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For example, the Hydrogen peroxide molecule has exactly 2 hydrogens and 2 oxygens

The Molecular formula is

This shows the actual number of atoms present

We will use hydrogen peroxide again

The 2 is common for both hydrogen and oxygen so we can divide each by 2

This will give us HO instead of H_{2}O_{2}

This gives us another obvious equation

This shows that n number of empirical formulas can fit inside the molecular formula

For different types of questions we need to use different methods

There are 2 ways of calculating the empirical formula depending on the values available. We will discuss it for each method:

Simplifying the percentages does not give the empirical formula

To understand this we will take an example

A compound contains 23% of Carbon and 10% of Hydrogen and 67% of Oxygen by mass

As it highlights the percentage of mass and not the number, use this method

If we were to take a sample of 100g. Then the masses would be 23g, 10 g and 67 g respectively

Then we need to divide it by the molar mass to get the number of moles for each element/atom

C H O 23/12 : 10/1 : 67/16 1.92 : 10 : 4.1875 1 : 5.2 : 2.2 5 : 26 : 11

Then divide each by the smallest number of moles in the list which is 1.92.

But as you need to find the whole number ratio you need to times it by a value for each to get a whole number for each

Then finally write the equation

We will see another more advanced way of finding the empirical formula

When a hydrocarbon reacts there is a fast way and a long way of finding the empirical formula or the formula of the compound

We will usually need to use the long method as working and the short method for checking

A hydrocarbon is burnt in excess of 50cm^{3}of Oxygen. 20cm^{3}of Oxygen is remaining. 50cm^{3}of CO_{2}is produced and 50cm^{3}of H_{2}O is produced. Finding the formula of the hydrocarbon

First thing to remember is that the volume of reactant used up is the same as the mole ratio

C_{x}H_{y}+ (x+y/4) O_{2}= XCO_{2}+ (Y/2) H_{2}O Only 30cm^{3}of Oxygen reacts: 30 50 50 3 5 5

Now we know that there must be 5 Carbon atoms and 10 Hydrogen atoms

Y/2 = 5 and X = 5 so the formula is C_{5}H_{10} and the empirical formula is CH_{2}

Always remember: X+(Y/4)

Then you can substitute x and y to find the number of moles for oxygen gas

This is the fast method and this is formula only works for hydrocarbons

We will see the other method which works in any situation

We will use the above example

Only 30cm^{3}of Oxygen reacts: 30 50 50 3 5 5 So 5 moles of CO_{2}formed: C : CO_{2}1 : 1 So one carbon atom is required to make one CO_{2}molecule SO C is 5 Then for Hydrogen and water 2 : 1 2H : H_{2}O X : 5 so X must be equal to 10

This is very similar to the first one but can be applied if the mole ratio is not even given. For example ( different values ):

C : CO_{2}Molar Masses 12 : 44 X : 1.8g So we can find the corresponding mass of the carbon atom only and then the same for Hydrogen: 2H : H_{2}O 2 : 18 X : 2.5g Then find the mass of Hydrogen also. Then find the empirical formula using the first method

The mass will be given by the question. This is especially useful when the organic compound has oxygen it because after we find the mass of the Carbon and Hydrogen atoms, the remaining is Oxygen

Sometimes they could give the mass/percentage of O, C or H present

If 48.7% is Carbon and 8.1% is Hydrogen, the rest is Oxygen. Calculate the formula

C H O 48.7 8.1 43.2 This could represent the mass of each element present in 100g Now we divide each by their corresponding Relative atomic Masses 4.058 8.1 2.7 Then we divide each by the smallest ratio value which is 2.7 1.5 3 1 3 6 2 so the formula is C_{3}H_{6}O_{2}

What we need to understand is that Molecular formula is a multiple of the Empirical formula

But how do we find "n" or this scalar product?

They will usually give the Molar mass of the Molecular formula. This means the mass of the Empirical formula must be smaller than this and many combined would give the exact figure

For example:

The molar mass of the compound is 65 and the Empirical formula is CH

Find the Molecular formula

So CH has a mass of 13 and 65/13 will give us 5!

So we need to multiple each atom by 5