A gaseous compound has the following composition by mass: C, 25.0%; H, 2.1%; F, 39.6%; O, 33.3%. Its molecular weight is 48 g/mol
And you want to know what? The formula? empirical formula? molecular formula?
Take a 100 g sample which will give you
25.0 g C
2.1 g H
39.6 g F
33.3 g O
Now convert each to moles, then find the molar ratios of the elements to each other with the smallest being 1.00. The easy way to do that is to divide the smallest number by itself (assuring you that this number will be 1.000), then divide the other numbers by the same small number. Round to whole numbers and that will be the empirical formula.
To deduce the molecular formula, add up the empirical formula mass and
48/(empirical formula mass) = ?? and you round this to a whole number. It will be 1.00 or 2.00 or 3.00 (when rounded). This indicates the number of empirical formula units it takes to make a molecular formula; i.e,
(CxHyOzFw)n
Post your work if you get stuck.
To determine the molecular formula of the compound, we need to find the empirical formula first and then use the molecular weight to calculate the molecular formula. The empirical formula gives us the simplest whole-number ratio of atoms in a compound.
1. Determine the number of moles of each element:
- Carbon (C):
Mass of C = 25.0% of total mass = 25.0% x mass of compound
Moles of C = mass of C / molar mass of C
- Hydrogen (H):
Mass of H = 2.1% of total mass = 2.1% x mass of compound
Moles of H = mass of H / molar mass of H
- Fluorine (F):
Mass of F = 39.6% of total mass = 39.6% x mass of compound
Moles of F = mass of F / molar mass of F
- Oxygen (O):
Mass of O = 33.3% of total mass = 33.3% x mass of compound
Moles of O = mass of O / molar mass of O
2. Divide all the mole values by the smallest moles obtained in step 1 to get the simplest whole-number ratio.
- Check the mole values obtained in step 1 and divide all by the smallest moles
3. Round the values to the nearest whole number to get the subscripts of the elements.
- After dividing the moles, round the values to the nearest whole number
4. Write the empirical formula using the elements and their subscripts obtained in step 3.
Now, that we have the empirical formula, we can use the molecular weight of the compound to find the molecular formula.
1. Calculate the empirical formula mass by summing the atomic masses of the elements in the empirical formula.
- Multiply the atomic mass of each element by the subscript and sum them up.
2. Divide the molecular weight by the empirical formula mass to find the whole-number factor.
- Whole-number factor = molecular weight / empirical formula mass
3. Multiply the subscripts in the empirical formula by the whole-number factor obtained in step 2 to obtain the molecular formula.
Using the given information, we can now calculate the empirical formula:
1. Calculate the moles of each element using the percentages and molecular weight given.
C:
Mass of C = 25.0% x molecular weight = 0.250 x 48 g/mol = 12 g
Moles of C = 12 g / molar mass of C = 12 g / 12.01 g/mol = 0.998 moles
H:
Mass of H = 2.1% x molecular weight = 0.021 x 48 g/mol = 1.01 g
Moles of H = 1.01 g / molar mass of H = 1.01 g / 1.008 g/mol = 1.00 moles
F:
Mass of F = 39.6% x molecular weight = 0.396 x 48 g/mol = 19.0 g
Moles of F = 19.0 g / molar mass of F = 19.0 g / 18.99 g/mol = 1.00 moles
O:
Mass of O = 33.3% x molecular weight = 0.333 x 48 g/mol = 16.0 g
Moles of O = 16.0 g / molar mass of O = 16.0 g / 16.00 g/mol = 1.00 moles
2. Divide all the moles by the smallest moles obtained (0.998), giving us the simplest whole-number ratio:
C: 0.998 / 0.998 ≈ 1
H: 1.00 / 0.998 ≈ 1
F: 1.00 / 0.998 ≈ 1
O: 1.00 / 0.998 ≈ 1
3. Round the values to the nearest whole number to get the subscripts:
C: 1
H: 1
F: 1
O: 1
The empirical formula of the compound is CHFO.
Lastly, we can find the molecular formula:
1. Calculate the empirical formula mass:
Empirical formula mass = (subscript of C x molar mass of C) + (subscript of H x molar mass of H) + (subscript of F x molar mass of F) + (subscript of O x molar mass of O)
Empirical formula mass = (1 x 12.01 g/mol) + (1 x 1.008 g/mol) + (1 x 18.99 g/mol) + (1 x 16.00 g/mol)
Empirical formula mass = 48.01 g/mol
2. Divide the molecular weight by the empirical formula mass to find the whole-number factor:
Whole-number factor = molecular weight / empirical formula mass = 48 g/mol / 48.01 g/mol ≈ 1
3. Multiply the subscripts in the empirical formula by the whole-number factor to obtain the molecular formula:
C: 1 x 1 = 1
H: 1 x 1 = 1
F: 1 x 1 = 1
O: 1 x 1 = 1
The molecular formula of the compound is also CHFO.