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