__Empirical Formula:__Gives the lowest whole number ratio of atoms in a the formula.

__Molecular Formula:__A multiple of the empirical formula that shows the actual number of atoms that combine to form a molecule. (I will show you how to determine what the multiple is later on.)

Let's being with the empirical formula...

*What is the empirical formula of a compound consisting of 60.0% C and 40.0% H?*

We are going to assume that we have 100g of the compound, because 80% + 20% works out to 100%. It is the ratio that counts, so it's okay if we don't really have 100g- the ratio will still be the same as long as the percentages are the same.

1) Find the number of moles of each element...

80g of C x 1mole of O

-------------- = 6.67 moles of oxygen

16.0g

20g of H x 1 mole of H

--------------- = 20 moles of H

1.0g

2) Now divide each of the values of moles by the smallest value of mole that you've calculated. The number that you get will be that elements ratio.

6.67

For Oxygen: -------- =

**1** 6.67

20

For Hydrogen: ------ =

**3** 6.67

Therefore, your empirical formula is

**CH3**.If you ever get ratios that are like 2.50 or 5.66, just multiply all the ratios by a number that will give you a whole number. For example, if you had:

5.66

3

1

You would multiply each of those ratios by 3 because that would give you whole numbers:

5.66 x 3 = 17

3 x 3 = 9

1 x 3 = 3

Voila!

Now moving on to molecular formula...

It requires you to find the empirical formula, then multiply it by a multiple.

To find that multiple, you use this formula:

molar mass of compound

multiple = -------------------------------------------

molar mass of the empirical formula

*A molecule has an empirical formula C2H5, and a molar mass of 58g/mol.*

Molar mass of empirical: (2x12)+(5x1) = 29g/mol

Molar mass of compound: 58g/mol (given to us in the question)

58g/mol

------------- = 2

29g/mol

Now we know that our multiple is 2, so we need to multiply our empirical formula by 2.

2(C2H5) =

**C4H10**

C4H10 is our molecular formula.

## No comments:

## Post a Comment