Determine the empirical formula for the following compounds with these compositions (by mass):
a) Cr 47.7% F 52.3%
b) Na 42.1% P 18.9% O 39.0%
Both are solved the same way. I'll do the first one.
Take a 100 g sample which gives you
47.7 g Cr
52.3 g F
Convert to mols. mol = g/atomic mass
mols Cr = 47.7/52 = about 0.92
mols F = 52.3/19 = about 2.75
Now find the ratio of one element to the other with the smallest number being 1.0. The easy way to do that is divide both numbers by the smaller one.
0.92/0.92 = 1.00
2.75/0.92 = 2.99 which rounds to a whole number of 3.0
So the formula is Cr1.00F3.00 but you don't need the 1.00 and the formula we usually write is CrF3.
Post your work if you want me to check your work on part b.
Na3PO4
To determine the empirical formula for the compounds, we need to find the ratio of the elements present in each compound.
a) Cr 47.7% F 52.3%:
Step 1: Convert the percentage values to grams.
Cr: 47.7 g
F: 52.3 g
Step 2: Convert the mass of each element to moles using their respective atomic masses.
Atomic mass of Cr = 52.0 g/mol
Atomic mass of F = 19.0 g/mol
Moles of Cr = 47.7 g / 52.0 g/mol = 0.916 mol
Moles of F = 52.3 g / 19.0 g/mol = 2.748 mol
Step 3: Divide the moles of each element by the smallest number of moles to find the simple whole number ratio.
Dividing by 0.916 moles (the smallest number):
Moles of Cr = 0.916 mol / 0.916 mol = 1
Moles of F = 2.748 mol / 0.916 mol = 3
The empirical formula for the compound is CrF3.
b) Na 42.1% P 18.9% O 39.0%:
Step 1: Convert the percentage values to grams.
Na: 42.1 g
P: 18.9 g
O: 39.0 g
Step 2: Convert the mass of each element to moles using their respective atomic masses.
Atomic mass of Na = 23.0 g/mol
Atomic mass of P = 31.0 g/mol
Atomic mass of O = 16.0 g/mol
Moles of Na = 42.1 g / 23.0 g/mol = 1.83 mol
Moles of P = 18.9 g / 31.0 g/mol = 0.61 mol
Moles of O = 39.0 g / 16.0 g/mol = 2.44 mol
Step 3: Divide the moles of each element by the smallest number of moles to find the simple whole number ratio.
Dividing by 0.61 moles (the smallest number):
Moles of Na = 1.83 mol / 0.61 mol = 3
Moles of P = 0.61 mol / 0.61 mol = 1
Moles of O = 2.44 mol / 0.61 mol = 4
The empirical formula for the compound is Na3PO4.
To determine the empirical formula for a compound, you need to find the simplest ratio of atoms in the compound. This can be achieved by first converting the percentages to grams and then calculating the number of moles of each element.
a) For the compound with 47.7% Cr and 52.3% F:
1. Assume we have 100g of this compound. Therefore, there would be 47.7g of Cr and 52.3g of F.
2. To find the number of moles, divide the mass of each element by its molar mass, which can be obtained from the periodic table:
- Molar mass of Cr = 52 g/mol
- Molar mass of F = 19 g/mol
Moles of Cr = 47.7g / 52 g/mol = 0.917 moles
Moles of F = 52.3g / 19 g/mol = 2.75 moles
3. Divide both moles by the smallest value to obtain the simplest whole number ratio:
Cr: F = 0.917 / 0.917 : 2.75 / 0.917
Cr: F = 1 : 3
Therefore, the empirical formula for this compound is CrF3.
b) For the compound with 42.1% Na, 18.9% P, and 39.0% O:
1. Assume we have 100g of this compound. Therefore, there would be 42.1g of Na, 18.9g of P, and 39.0g of O.
2. To find the number of moles, divide the mass of each element by its molar mass:
- Molar mass of Na = 23 g/mol
- Molar mass of P = 31 g/mol
- Molar mass of O = 16 g/mol
Moles of Na = 42.1g / 23 g/mol = 1.83 moles
Moles of P = 18.9g / 31 g/mol = 0.61 moles
Moles of O = 39.0g / 16 g/mol = 2.44 moles
3. Divide the moles by the smallest value to obtain the simplest whole number ratio:
Na: P: O = 1.83 / 0.61 : 0.61 / 0.61 : 2.44 / 0.61
Na: P: O = 3 : 1 : 4
Therefore, the empirical formula for this compound is Na3PO4.