This Empirical Formula Calculator finds an empirical formula corresponding to the given compound chemical composition. Enter in the corresponding fields of the calculator the symbol of the chemical element that is part of the compound under study and its mass. In case of more then one element you can click the “+” symbol on the right hand side to get additional fields. You can also click the “–” symbol to delete the redundant lines. The masses of the elements must be presented in the same mass units or as percentages.
How to Find Empirical Formula
The empirical formula of a chemical compound is the simplest whole number ratio of atoms present in the compound. The empirical formula, in most cases, is not unique and is not associated with only one particular substance.
For example, two substances – acetylene (C2H2) and benzene (C6H6) have the same empirical formula CH. This is a consequence of the fact that the ratio of carbon and hydrogen atoms in these substances is the same and equals 1:1. Thus, in order to determine the empirical formula it is necessary to know the chemical composition of the substance.
The most common approach to determining a compound’s chemical composition is to measure first the masses of its constituent elements. So, any experimental data involving masses can be used to derive the corresponding numbers of atoms in the compound. The easiest way to express the number of atoms is in moles.
For this, we have to use molar masses to convert the mass of each element to the number of moles. Then we consider the number of moles of each element relative to each other, converting these numbers into a whole-number ratio that can be used to derive the empirical formula of the substance.
Consider, for example, a sample of compound determined to contain 2 g of carbon, 0.336 g of hydrogen and 2.67 g of oxygen. We know the molar masses of these elements: C – 12.011 g/mol, H – 1.008 g/mol, O – 16 g/mol. The corresponding numbers of moles are:
0.336 g H / 1.008 (g/mol) H = 0.3333 mol H
2.67 g O / 16 (g/mol) O = 0.1669 mol O
If we now divide the found numbers of moles by the minimum value of 0.1665 and round to whole values, we get the following indices for the empirical formula: 1, 2, 1. So the empirical formula will be CH2O.
Note that rounding is necessary because errors are inevitable when measuring masses, and index values in chemical formulas can only be integers.
To calculate the empirical formula we can also use percentages instead of masses in grams. In our case, if we know that the sample contains 40% C, 6.7% H and 53.3% O, then we can assume that a 100 g sample contains 40 g of carbon, 6.7 g of hydrogen and 53.3 g of oxygen. The corresponding numbers of moles in this case are:
6.7 g H / 1.008 (g/mol) H = 6.647 mol H
53.3 g O / 16 (g/mol) O = 3.331 mol O
Then we divide the found numbers of moles by the minimum value of 3.330 and round to whole values. As a result we get the same indices for the empirical formula: 1, 2, 1. Thus, as in the previous example, the empirical formula would be CH2O.
Molecular Formula vs Empirical Formula
The empirical formula tells us the relative ratios of different atoms in a compound. The molecular formula, on the other hand, shows the exact number of each type of atom in a molecule.
In the above example, the empirical formula CH2O coincides with the actual molecular formula for formaldehyde. But such substances as Glucose (C6H12O6), Ribose (C5H10O5) and Acetic acid (C2H4O2) all have different molecular formulas but the same empirical formula CH2O.
Thus, the molecular formula of a substance is the empirical formula of that substance whose indices are all multiplied by the same integer n, which can be symbolically expressed as the following formula:
In order to derive the molecular formula of a substance, in addition to the chemical composition, it is also necessary to know the molar mass of that substance. To perform these calculations, we recommend using our Molecular Formula Calculator.
Using Empirical Formula Calculator
The symbols of chemical elements should be entered using the upper case for the first character and the lower case for the second character (if any). Compare: Co – cobalt is a chemical element and CO – carbon monoxide is a chemical compound.
Using the appropriate drop-down menu you can choose an output format for the empirical formula:
• Html – The empirical formula is represented using html tags for indices. For example, CO2. Clicking the ‘Copy to clipboard’ button ( ) you can copy the result ‘as is’, including all the tags, and then you can paste it to any html-page. However, clicking Ctrl-A and Ctrl-C you can copy the result without the tags and paste it to a DOC document keeping duly formatted indices.
• Small indices – The empirical formula is represented using ‘tiny’ symbols for indices. For example, CO₂ where unicode character is used: ₂ = (\u2082).
• Normal – The empirical formula is represented using ‘normal’ numbers for indices. For example, CO2.
Examples of Empirical Formula Calculation
Example 1. Caffeine is known to have the following percentage composition of chemical elements: 49.48% carbon, 5.19% hydrogen, 16.48% oxygen and 28.85% nitrogen. What is its empirical formula?
Solution. The molar masses of the constituent elements are: C – 12.01 g/mol, H – 1.008 g/mol, O – 16 g/mol, N – 14.007 g/mol. Dividing the given percentages (we assume that we have 100 g of сaffeine) by the corresponding molar masses we will have:
Carbon: 49.48 ÷ 12.011 = 4.12
Hydrogen: 5.19 ÷ 1.008 = 5.149
Oxygen: 16.48 ÷ 16.000 = 1.03
Nitrogen: 28.85 ÷ 14.007 = 2.06
Then, dividing the found numbers of moles by the minimum value of 1.03 and rounding to whole values we will have the following indices for the empirical formula: C = 4, H = 5, O = 1, N = 2. So the empirical formula of caffeine is C4H5N2O.
Example 2. When 1.25 g of magnesium is heated in a nitrogen environment, a chemical reaction occurs. The reaction produces 1.73 g of a magnesium and nitrogen compound. Calculate the empirical formula of the formed substance.
Solution. We know the mass of magnesium involved in the reaction. We also know the mass of the magnesium-nitrogen compound formed as a result of the reaction. It is not difficult to find the mass of the reacted nitrogen: 1.73 g – 1.25 g = 0.48 g.
The molar masses of the elements taking part in the reaction are: Mg – 24.305 g/mol, N – 14.007 g/mol. Dividing the masses of the substances taking part in the reaction by the corresponding molar masses we will have:
Magnesium: 1.25 ÷ 24.305 = 0.0514
Nitrogen: 0.48 ÷ 14.007 = 0.0343
Now, dividing the found numbers of moles by the minimum value of 0.0343 we will get the the following result: Mg = 1.5, N = 1. The factor 1.5 for Mg is not close to a whole number, so we can’t just round it up or down.
In this case, we must choose a coefficient to multiply all the factors found to get integer values. It is easy to see that such a coefficient is 2. Thus, we will have the following indices for the empirical formula: Mg = 3, N = 2. And the empirical formula of reaction product is Mg3N2 (magnesium nitride).
As you can see, all these calculations are not very complicated, but it is much easier to get the result using our Empirical Formula Calculator. It can be done in a flash and without computational errors.
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