Question

Determine the empirical and molecular formula for chrysotile asbestos. Chrysolite has the following percent composition: 28.03% Mg, 21.20% Si, 1.16 % H, and 49.21% O. The molar mass of chrysotile is 520.8 g/mol.

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 by dividing all the values with the smallest values of the mole.. To find the molecular formula we have a formula that is n = $\dfrac{Molecular\,mass}{Empirical\,formula\,mass}$

Complete answer:
From your chemistry you have learned about the molecular formula, empirical formula and its relation with molecular weight.
Molecular formula represents the 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 empirical formula and the molecular formula of the chrysotile
Now to calculate the empirical formula firstly we have to convert the percentage composition of given atoms into moles by dividing the mass % by their molecular masses.
Step 1: Element: Mg Si H O
% composition: 28.03% 21.60% 1.16% 49.21%
Now the moles of Mg = $\dfrac{28.03}{24}=1.167$ (molar mass of Mg = 24)
moles of Si = $\dfrac{21.60}{28}=0.77$ (molar mass of Si = 28)
moles of H = $\dfrac{1.16}{1}=1.16$ (molar mass of H = 1)
moles of O = $\dfrac{49.21}{16}=3.07$ (molar mass of O = 16)
So, the ratio of moles of the elements of the compound Mg : Si : H : O = 1.167 : 0.77 : 1.16 : 3.07
Step 2: Now, divide these values with the smallest one to find the whole ratio, we will get:
\[\dfrac{1.167}{0.77}:\dfrac{0.77}{0.77}:\dfrac{1.16}{0.77}:\dfrac{49.21}{1.16}\]
So, the ratio of the element present in a compound will be 1.5 : 1 :1.5 : 3.9,
Thus the empirical formula will be $M{{g}_{1.5}}S{{i}_{1}}{{H}_{1.5}}{{O}_{3.9}}$$\simeq M{{g}_{3}}S{{i}_{2}}{{H}_{3}}{{O}_{8}}$
Now, we have to calculate the molecular formula so for this we will use the formula:
\[n\text{ }=\dfrac{Molecular\,mass}{\,Empirical\,formula\,mass}\] ………………(1)
So, the empirical formula mass will be calculated by finding the molar mass of the compound $M{{g}_{3}}S{{i}_{2}}{{H}_{3}}{{O}_{8}}$ and the molar mass will be 259. Molecular mass of chrysotile is given as 520.8 in the question.
Now put the values in equation (1), we will get
\[n=\dfrac{520.8}{259}=2\]
Therefore, the molecular formula will be double of the empirical formula because the value of n is 2 and if we multiply this value with the empirical one it will give molecular formula as $M{{g}_{6}}S{{i}_{4}}{{H}_{6}}{{O}_{16}}$.
Thus the empirical formula will be $M{{g}_{3}}S{{i}_{2}}{{H}_{3}}{{O}_{8}}$ and molecular formula is $M{{g}_{6}}S{{i}_{4}}{{H}_{6}}{{O}_{16}}$

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.