Question

How would you determine the empirical and molecular formula for hydrogen peroxide, which is $$5.94\mathrm{\%}$$ H and $$94.06\mathrm{\%}$$ O and has a molar mass of $\text{34}\text{.01g/mol}$?

Answer 1

Hint: Empirical formulae is the smallest base unit that makes up the molecular formulae.
The empirical formula is related to mass formulae as,
$\text{Molecular}\text{formula=n }\times \text{ Empirical}\text{formula}$

Complete answer:
So in the question, it is asked how we will determine the empirical and molecular formula for hydrogen peroxide if the percentage of H and O is given and the molar mass of the compound i.e. hydrogen peroxide is given.
Let’s first understand what is the difference between an empirical formulae and the molecular formula of a compound?
 The empirical formula of a compound gives the simplest whole number ratio of the atoms present in the compound whereas in molecular formulae it tells exactly about all the kinds of atoms present in the formulae.
So here let’s solve for empirical formulae first from which we could find the molecular formulae of the compound.
Since we know that the percentage ranges till 100% and we would take 100-x to get the mass of the sample in g.
The percentage composition of H and O will make up 100%.
Now convert the percentage of atoms given to grams.
$\text{5}\text{.94 }$
$\text{94}\text{.06 }$
Now we have to find the number of moles of H and O. For calculating the number of moles of each element we will divide the given mass of the element with its atomic mass in $\text{g/mol}$
Therefore, the moles of H =$\text{5}\text{.94g}\text{of}\text{H }\times {\displaystyle \frac{\text{1}\text{mol}\text{of}\text{H}}{\text{1}\text{.00794g}\text{of}\text{H}}}\text{=5}\text{.89mol}\text{of}\text{H}$
Moles of O =$\text{94}\text{.06g}\text{of}\text{O }\times {\displaystyle \frac{\text{1}\text{mol}\text{of}\text{O}}{\text{15}\text{.999g}\text{of}\text{H}}}\text{=5}\text{.88mol}\text{of}\text{O}$
Since both the H and O have almost the same number of moles there is the same number of H and O in the compound and we could write the empirical formulae as HO , in which the H and O is having the simplest whole number.
Now we have to find the molecular formulae of the compound, we should know the multiplying factor ‘n’ in the equation,
$\text{Molecular}\text{formula=n }\times \text{ Empirical}\text{formula}$
For finding n we have to know the empirical mass.
Empirical mass is the total mass of the empirical formulae and it is the summation of the atomic mass of the atoms in the empirical formulae.
The empirical formula is HO and the atomic mass of H is 1.00794g and that of O is 15.999g.
$\text{Empirical}\text{mass=1 }\times \text{ 1}\text{.00794+1 }\times \text{ 15}\text{.999=17}\text{.0069g/mol}$
We can get the multiplying factor n by the equation,
$\text{n=}{\displaystyle \frac{\text{Molecular}\text{mass}}{\text{Empirical}\text{mass}}}$
Molecular mass is given as $\text{34}\text{.01g/mol}$
$\text{n=}{\displaystyle \frac{\text{34}\text{.01g/mol}}{\text{17}\text{.0069g/mol}}}\text{=1}\text{.9999=2}$
Now find the molecular formulae using the equation,$\text{n=}{\displaystyle \frac{\text{34}\text{.01g/mol}}{\text{17}\text{.0069g/mol}}}\text{=1}\text{.9999=2}$
$\text{Molecular}\text{formula=n }\times \text{ Empirical}\text{formula}$
$\text{Molecular}\text{formula=2 }\times \text{ HO=}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$
So finally we got the molecular formulae of hydrogen peroxide as ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$

Note:
Since there are few equations which may confuse us between the numerator and denominator terms, it is always better to study a standard solution and rearrange the standard equations according to various questions asked.
The empirical formula is just the simplest whole number ratio; it only tells us a possibility of the chemical combinations taking place between the atoms when there is no data available. The molecular formulae too will have the whole number as its sub-script.