Percentage Composition:

is the percent mass of each element present in a compound.

Percent Composition

Can be calculated if given:

the chemical formula OR

masses of elements in compound

By Chemical Formula

% mass =

molar mass of an element X 100%

total molar mass of the compound

Example: What is the % composition of CaCO3?

Step 1: Find the molar mass of CaCO3 : Ca x 1 = 40.1 g/mol

C x 1 = 12.0 g/mol

O x 3 = 48.0 g/mol

CaCO3 = 100.1 g/mol

CHEMICAL FORMULA

Step 2: Find the % composition:

% Ca = 40.1 g/mol x 100 % = 40.1 % Ca 100.1 g/mol

% C = 12.0 g/mol x 100% = 12.0 % C 100.1 g/mol

% O = 48.0 g/mol x 100 % = 48.0 % O 100.1 g/mol

Example: Calculate the percent composition of the compounds that is formed from this reaction: 29.0g of Ag combines completely with 4.30g of S.

STEP 1: find the total mass of the elements. 29.0g + 4.30g = 33.30g

STEP 2: find the % composition

Ag = 29.0g x 100% = 87.1% 33.30g

S = 4.30g x 100% = 12.9% 33.30g

Masses of elements in compound

Try These:1) Find the percent composition of

KMnO4.

2) Calculate the % composition of the compound that results from 9.03g Mg reacting completely with 3.48g N.

ANSWERS:

1) K = 24.7%Mn = 34.8%

O = 40.5%

2) Mg = 72.2%

N = 27.8%

1) Do Problem 17 on page 131

2) Do Problem 18 & 19 on page 133

CHECK YOUR ANSWERS!

Example 2 How much carbon is present in 15.2 g of carbon dioxide gas?

% carbon = 12.0 g/mol x 100 % = 27.3 %

44.0 g/mol

Xg carbon = 15.2 g of CO2 X 27.3 g C 100.

g CO2

= 4.15 g of Carbon

Percent Composition

Can be used to: calculate the mass of elements in a compound

determine the empirical formula of a compound

determine the molecular formula of a compound

Empirical Formula

shows the simplest mole ratio of the elements.

CO is a 1:1 ratio of carbon to oxygen H2O is a 2:1 ratio

CO2 is a 1:2 ratio

Empirical formulas can’t be reduced.

Molecular Formula

shows the actual number of atoms in a molecule.

The molecular formula for hydrogen peroxide is H2O2. Its empirical formula would be HO.

Often the molecular formula is the same as the empirical formula: H2O, CO2

Empirical?

CH4O– yes, cannot be reduced further

C2H6

– no, empirical would be CH3

C3H10O– yes

C6H6O2

– no. What would empirical be?– C3H3O

Calculating Empirical Formulas

A chemist with an unknown compound can easily figure out its percent composition, but it is much more

meaningful to know its formula.

EXAMPLE: What is the empirical formula for a compound that is 25.9% nitrogen and 74.1% oxygen?

Method1. Write the mass (g) of each

element in the compound. So….we assume that it is a 100g sample:

25.9% N = 25.9g

74.1% O = 74.1g

2. Convert the mass of each element to moles, by dividing by the molar mass.

N = 25.9g = 1.85 mol

14.0g/mol

O = 74.1g = 4.63 mol

16.0g/mol

3. Calculate the simplest whole number ratio by dividing the number of moles by the smallest number of moles.

1.85 : 4.63 = 1 : 2.5 1.85 1.85

(If the result is not within 0.1 of a whole number, multiply all numbers by a whole number)

2 ( 1 : 2.5) = 2 : 5

4. Write the empirical formula using the numbers you obtained. N2 O5

NOTE: For inorganic compounds, write

the most positive element first.

For organic compounds, write C first, H second and all others alphabetically.

A special present just for you……..

Page 135, Problems #20 & 21

Check your answers

Molecular FormulaGiven the empirical formula and

the gram formula mass (gfm)

OR

Given the percent composition and the gram formula mass (gfm)

Example #1Calculate the molecular formula for NaO

having a gfm of 78g.

Determine the efm (empirical formula mass).

NaO = 23.0g + 16.0g = 39.0Divide the efm into the gfm.

78.0 = 2

39.0This is the conversion factor used to

determine the molecular formula. Na2O2

Example #2Find the molecular formula for a

compound having a composition of 58.8% C, 9.8% H and 31.4% O and a gmm of 102g/mol.

Determine the mass of each component.

C = 102g/mol x 58.8% = 60.0g/mol

H = 102g/mol x 9.8% = 10.0g/mol

O = 102g/mol x 31.4% = 32.0g/mol

convert to moles

C = 60.0g/mol = 5

12.0g

H = 10.0g/mol = 10 1.0g

O = 32.0g/mol = 2

16.0g

Use moles as subscripts for components of compound

C5H10O2

Check the gmm of this compound…does it equal 102.0g/mol?

5(12.0) + 10(1.0) + 2(16.0) = 102.0g/molYES!

And Now…..Oh Yeah! And

there’s more…

Page 136, Problems #22 & 23

Now Try page 139, #41 44