A compound contains 87.5% by mass of nitrogen and 12.5% by mass of hydrogen. Determine the empirical formula of this compound.
A compound contains 87.5% by mass of nitrogen and 12.5% by mass of hydrogen. Determine the empirical formula of this compound.
Determination of Empirical Formula:
The empirical formula can be determined from the percentage composition of the compound as follows:
1. The percentage weight of oxygen, carbon or hydrogen (necessary elements in every organic compound) is sometimes not given. Hence, it is first determined by subtracting the sum of other elemental percentages from 100.
2. The percentage weight of each element is divided by its atomic weight. This gives the ratio of the number of atoms in a molecule of the compound.
4. To get a whole number ratio of atoms of different elements and water molecules, the figures obtained in step 2 is divided by the smallest number.
5. The empirical formula is now derived by writing the symbols of various elements side by side with the number of atoms of each one as the subscript to the lower right of its symbol.
Element Atomic mass Percentage Relative number of moles Simplest mole ratio Whole number ratio N 14 87.5 87.5/14= 6.25 6.25/6.25=1 1 H 1 12.5 12.5/1=12.5 12.5/6.25=2 2
Therefore, the empirical formula of this compound is NH2.
Determination of Empirical Formula:
The empirical formula can be determined from the percentage composition of the compound as follows:
1. The percentage weight of oxygen, carbon or hydrogen (necessary elements in every organic compound) is sometimes not given. Hence, it is first determined by subtracting the sum of other elemental percentages from 100.
2. The percentage weight of each element is divided by its atomic weight. This gives the ratio of the number of atoms in a molecule of the compound.
4. To get a whole number ratio of atoms of different elements and water molecules, the figures obtained in step 2 is divided by the smallest number.
5. The empirical formula is now derived by writing the symbols of various elements side by side with the number of atoms of each one as the subscript to the lower right of its symbol.
| Element | Atomic mass | Percentage | Relative number of moles | Simplest mole ratio | Whole number ratio |
| N | 14 | 87.5 | 87.5/14= 6.25 | 6.25/6.25=1 | 1 |
| H | 1 | 12.5 | 12.5/1=12.5 | 12.5/6.25=2 | 2 |
Therefore, the empirical formula of this compound is NH2.
