Question

How do you find the percent composition of all the elements in sulfur trioxide?

Answer 1

Hint: The chemical formula of sulfur trioxide is $S{O_3}$. It is formed of one sulfur atom and three oxygen atoms. The formula of the percent composition is the same as the formula of mass percent where the mass of the atom is divided by the total mass of the compound multiplied by 100.

Complete step by step answer:
The sulfur trioxide is the chemical compound with molecular formula or chemical formula $S{O_3}$. In every sample of substance there will be the same number of sulfur atoms and oxygen atoms.
The percent composition is calculated by dividing the mass of the atom by the total mass of the compound or the molecular weight multiplied by 100.
The formula to calculate the percentage composition is shown below.
$C\% = \dfrac{{{M_A}}}{{{M_T}}} \times 100$
Where,
$C\% $ is the percentage composition
${M_A}$ is the mass of atom
${M_T}$ is the total mass of the compound.
The atomic weight of sulphur is 32.059 g/mol. The atomic weight of oxygen is 16.0.
The total mass is $32.059 + 3 \times 16 = 80.059$.
The molecular weight of sulfur trioxide is 80.059 g/mol.
To calculate the percent composition of sulfur, substitute the values in the given equation.
$ \Rightarrow C\% = \dfrac{{32.059}}{{80.059}} \times 100$
$ \Rightarrow C\% = 40.044\% $
Therefore, the percent composition of sulfur in sulfur trioxide is 40.044 %.
To calculate the percent composition of oxygen, substitute the values in the given equation.
$ \Rightarrow C\% = \dfrac{{16.00}}{{80.059}} \times 100$
$ \Rightarrow C\% = 19.98\% $

Therefore, the percentage composition of oxygen atoms in sulfur trioxide is 19.98 %.

Note: The mass percent or the percent composition of the atom helps to determine the empirical formula of the chemical compound. The empirical formula is the simplest ratio of the element present in the compound.