The percentage composition of sodium phosphate as determined by analysis is 42.1% sodium, 18.9% phosphorus, and 39% oxygen. Find the empirical formula of the compound (H=1, N=14, O=16, Cl= 35.5, Pt= 195)
The percentage composition of sodium phosphate as determined by analysis is 42.1% sodium, 18.9% phosphorus, and 39% oxygen. Find the empirical formula of the compound (H=1, N=14, O=16, Cl= 35.5, Pt= 195)
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.
3. To get a whole number ratio of atoms of different elements and water molecules, the figures obtained in step 2 are divided by the smallest number.
4. 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 Percentage weight Atomic mass Relative no. of moles Simple ratio of atoms Whole number ratio Na 42.1 23 42.1/23 = 1.83 1.83/0.61=3 3 P 18.9 31 18.9/31 = 0.61 0.61/0.61=1 1 O 39 16 39.0/16 = 2.44 2.44/0.61=4 4
So, the simple ratio is Na: P: O= 3:1:4
The empirical formula= Na3PO4
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.
3. To get a whole number ratio of atoms of different elements and water molecules, the figures obtained in step 2 are divided by the smallest number.
4. 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 | Percentage weight | Atomic mass | Relative no. of moles | Simple ratio of atoms | Whole number ratio |
| Na | 42.1 | 23 | 42.1/23 = 1.83 | 1.83/0.61=3 | 3 |
| P | 18.9 | 31 | 18.9/31 = 0.61 | 0.61/0.61=1 | 1 |
| O | 39 | 16 | 39.0/16 = 2.44 | 2.44/0.61=4 | 4 |
So, the simple ratio is Na: P: O= 3:1:4
The empirical formula= Na3PO4
