A sample of Freon was found to contain 3.21 g of C, 19.0 g of Cl, and 10.2 g F. What is the empirical formula of this compound?
C0.0352Cl0.0705F0.0705
C0.5ClF
C2Cl4F4
CCl2F2
CClF
mols C = 3.21/atomic mass C = ?
moles Cl = 19.0/atomic mass Cl = ?
moles F = 10.2/atomic mass F = ?
Now find the ratio of the elements to each other with the smallest number being 1.000. The easy way to do that is to divide the smallest number by itself which automatically makes it 1.000, then divide the other numbers by the same small number, then round to whole number. That will give you the empirical formula.
To determine the empirical formula of the compound, we need to find the simplest whole number ratio of the elements C, Cl, and F in the sample.
Step 1: Determine the moles of each element.
The molar mass of carbon (C) is 12.01 g/mol.
The molar mass of chlorine (Cl) is 35.45 g/mol.
The molar mass of fluorine (F) is 18.998 g/mol.
Moles of C = Mass of C / Molar mass of C
= 3.21 g / 12.01 g/mol
= 0.267 mol C
Moles of Cl = Mass of Cl / Molar mass of Cl
= 19.0 g / 35.45 g/mol
= 0.536 mol Cl
Moles of F = Mass of F / Molar mass of F
= 10.2 g / 18.998 g/mol
= 0.537 mol F
Step 2: Determine the mole ratio.
Divide the moles of each element by the smallest number of moles.
In this case, the smallest number of moles is 0.267 mol.
Mole ratio of C = Moles of C / Smallest number of moles
= 0.267 mol / 0.267 mol
= 1
Mole ratio of Cl = Moles of Cl / Smallest number of moles
= 0.536 mol / 0.267 mol
= 2
Mole ratio of F = Moles of F / Smallest number of moles
= 0.537 mol / 0.267 mol
= 2
Step 3: Write the empirical formula.
The empirical formula is the smallest whole number ratio of the elements.
Empirical formula = C1Cl2F2
Therefore, the empirical formula of this compound is CCl2F2.
To find the empirical formula of a compound, you need to determine the ratio of elements present in the compound.
1. Start by converting the given masses of each element to moles using their respective molar masses. The molar mass of carbon (C) is 12.01 g/mol, chlorine (Cl) is 35.45 g/mol, and fluorine (F) is 18.998 g/mol.
- Moles of C = 3.21 g C / 12.01 g/mol = 0.267 mol C
- Moles of Cl = 19.0 g Cl / 35.45 g/mol = 0.536 mol Cl
- Moles of F = 10.2 g F / 18.998 g/mol = 0.538 mol F
2. Divide each element's moles by the smallest number of moles to get the simplest, whole number ratio.
- C:Cl:F = 0.267 mol / 0.267 mol : 0.536 mol / 0.267 mol : 0.538 mol / 0.267 mol
- Simplifying, we get C:Cl:F = 1:2:2
3. Based on the ratio obtained, we can determine the empirical formula.
The empirical formula of the compound is CCl2F2 (Choice D).