% Composition & Empirical Formula

V.4 PERCENTAGE COMPOSITION

• Percentage composition is the percentage (by mass) of the species in a chemical formula.

How to calculate it…For MgO

1) Find the total molar mass of your compound.

2) Find the specific molar mass of the each element you have.

3) Divide the specific molar mass / total mass

4) Multiply by 100%

Example #1: What is the percent composition of ?

a) MgO (what % of the mass is magnesium, what % of the mass is oxygen?)

b) FeCl2

c) (NH4)3PO4

Example #2: What is the percent composition of the bold species? NiSO47H20

Empirical Formula

V.5 EMPIRICAL & MOLECULAR FORMULA

Empirical Formula is the SMALLEST ‘whole number ratio’ of atoms which represents the molecular make-up of a compound.

CH2, C2H4, C3H6, C4H8, C5H10 --- all contain twice as many H’s as C’s

 Therefore, the empirical formula (or simplest ratio) is CH2

 In this case, all the formulae are whole-number multiples of CH2

C2H4 = 2 x CH2

C3H6 = 3 x CH2

C4H8 = 4 x CH2

The empirical formula is the simplest whole number ratio between atoms in a compound.

It is determined experimentally by measuring the mass of the elements that combine to form a

compound.

Example: A compound was found to be composed of 38.7 g C, 9.68 g H and 51.6 g O. Calculate the empirical formula.

A compound was found to be composed of 38.7 g C, 9.68 g H and 51.6 g O.

Calculate the empirical formula.

CH3O

= 1

= 3

= 1

38.7 g C x 1 mole = 3.225 mol 12.0 g

9.68 g H x 1 mole = 9.584 mol 1.01 g

51.6 g O x 1 mole = 3.225 mol 16.0 g

3.225 mol

3.225 mol

3.225 mol

Change grams to moles first !

Divide by the smallest number to get whole numbers!

Write the formula!

A compound is found to contain 63.55 % Ag, 8.23 % N and 28.24 % O. Calculate the empirical formula.

0.5879 mol

0.5879 mol

0.5879 mol= 1.765 mol

= 0.5890 mol

28.24 g O x 1 mole

16.0 g

AgNO3

= 3

= 1

= 163.55 g Ag x 1 mole

107.9 g8.23 g N x 1 mole

14.0 g

= 0.5879 mol

Assume that you have 100 g- that means 63.55 g Ag, 8.23 g N, and 28.24 g O

Change grams to moles using molar mass!

A compound is found to contain 50.07 % Cu, 16.29 % P and 33.64% O. Calculate the empirical formula.

Change % into grams first, then into moles.

Only do this for ionic compounds- rearrange it as a polyatomic ion

Not for covalent!

Cu3(PO4)2

0.5255 mol

= 4

= 1.5

= 1

= 2.103 mol

= 0.7885 mol50.07 g Cu x 1 mole

63.5 g

= 0.5255 mol16.29 g P x 1 mole

31.0 g

33.64 g O x 1 mole 16.0 g

0.5255 mol

0.5255 mol

Cu3P2O8

Double

= 3

= 2

= 8

Rearrange

What is the empirical formula of a compound containing 39.0 % Si and 61.0 % O?

Page 93 – Common Fractions → Decimal Conversions– Helpful to be able to recognize these! Saves you time!

2.67, 1.33, 5.67, 3.33, etc involve THIRDS ( x 3 to clear fraction)

1.75, 2.25, 3.75, etc involve QUARTERS ( x 4 to clear fraction)

IMPORTANT: Don’t round intermediate values, keep 3 or 4 decimals...

Homework Percentage composition questions

Page 91#44 a, d, g, I, k, n

#45 a, b, f

Empirical Formula questions

Page 93 # 46 (odd)