g\,of\,hydrogen$
$1\,g\,of\,water\,contains \to \,\dfrac{{11.19\,g\,of\,hydrogen}}{{100}}$
$23.5\,g\,of\,water\,contains \to \,\dfrac{{11.19\,g\,of\,hydrogen}}{{100}} \times 23.5$ = $2.63g\,of\,hydrogen$
If we try to find the percentage composition of oxygen, it will be like same as hydrogen. Dividing molar mass of oxygen with total molar mass of water and then multiplying it with $100$ to convert it into percentage.
$\dfrac{{16\,g\,mo{l^{ - 1}}}}{{18.015\,g\,mo{l^{ - 1}}}} \times 100$ = $88.8\,\%$ Now we can easily calculate amount of oxygen.

Note: To find the amount of hydrogen in $23.5\,g\,of\,water$ firstly you get to know the amount in $100\,g$ and then in $1\,g\,$ by dividing that amount by $100$. For finding the amount of hydrogen in $23.5\,g\,of\,water$ then you have to multiply the calculation with $23.5\,g\,$. These multiplication steps are more familiar because we used to do these in our basic mathematics.