Empirical Formulas!

We're back from Christmas break and we're starting the New Year with some chemistry. Yay!!

Empirical formulas are the simplest formula of a compound. In fact, they show only the simplest ratios rather than the actual number of atoms (that would be a molecular formula). The empirical formula for Chlorine gas is simply Cl. This means that Dinitrogen tetraoxide would not be written as N₂O₄. Here are some examples of empirical formulas:

To determine the empirical formula we need to know the ratio of each element. To determine the ratio, fill in the table below for each problem:

If the smallest ratio is in decimals, you need to multiply everything by a common number. For example, multiply 0.5 by 2. Multiply 0.33 or 0.66 by 3. Multiply 0.25 or 0.75 by 4. Multiply 0.2, 0.4, 0.6, and 0.8 by 5.

EXAMPLES:
-A sample of an unknown compound is found to contain 8.4g of 'C', 2.1g of 'H' and 5.6g of 'O'. Determine the empirical formula.
**Note that some information filled out for each atom is not initially determined.
ATOM: C
-----MASS: 8.4
-----MOLAR MASS: 12.0
-----MOLES: 0.70
-----MOLE / SMALLEST MOLE: 2
-----RATIO: 2

8.4g x 1mol/12.0g = 0.70mol

ATOM: H
-----MASS: 2.1
-----MOLAR MASS: 1.0
-----MOLES: 2.1
-----MOLE / SMALLEST MOLE: 6
-----RATIO: 1

2.1g x 1mol/1.0g = 2.1mol

ATOM: O
-----MASS: 5.6
-----MOLAR MASS: 16.0
-----MOLES: 0.35
-----MOLE / SMALLEST MOLE: 1
-----RATIO: 1

5.6g x 1mol/16.0g = 0.35mol

C_0.7H₂₁O_0.35

Divide by the smallest mole, 0.35!
0.70/0.35 = 2
2.1/0.35 = 6
0.35/0.35 = 1

Therefore, the empirical formula is: C₂H₆O

-An unknown compound that has a total mass of 0.888g is made up of Carbon, Hydrogen and Oxygen. The mass of Carbon is 0.576g while the Hydrogen totals 0.120g. Determine the empirical formula.
**Note that some information filled out for each atom is not initially determined.
0.888 - (0.576 + 0.120) = 0.192g

ATOM: C
-----MASS: 0.576g
-----MOLAR MASS: 12.0
-----MOLES: 0.048
-----MOLE / SMALLEST MOLE: 4
-----RATIO: 4

0.576g x 1mol/12.0g = 0.048mol

ATOM: H
-----MASS:0.120g
-----MOLAR MASS: 1.0
-----MOLES: 0.12
-----MOLE / SMALLEST MOLE: 10
-----RATIO: 10

0.120g x 1mol/1.0g = 0.12mol

ATOM: O
-----MASS: 0.192
-----MOLAR MASS: 16.0
-----MOLES: 0.012
-----MOLE / SMALLEST MOLE: 1
-----RATIO: 1

0.192g x 1mol/16.0g = 0.012mol

Divide by the smallest mole, 0.012!
0.048/0.012 = 4
0.12/0.012 = 10
0.012/0.012 = 1

Therefore, the empirical formula is: C₄H₁₀O₁

-Determine the empirical formula of a compound that is 50.5% C, 5.26% H and 44.2% N.
**Note that some information filled out for each atom is not initially determined. We can assume that the percentages actually represent the mass.

ATOM: C
-----MASS: 50.5
-----MOLAR MASS: 12.0
-----MOLES: 4.21
-----MOLE / SMALLEST MOLE: 1.33...
-----RATIO: 4

50.5g x 1mol/12.0g = 4.21mol

ATOM: H
-----MASS: 5.26
-----MOLAR MASS: 1.0
-----MOLES: 5.26
-----MOLE / SMALLEST MOLE: 1.66
-----RATIO: 5

5.26g x 1mol/1.0g = 5.26mol

ATOM: N
-----MASS: 44.2
-----MOLAR MASS: 14.0
-----MOLES: 3.16
-----MOLE / SMALLEST MOLE: 1
-----RATIO: 3

44.2g x 1mol/14.0g = 3.16mol

Divide by the smallest mole, 3.16!
4.21/3.16 = 1.33
5.26/3.16 = 1.66
3.16/3.16 = 1

C_1.33H_1.66O₁

Multiply all of the subscripts by 3 because we don't want decimals!

Therefore, the empirical formula is C₄H₅O₃




 
Next time: Molecular Formulas!