The empirical formula of a compound gives the simplest ratio of the number of different atoms present, whereas the molecular formula gives the actual number of each different atom present in a molecule ¹. For example, the empirical formula of glucose is CH2O, which means that the ratio of carbon, hydrogen, and oxygen atoms in glucose is 1:2:1. The molecular formula of glucose is C6H12O6, which means that a molecule of glucose contains 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms ¹.
In general, the empirical formula is the simplest formula for a compound, while the molecular formula is a multiple of the empirical formula ¹. To calculate the empirical formula of a compound, you need to determine the mass of each element present in the compound, convert the mass of each element to moles using the element's molar mass, divide each mole value by the smallest number of moles calculated,
and write the empirical formula using the mole values as subscripts for each element.
To calculate the empirical formula of a compound, you need to follow these steps:
1. Determine the mass of each element present in the compound.
2. Convert the mass of each element to moles using the element's molar mass.
3. Divide each mole value by the smallest number of moles calculated in step 2.
4. If all the mole values are whole numbers (or very close), write the empirical formula using the mole values as subscripts for each element.
Here is an example to illustrate the steps:
Suppose you have a compound that contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. 1. Assume a 100 g sample of the compound so that the given percentages can be directly converted into grams.
- Carbon: 40.0 g
- Hydrogen: 6.7 g
- Oxygen: 53.3 g
2. Use each element's molar mass to convert the grams of each element to moles.
- Carbon: 40.0 g / 12.01 g/mol = 3.33 mol
- Hydrogen: 6.7 g / 1.01 g/mol = 6.63 mol
- Oxygen: 53.3 g / 16.00 g/mol = 3.33 mol
3. Divide the number of moles of each element by the smallest number of moles (3.33 mol).
- Carbon: 3.33 mol / 3.33 mol = 1
- Hydrogen: 6.63 mol / 3.33 mol = 1.99 ≈ 2
- Oxygen: 3.33 mol / 3.33 mol = 1
4. Write the empirical formula using the mole values as subscripts for each element.
- C1H2O1 or CH2O
Therefore, the empirical formula of the compound is CH2O.
The molecular formula gives the actual number of each different atom present in a molecule
To calculate the molecular formula of a compound, you need to know the empirical formula and the molar mass of the compound. You can calculate the molecular formula by dividing the molar mass of the compound by the empirical formula mass, which is the sum of the atomic masses of all the atoms in the empirical formula. The whole number ratio between the molecular formula mass and the empirical formula mass is the multiple by which the subscripts in the empirical formula should be multiplied to get the molecular formula Here are the steps to calculate the molecular formula:
1. Calculate the empirical formula of the compound using the steps I provided earlier.
2. Calculate the empirical formula mass (EFM), which is the sum of the atomic masses of all the atoms in the empirical formula.
3. Divide the molar mass of the compound by the EFM to get the whole number ratio between the molecular formula mass and the empirical formula mass.
4. Multiply the subscripts in the empirical formula by the whole number ratio to get the molecular formula.
Here's an example to illustrate the steps:
Suppose you have a compound with an empirical formula of CH2O and a molar mass of 180 g/mol.
1. The empirical formula mass is:
- (1 x 12.01 g/mol) + (2 x 1.01 g/mol) + (1 x 16.00 g/mol) = 30.03 g/mol
2. The whole number ratio between the molecular formula mass and the empirical formula mass is:
- 180 g/mol ÷ 30.03 g/mol = 5.996 ≈ 6
3. The molecular formula is:
- C6H12O6
Therefore, the molecular formula of the compound is C6H12O6.
https://sciencenotes.org/empirical-vs-molecular-formula/.
https://www.bbc.co.uk/bitesize/guides/zbysqp3/revision/1
https://sciencing.com/how-to-find-molecular-formula-from-empirical-formula-13712170.html.