With the Molecular and Empirical formula calculator you can determine the empirical and molecular formula of a compound just by knowing its percentage composition. In order to use it, you must follow these steps:
The empirical formula of a compound, also known as the minimal formula, gives the simplest ratio of whole numbers of atoms of each element present in the compound. It is called empirical because it is determined using data from experiments and, therefore, empirical.
NO2 → The empirical formula of nitrogen tetroxide once reduced.
C5H3N3 → The empirical formula and molecular formula of cyanopyrazine are the same, since the ratio of atoms cannot be simplified.
In other words, an empirical formula is a simple expression that gives the relative proportions of the elements in a compound, but not the actual numbers of atoms. It is based on the mass ratios of the elements in the compound, rather than the actual numbers of atoms.
For example, consider the compound methane (CH4). The empirical formula for methane is CH4, indicating that it is composed of one carbon atom and four hydrogen atoms. The actual molecular formula for methane is also CH4, which gives the actual numbers of atoms in the molecule. In this case, the empirical formula and the molecular formula are the same, but this is not always the case.
Empirical formulas are useful for representing the composition of compounds in a simple and concise way.
We will explain how to find the empirical formula of a compound through the following example:
Example 01: In an experiment, a compound was found to contain 32.65% sulfur, 65.3% oxygen, and 2.04% hydrogen. What is the empirical formula of the compound?
Step 01: We assume that we have a sample of 100 grams of the compound, with this we can directly convert the percentages to fallow deer.
32.65% → 32.65g S
65.3% → 65.3 g O
2.04% → 2.04 g of H
Step 02: Next we will divide all the given masses of each element by the molar mass.
32.65g of S / 32gm-1 = 1.0203 moles of S
65.3g of O / 16gm-1 = 4.08 moles of O
2.04g of H / 1.008gm-1 = 2.024 moles of H
Step 03: Next, we divide each of the molar masses obtained in step 2 by the smallest molar mass. If the result of the division is a decimal, it must be rounded to the nearest integer.
1.0203 moles of S / 1.0203 = 1
4.08 moles of O / 1.0203 = 3.998 ≈ 4
2.024 moles of H / 1.0203 = 1.984 ≈ 2
Step 04: Using the coefficients calculated in the previous step, they will be converted into subscripts of the chemical formula.
S = 1
OR = 4
H = 2
With this we obtain the empirical formula: H2SO4
The molecular formula describes the atoms within a single molecule. The ratio of atoms within a molecular formula is the same as in the empirical formula, but it is not reduced. Therefore, the subscripts of the molecular formula will always be integer multiples of those present in the empirical formula. The composition of a molecular formula does not have to coincide with that of the empirical formula. It will be necessary to know the molecular mass of the compound in order to calculate the molecular formula.
H2O → Molecular formula of water
C4H10 → Molecular formula of butane
C8H10N4O2 → Molecular formula of caffeine
The molecular formula is a chemical formula that gives the exact number of atoms of each element in a molecule. It represents the actual composition of a molecule and can be used to identify a specific chemical compound. The molecular formula is different from the empirical formula, which gives the relative proportions of the elements in a compound but not the actual numbers of atoms. The molecular formula is more specific and gives more information about the composition of a compound.
As we did with the empirical formula, we will proceed to explain how to calculate the molecular formula through the following example:
Example 02: Knowing that the empirical formula of a compound containing hydrogen and boron is BH3, and having a molar mass of 27.7 g/ mole, find its molecular formula.
Step1: We calculate the molar mass of the empirical formula.
|(g / mol)||(%)||(g / mol)|
|Total Molecular Mass:||13.83 g / mol|
(27.7 g / mol) / (13.83g / mol) ≈ 2