Question

[Co = 59 amu]. The mass percent composition of a substance is 24.7% Ca, 1.24% H, 14.8% C and 59.3% O. Calculate the molecular formula of the substance if its molecular mass is 162 u.

Answer 1

Hint: To solve this firstly you have to convert the percent composition into the moles by dividing it with their molecular mass and then you have to find the empirical formula of the compound. To find the molecular formula we have a formula that is
$n= \dfrac{Empirical\,formula\,mass}{Molecular\,mass}$.

Complete step by step solution:
From your chemistry you have learned about the molecular formula, empirical formula and its relation with molecular weight. Molecular formula represents actual number of atoms of constituent elements present in a single molecule of a compound while Empirical formula represents a simple ratio of number of atoms of different elements present in a single molecule of a compound. In this question we have to find the molecular formula but before that we have to find the empirical formula of the compound. Now to calculate the empirical formula firstly we have to calculate the percentage composition of given atoms into moles by dividing the mass % by their molecular masses.
Step 1 : Element: Calcium Hydrogen Carbon oxygen
% composition: 24.7% 1.24% 14.8% 59.3%
Now the moles of calcium = $\dfrac{24.7}{40}=0.6175$
moles of Hydrogen = $\dfrac{1.24}{1}=1.24$
moles of carbon = $\dfrac{14.8}{12}=1.23$
moles of oxygen = $\dfrac{59.3}{16}=3.70625$
So, the ratio of moles of the elements of the compound Ca : H : C : O = 0.6175 : 1.24 : 1.23 : 3.70625
 Step 2: Now, divide these values with the smallest one to find the whole ratio, we will get:
\[\dfrac{0.6175}{0.6175}:\dfrac{1.24}{0.6175}:\dfrac{1.23}{0.6175}:\dfrac{3.70625}{0.6725}\]
So, the ratio of the element present in a compound will be 1 : 2 : 2 : 6,
Thus the empirical formula will be $Ca{{H}_{2}}{{C}_{2}}{{O}_{6}}$
Now, we have to calculate the molecular formula so for this we will use the formula:
\[n\text{ }=\dfrac{Empirical\,formula\,mass}{molecular\,mass}\]……………. (1)
So, the empirical formula mass will be calculated by finding the molar mass of the compound $Ca{{H}_{2}}{{C}_{2}}{{O}_{6}}$ and the molar mass will be 162. Molecular weight/ mass of substance is given as 162 in the question.
Now put the values in equation (1), we will get
\[n=\dfrac{162}{162}=1\]
Therefore, the molecular formula will be the same as the empirical formula because the value of n is 1 and if we multiply this value with the empirical one it will remain the same.

Thus the molecular formula is $Ca{{H}_{2}}{{C}_{2}}{{O}_{6}}$

Note: Molecular formula will be equal to n multiplied with empirical formula. If we are asked to find the percentage composition of each of the elements in a compound then we will use the formula $%\,composition=\dfrac{Atomicity\times Atomic\,mass}{Molar\,mass}\times 100$, where atomicity refers to no. of atoms of an element in a compound.