Question

Calculate the mass of oxygen in a molecule of $C{O_2}$ by using percentage composition?

Answer 1

Hint: Percentage composition is a way to determine the percentage of every element which makes up a particular compound. Mathematically, percentage composition is defined as the ratio of the amount of every element to the total amount of individual elements present in a compound multiplied by \[100\].

Complete answer:
Percentage composition of any compound is easily calculated by following some basic rules-
$ \to $ We have to find out the molar mass of all the elements present in a given compound.
$ \to $Then find out the molar mass of a particular element whose percentage composition we need to find.
$ \to $ Now calculate the molar mass of that element for one mole of compound by dividing the molar mass of the element by the molar mass of the compound given.
$ \to $ Now to obtain the result in percentage multiply the value by \[100\].
The general formula used to calculate percentage composition of any element is: -
$\% Pc = \dfrac{{Me}}{{Mc}} \times 100$
Where, $\% Pc$ is percentage composition of particular element
$Me$, is total mass of element present in given compound
$Mc$, is total mass of compound given
A single molecule of $C{O_2}$ contain one carbon atom and two oxygen atoms. Therefore, when we express the amount of atom in a mole, we say that one mole of carbon dioxide contains one mole of carbon and two moles of oxygen. We know that the atomic mass of a carbon atom is $12g$ and oxygen has an atomic mass of $16g$. So total molecular mass of $C{O_2}$ become-
$Mc = 12 + 2 \times 16$
After solving the above equation, we get
$Mc = 44g$
Therefore, the molecular mass of one mole $C{O_2}$ is $44g$.
to find the percentage composition of oxygen we have to calculate the total mass of oxygen present in one mole of $C{O_2}$. Since, atomic mass of oxygen is $16g$and total two oxygen atoms are present in one molecule, so total mass of oxygen in one mole of $C{O_2}$ will become-
$Me = 16 \times 2$
After solving the above equation, we get
$Me = 32g$
Now put all the value in the formula mentioned above-
$\% Pc = \dfrac{{Me}}{{Mc}} \times 100$
$\% Pc = \dfrac{{32}}{{44}} \times 100$
After solving the above equation, we get
$\% Pc = 72.72g$
It means that $100g$of $C{O_2}$ contain $72.72g$ of oxygen, so amount of carbon present in compound is-
$\left( {100 - 72.72} \right)g$ of carbon
$ = 27.28g$ of carbon
Hence, we conclude that every $100g$ of $C{O_2}$ will contain $72.72g$ of oxygen and $27.28g$ carbon.
$ \Rightarrow $ The mass of oxygen in a molecule of$C{O_2}$ by using percentage composition is $72.72g$.

Note:
Percentage composition is just a way to express all the constituents of a particular compound in terms of percentage. The basic significance of determining percentage composition is observed during the chemical analysis. Mole-mole ratio of any compound is also calculated with the help of percentage composition.