Empirical and Molecular formulas

• Empirical – lowest whole number ratio of elements in a compound

• Molecular – some multiple of the empirical formula

• Examples:CH4 C6H12O6 Na2SO4 C3H6

Can an empirical formula also be the molecular formula? YES C12H22O11

E – 1:4 M- 6:12:6 reduced to 1:2:1

E – 2:1 M -3:6 reduced to 1:2

Steps for determining empirical formulas.

1) Assume a 100 g sample when given percents. This makes the 10.3 % Z = 10.3 g Z

2) Change grams into moles for each element.3) Divide the all the moles by smallest number

of moles to get the lowest whole number ratio.

4) Write the empirical formula.

A compound was found to contain 36.11 % calcium and 63.89 % chlorine by mass. What is its empirical

formula? What assumption did you make?

100 g sample

36.11 % Ca = 36.11 g Ca x

63.89 % Cl = 63.89 g Cl x

= 0.9009 mol Ca

= 1.802 mol Cl

0.9009

0.9009

= 1 mol Ca

= 2 mol Cl

Therefore the empirical formula is CaCl2

Step 1 Step 2 Step 3

Step 4

GOOFY MATH( .33 = 1/3 .5 = ½ .67 = 2/3)

A compound was found to contain 68.42 % chromium and the rest oxygen by mass. What is its empirical formula? What assumption did you make? 100 g sample

68.42 g Cr

31.58 g O

= 1.316 mol Cr

= 1.974 mol O

1.316

1.316

= 1 mol Cr

= 1.5 mol O

X 2

X 2

= 2 Cr

= 3 O

Cr2O3

Way to much to round off so you have to get rid of the fraction.

Ethene is a compound containing only carbon and hydrogen. It contains 85.63 % carbon and the rest hydrogen by mass. It also has a molar mass of 28.05 g/mol. What are empirical and molecular formulas of this compound? What assumption did you make?

85.63 g C x

14.37 g H x

100 g sample= 7.130 mol C

= 14.26 mol H

7.130

7.130

= 1 mol C

= 2 mol H

Empirical formulaCH2

Remember molecular is some multiple of the empirical. So take the molar mass of the compound and divide it by the molar mass of the empirical formula.

28.05/14.03 = 2

So take 2 x CH2 = C2H4