Question

How can I calculate the percent composition of ${{C}_{4}}{{H}_{4}}O$?

Answer 1

Hint: Percent composition of any compound can be defined as the ratio of amount of every individual element to the total amount of individual elements present in the compound multiplied by 100. This helps in chemical analysis of given compounds.

Complete answer:
Hence to find out the percentage composition of ${{C}_{4}}{{H}_{4}}O$ we have to divide the total mass of each atom by the molecular mass and multiply it by 100. The formula used is shown as:
Percentage by mass = $\dfrac{mass\text{ of component}}{total\ \text{mass}}\times 100$
Mass of 4 carbon atoms = $4\ \text{C atoms }\times \ \dfrac{12.01}{1\ \text{C atom}}=48.04u$
Mass of 4 hydrogen atoms = $4\ H\text{ atoms }\times \ \dfrac{1.008}{1\ H\text{ atom}}=4.032u$
Mass of 1 oxygen atom = \[1\ O\text{ atom }\times \ \dfrac{16}{1\ O\text{ atom}}=16u\]
Total mass of ${{C}_{4}}{{H}_{4}}O$ molecule = Mass of 4 carbon atoms + Mass of 4 hydrogen atoms + Mass of one oxygen atom
= 48.04 u + 4.032 u + 16 u = 68.07 u
Hence percentage of atoms can be calculated by the following manner
% of C = $\dfrac{mass\ \text{of C}}{total\ \text{mass}}\times 100%=\dfrac{48.04}{68.07}\times 100%=70.57%$
% of H = $\dfrac{mass\ \text{of H}}{total\ \text{mass}}\times 100%=\dfrac{4.032}{68.07}\times 100%=5.923%$
% of O = $\dfrac{mass\ \text{of O}}{total\ \text{mass}}\times 100%=\dfrac{16}{68.07}\times 100%=23.50%$
Hence from the above calculation we can consider that percentage of carbon atoms in ${{C}_{4}}{{H}_{4}}O$ is 70.51%, percentage of hydrogen atom in given compound is 5.923 and percentage of oxygen atom in ${{C}_{4}}{{H}_{4}}O$ is 23.50% and the overall percentage is almost 100.
In this way we can calculate the percentage composition.

Note:
${{C}_{4}}{{H}_{4}}O$is five-membered aromatic ring with four carbon atoms and one oxygen atom and this is also kept in the category of heterocyclic compounds and known by the name furan. This is colorless, flammable and highly volatile in nature.