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Empirical & Molecular Formula Simplified Revision Notes

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1.2.2 Empirical & Molecular Formula

Understanding the empirical and molecular formulas of compounds is essential for determining the composition and structure of molecules.

Empirical Formula

The empirical formula of a compound represents the simplest whole-number ratio of atoms of each element present. It does not necessarily show the actual number of atoms in a molecule but provides the simplest ratio.

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Example: For glucose C6H12O6C_6H_{12}O_6, the empirical formula is CH2OCH_2O, which shows the simplest 1:2:1 ratio of carbon, hydrogen, and oxygen atoms.

Molecular Formula

The molecular formula provides the actual number of atoms of each element in a molecule. It is often a whole-number multiple of the empirical formula.

infoNote

Example: For glucose C6H12O6C_6H_{12}O_6, the molecular formula shows the actual number of atoms in the molecule. In this case, the molecular formula is 6 times the empirical formula CH2OCH_2O.

Relationship Between Empirical and Molecular Formula

The molecular formula is always a whole-number multiple of the empirical formula. This relationship can be expressed as:

Molecular Formula=(Empirical Formula)×n\text{Molecular Formula} = (\text{Empirical Formula}) \times n

Where nn is the ratio of the molar mass of the compound to the molar mass of the empirical formula.

n=Mr(Molecular)Mr(Empirical)n = \frac{M_r (\text{Molecular})}{M_r (\text{Empirical})}

Calculating Empirical Formula

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Example: How to calculate the empirical formula from given data

Step 1: Convert masses to moles:

Use the formula:

Moles=MassAr\text{Moles} = \frac{\text{Mass}}{A_r}

Where ArA_r is the relative atomic mass of each element.


Step 2: Divide each mole value by the smallest number of moles:

This gives the simplest ratio.


Step 3: Adjust the ratio to whole numbers:

If the resulting ratio isn't a whole number, multiply all values by a suitable factor to obtain whole numbers (e.g., multiply by 2 if one value is 1.5).

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Example: A compound contains 2.4 g of carbon and 0.4 g of hydrogen.

Calculate the empirical formula.


Step 1: Convert masses to moles:

Moles of C=2.412=0.2\text{Moles of C} = \frac{2.4}{12} = 0.2Moles of H=0.41=0.4\quad \text{Moles of H} = \frac{0.4}{1} = 0.4

Step 2: Divide each mole value by the smallest number of moles:

0.20.2=1\frac{0.2}{0.2} = 10.40.2=2\quad \frac{0.4}{0.2} = 2

Step 3: Adjust the ratio to whole numbers:

The ratio is 1:2, so the empirical formula is CHâ‚‚.

Calculating Molecular Formula

To calculate the molecular formula, you need:

  1. The empirical formula.
  2. The molar mass (MrM_r) of the compound.

Steps:

  1. Find the molar mass of the empirical formula.
  2. Divide the molar mass of the molecular formula by the empirical formula mass to find nn.
  3. Multiply the empirical formula by nn to obtain the molecular formula.
infoNote

Example: The empirical formula of a compound is CH2OCH_2O and its molar mass is 180 g/mol

Determine the molecular formula.


Step 1: Find the molar mass of the empirical formula.

Molar mass of

CH2O=12+2+16=30g/mol.CH_2O = 12 + 2 + 16 = 30 g/mol.

Step 2: Divide the molar mass of the molecular formula by the empirical formula mass to find nn.

Find nn:

n=18030=6n = \frac{180}{30} = 6

Step 3: Multiply the empirical formula by nn to obtain the molecular formula.

Multiply the empirical formula by 6:

Molecular formula=C6H12O6\text{Molecular formula} = C_6H_{12}O_6
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