Question

What is the mathematical formula for calculating the mass percent composition from a chemical formula?

Answer 1

Hint: Let us first see what a mole is. Mole is the SI unit of measurement and is used to determine the amount of a substance. A mole of any substance has exactly $6.022\times {{10}^{23}}$ particles which can be ions, atoms, electrons, or molecules.

Complete answer:
Now, the mass of one mole of a substance is equivalent to the sum of the mass of all the particles ($6.022\times {{10}^{23}}$) contained in one mole of a substance.
The molar mass of a substance is defined as the mass of one mole of a substance. Its usually expressed in g/mol but its SI base unit is kg/mol.
Also, the atomic weight or atomic mass of an element can be calculated by taking the weighted average of all the isotopes of that element that exist in nature, considering their abundance.
The mass percentage composition is used to determine the composition of an element in a compound or a molecule.
To determine the mass percentage composition, we first need to determine the molecular mass of the compound using its chemical formula.
Now, the molecular mass of one mole of a compound can be calculated by taking the sum of atomic masses of all the elements present in the molecule.
Also, the mass of an element present in one mole of the compound needs to be calculated using the chemical formula of the compound.
So, the formula to calculate the mass percentage composition of a constituent in a compound is given by:
\[\text{percentage composition = }\dfrac{\text{mass of molecule in 1 mole of compound}}{\text{molecular mass of compound}}\times 100\]

Note:
Let us use the example of sulfuric acid ${{H}_{2}}S{{O}_{4}}$ to determine mass percentage composition.
Now, the approximate atomic masses of the elements in the compound are
H = 1 u
S = 32 u
O = 16 u
So, the compound's molecular mass will be
\[\begin{align}
  & {{M}_{{{H}_{2}}S{{O}_{4}}}}=(2\times {{M}_{H}})+{{M}_{s}}+(4\times {{M}_{O}}) \\
 & {{M}_{{{H}_{2}}S{{O}_{4}}}}=(2\times 1)+32+(4\times 16) \\
 & {{M}_{{{H}_{2}}S{{O}_{4}}}}=98\text{ g/mol} \\
\end{align}\]
Now the percentage compositions of the elements will be
\[(O)=\dfrac{4\times 16}{98}\times 100=65.31\]%
\[(H)=\dfrac{2\times 1}{98}\times 100=2.04\]%
\[(S)=\dfrac{32}{98}\times 100=32.65\]%