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### What do you want to do? # Empirical & Molecular Formula

## What is Molecular Formula?

### The actual whole number ratio of atoms and elements in a formula of a compound

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

The Molecular formula is

#### H2O2

This shows the actual number of atoms present

## What is Empirical formula?

### It is the simplest whole number ratio of atoms and elements in a formula of a compound

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 H2O2

This gives us another obvious equation

#### Molecular formula = n(Empirical formula)

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

## How do we calculate the Empirical 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:

• If they give the percentage of each element present in a compound
• 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

#### C5H26O11

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

## Calculating the Empirical Formula using Combustion data

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 50cm3 of Oxygen. 20cm3 of Oxygen is remaining

. 50cm3 of CO2 is produced and 50cm3 of H2O 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

```CxHy+ (x+y/4) O2 = XCO2 + (Y/2) H2O

Only 30cm3 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 C5H10 and the empirical formula is CH2

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 30cm3 of Oxygen reacts:

30    50    50

3    5      5

So 5 moles of CO2 formed:

C   :   CO2

1   :      1

So one carbon atom is required to make one CO2 molecule

SO C is 5

Then for Hydrogen and water

2  :   1

2H  : H2O

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    :  CO2

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    :  H2O

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

## Combustion Data with Percentages

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 C3H6O2

```

## Finding the Molecular Formula

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

#### Molecular Formula = n(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