Formulas-part+2

//**Empirical formula **//

This is a formula that shows the simplest whole number ratio of the atoms in a compound. (It does not show the actual number of atoms). Let’s take an example:

You have a compound with the formula: C 6 H 12 O 6

This formula shows that in one molecule of the above compound there are 6 carbon, 12 hydrogen and 6 oxygen atoms.

The empirical formula for this compound will be CH2O. This one shows the ratios of the atoms. In this molecule for every 1 carbon atom there has to be 2 hydrogen and 1 oxygen atoms.

//So, How so we get the empirical formula of a compound? //

It is pretty simple. You will usually be given some experimental values and based on that you can find the empirical formula for a compound.

Look at this example:

24g of Magnesium (Mg) is burnt in air and 40g of Magnesium oxide is formed. Find the empirical formula of magnesium oxide?

We know that the mass of Mg reacted is 24g

The mass of Oxygen reacted is= 40-24 =16g

To find the empirical formula:



Another example of finding the empirical formula: A hydrocarbon is 80% Carbon and 20% Hydrogen. Find the empirical formula.

If we had 100g of the hydrocarbon, then 80g would be carbon and 20g hydrogen. So we can calculate the empirical formula as follows:



<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">We now know the empirical formula of the compound is CH 3. This just tells us that for every carbon atom there are 3 hydrogen atoms but doesn’t show us the actual number of atoms present in a molecule.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">To do that we need to know the molecular mass of the compound.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Supposing in this case the molecular mass is 30.626 g.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">We do the following:

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">The mass of the empirical formula = (12.01)+(3x1.01)=15.313g

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Therefore



<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">That means that the mass of the molecule is twice that of the empirical formula. So <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">We have 2x(CH 3 )=>C 2 H 6 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">The molecular formula is C 2 H 6

//<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">So far so good. Things have been going well. But then things can turn nasty at times. Sometimes questions aren’t as straight forward and you have to do some thinking to get to the answer. Look at the example below: //

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">0.493g of an organic compound gave 0.881g of CO2 and 0.359g of H2O on complete combustion. Determine the empirical formula of the organic compound?? // L <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Confused?! Let’s take this step by step. //

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">An organic compound is a compound that contains Carbon and Hydrogen and some times oxygen. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Write down the equation of the reaction as follows:



<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">The carbon dioxide came as a result of the Carbon in the compound reacting with oxygen:



<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">The water is formed as a result of the hydrogen in the compound reacting with oxygen:



<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Therefore the mass of hydrogen plus mass of Carbon =0.240+0.0399=0.2799g

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">However, the compound weighs 0.493g which means that the compound must also be containing oxygen:

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Mass of Oxygen= 0.493-0.2799=0.2131g

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Now that you have the mass of all three elements in the compound, you can find its empirical formula:



By the end of this lesson you should be able to:
 * Distinguish between the empirical formula and the molecular formula
 * Determine the empirical formula of a compound given its percentage composition or experimental values
 * To find the molecular formula of a compound when given its empirical formula and some experimental values

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