Question

A compound of relative molecular mass 34 and empirical formula HO has the molecular formula:
A.${\rm{H}}{{\rm{O}}_{\rm{2}}}$
B.HO
C.${{\rm{H}}_{\rm{2}}}{\rm{O}}$
D.${{\rm{H}}_{\rm{2}}}{{\rm{O}}_2}$

Answer 1

We know that a brief representation of the molecule of a substance in terms of the symbols of the various elements present in it is termed as chemical formula. Chemical formula is of two types, empirical formula and molecular formula. The compounds are known by their molecular formula only and the empirical formulae have only theoretical importance.

Complete step by step answer:
Let’s first discuss the empirical formula of the compound. It is the formula of the simplest whole number ratio of the atoms of different elements present in one molecule of the compound For example, The empirical formula of ${{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}$ is ${\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{O}}$.

Here, first we have to calculate the ratio of relative molecular mass to the empirical formula mass and then the value obtained is to be multiplied with the empirical formula.

The empirical formula is given as HO.

The empirical formula mass of HO$= 1 + 16 = 17\,{\rm{g}}$

The relative molecular mass is given as 34.

The ratio of empirical formula mass to relative molecular is,

$\Rightarrow {\rm{Ratio}} = \dfrac{{34}}{{17}} = 2$

Now, we have to multiply 2 to the empirical formula. That means, there are 2 hydrogen and 2 oxygen atoms in the compound of HO. This indicates that the molecular formula of the compound is ${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}$.

So, the correct answer is Option D.

Note:
Always remember that, molecular formula is equal to the multiplication of n-factor and empirical formula, that is, ${\rm{Molecular}}\,{\rm{formula}} = n \times {\rm{empirical}}\,{\rm{formula}}$, Here n is the common factor and its value can be 1,2,3,4…..etc. In the case of n=1, the molecular formula is equal to the empirical formula.