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The three most commonly used types of chemical formulas are empirical, molecular, and structural formulas. The structural ones serve to detail the way in which the atoms of the molecules of each chemical compound are held together. Of course, all this in those compounds that have molecules and not crystals.
On the other hand, there are the empirical and molecular formulas, which we will work on in this article.
The empirical formula (also called the minimal or condensed formula) indicates the proportional relationship between the number of atoms of each element present in the molecule without this relationship indicating exactly the number of atoms. Sometimes it can match the molecular formula.
The molecular formula shows exactly the relationship between the atoms that make up the molecule of an element or chemical compound . It is a multiple of the empirical formula and therefore can be determined by knowing the molecular weight of the compound and the molecular formula weight. One can only speak of a molecular formula in the case that the element or compound is made up of molecules; if they are crystals, the empirical formula is used.
Utility of the empirical and molecular formula
Thanks to the fact that the empirical formula tells us the proportion of atoms present in the molecule, it can help us to know what type of molecule it is, such as a protein or a lipid.
The molecular formula is used to know how much of each element is present in the formula and is often useful for equations.
The limitation that these types of formulas would have is that they are not used to know how the atoms are arranged in the molecule in question. This function is fulfilled by the structural formula and it would help us if, for example, we needed to know which simple sugar we are facing if we have the molecule C 6 H 12 O 6.
Example and instructions to solve a problem using the empirical and molecular formulas
A molecule with a molecular weight of 180.18 g/mol is analyzed and found to contain 40.00% carbon, 6.72% hydrogen, and 53.28% oxygen.
How to find the solution
Finding the empirical and molecular formula is basically the reverse process used to calculate mass percent or mass percent.
Step 1: Find the number of moles of each element in a sample of the molecule.
Our molecule contains 40.00% carbon, 6.72% hydrogen, and 53.28% oxygen. This means that a 100 gram sample contains:
40.00 grams of carbon (40.00% of 100 grams)
6.72 grams of hydrogen (6.72% of 100 grams)
53.28 grams of oxygen (53.28% of 100 grams)
- Note: 100 grams is used for a sample size just to make the math easier. Any sample size can be used, the proportions between items will remain the same.
Using these numbers, we can find the number of moles of each element in the 100-gram sample. Divide the number of grams of each element in the sample by the element’s atomic weight to find the number of moles.
moles C = 40.00 gx 1 mol C / 12.01 g/mol C = 3.33 moles C
moles H = 6.72 g x 1 mol H / 1.01 g/mol H = 6.65 moles H
moles O = 53.28 gx 1 mol O / 16.00 g/mol O = 3.33 moles O
Step 2: Find the ratios between the number of moles of each element.
Select the element with the largest number of moles in the sample. In this case, the 6.65 moles of hydrogen are the largest. Divide the number of moles of each element by the largest number.
The simplest molar ratio between C and H: 3.33 mol C / 6.65 mol H = 1 mol C / 2 mol H
The ratio is 1 mole of C for every 2 moles of H
The simplest relationship between O and H: 3.33 moles of O / 6.65 moles of H = 1 mol O / 2 moles of H
The ratio between O and H is 1 mol O for every 2 moles of H
Step 3: Find the empirical formula.
We have all the information we need to write the empirical formula . For every two moles of hydrogen, there is one mole of carbon and one mole of oxygen.
The empirical formula is CH 2 O.
Step 4: Find the molecular weight from the empirical formula.
We can use the empirical formula to find the molecular formula using the molecular weight of the compound and the molecular weight of the empirical formula.
The empirical formula is CH 2 O. The molecular weight is:
CH 2 O molecular weight = (1 x 12.01 g/mol) + (2 x 1.01 g/mol) + (1 x 16.00 g/mol)
CH 2 O molecular weight = (12.01 + 2.02 + 16.00) g/mol
CH 2 O molecular weight = 30.03 g/mol
Step 5: Find the number of empirical formula units in the molecular formula.
The molecular formula is a multiple of the empirical formula. We were given the molecular weight of the molecule, 180.18 g/mol. Divide this number by the empirical formula molecular weight to find the number of empirical formula units that make up the compound.
Number of empirical formula units in the compound = 180.18 g/mol / 30.03 g/mol
Number of empirical formula units in the compound = 6
Step 6: Find the molecular formula.
It takes six empirical formula units to make the compound, so multiply each number in the empirical formula by 6.
molecular formula = 6 x CH 2 O
molecular formula = C (1 x 6) H (2 x 6) O (1 x 6)
molecular formula = CH2O
Solution:
The empirical formula of the molecule is CH 2 O.
The molecular formula of the compound is C 6 H 12 O 6 .
References
Khan Academy (undated). Empirical, molecular and structural formulas. Available at: https://es.khanacademy.org/science/ap-chemistry/atoms-compounds-ions-ap/compounds-and-ions-ap/v/empirical-molecular-and-structural-formulas
ICT resources (undated). Empirical and molecular formulas. Available at: http://recursostic.educacion.es/secundaria/edad/3esofisicaquimica/3quincena7/3q7_contenidos_4b.htm
