Question

A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?

Answer 1

We can say the empirical formula of a compound is the simplest whole number ratio of each type of atom in a compound. It can be the same as the compound’s molecular formula but not always. We can calculate an empirical formula from information about the mass of each element in a compound or from the percentage composition. We can say the molecular formula is the expression of the number of atoms of each element in one molecule of a compound

Complete step by step answer:
The percentage composition of hydrogen is 4.07%.
The percentage composition of carbon is 24.27%.
The percentage composition of oxygen is 17.65%.
We can write the steps for determining the empirical formula of a compound as follows:
Obtain the mass of each element present in grams.
Determine the number of moles of each atom present.
Divide the number of moles of each element by the smallest number of moles.
Convert the numbers to whole numbers. The set of whole numbers are the subscripts in the empirical formula.
Now, let’s calculate the moles of elements in compounds using their moles mass.
The molar mass of hydrogen is $1\,g/mol$.
The molar mass of carbon is $12\,\,g/mol$.
The molar mass of oxygen is $35.5\,g/mol$.
The moles of the element are calculated using the mass divided by their molar mass. We can write the expression to calculate of moles of an element as,
Moles of element=$\dfrac{{{\text{Mass}}}}{{{\text{Molecular}}\,{\text{mass}}}}$
We can calculate the moles of each element now,
The moles of hydrogen atom=$\dfrac{{4.07\,g}}{{1\,g/mol}} = 4.07\,moles$
The moles of carbon atom = $\dfrac{{24.27\,g}}{{12\,g/mol}} = 2.02\,moles$
The moles of oxygen atom = $\dfrac{{71.65\,\,g}}{{35.5\,g/mol}} = 2.01\,moles$
The moles of hydrogen, carbon and oxygen are $4.07\,moles$, $2.02\,moles$ and $2.01\,moles$ respectively.
Let us now divide all values with least value obtained
$H = \dfrac{{4.07}}{{2.01}} = 2 \\\ C = \dfrac{{2.02}}{{2.01}} = 1 \\\ Cl = \dfrac{{2.01}}{{2.01}} = 1 \\\$
We can write the empirical formula of the compound as $C{H_2}Cl$.
Let us calculate the weight of empirical formulas to calculate the molecular formula.
Weight of empirical formula = $12 + 2 + 35.5 = 49.5\,g$

The weight of the empirical formula is $49.5\,g$.
The molecular weight of the compound is $98.96\,g$.
If the molar mass value we can calculate the molecular formula by the empirical formula.
$n\,{\text{ = }}\dfrac{{{\text{Molar}}\,{\text{mass}}}}{{{\text{Mass}}\,{\text{of}}\,{\text{the}}\,{\text{empirical}}\,{\text{formula}}}}$
$n\, = \dfrac{{98.96\,g}}{{49.5\,g}} \\\ n = 2 \\\$
The molecular formula of the compound is ${\left( {C{H_2}Cl} \right)_n}$.
Here, we have calculated the value of n as two.
Therefore, we can write the molecular formula of the compound as ${\left( {C{H_2}Cl} \right)_2} = {C_2}{H_4}C{l_2}$.
The molecular formula of the compound is ${C_2}{H_4}C{l_2}$.
Therefore, we have the empirical formula of the compound is $C{H_2}Cl$ and the molecular formula of the compound is ${C_2}{H_4}C{l_2}$.

Note:
An empirical formula does not mention the arrangement or number of atoms. It is standard for many ionic compounds such as calcium chloride $\left( {CaC{l_2}} \right)$ , and for macromolecules like silicon dioxide $\left( {Si{O_2}} \right)$. The arrangement of the molecule is structural formula. The number of each type of atom in a molecule is called a molecular formula. We can determine the percent of a specific element in the sample using elemental analysis. For compounds like glucose, acetic acid, formaldehyde have the same empirical formula $C{H_2}O$ but different molecular formulas.