Percent Composition
The total amount or % of an
element in its compound.
Percent Composition
Three STEPS
1) Formula determination
 using Criss-cross determine subscripts
2)
Formula mass
 using periodic table, total all elements
3) % Calculation
 element mass/formula mass x 100
Percent Composition
Find the percent composition of Iron(II)Nitrate
Step 1 Determine the formula 
Fe(NO3)2
Step 2
Determine the formula mass
1 Fe atom 1 x 55.8 amu
2 N atoms 2 x 14 amu
6 O atoms 6 x 16 amu

= 55.8 amu
= 28 amu
= 96 amu
179.8 amu
Percent Composition
Step 3. Calculate % of total mass
1 Fe atom = 55.8 amu Fe
x 100 = 31.0 %
179.8 amu Fe(NO3)2
2 N atoms = 28 amu N
179.8 amu Fe(NO3)2
x 100 = 15.6%
6 O atom = 96 amu O
179.8 amu Fe(NO3)2
x 100 = 53.4 %
Percent Composition
Problems:
1. Determine the % composition of Na and
C in Sodium Carbonate.
Formula
Na2CO3
Total
mass
106
amu
Math
%
Na =46/106 x100
Na= 43.4%
C =12/106 x100
C= 11.3%
O =48/106 x100
O= 45.3%
Percent Composition
2.
Formula
Al2(SO4)3
Determine the % composition of all the
elements in Aluminum Sulfate.
Total
mass
Math
%
Al =54/342.3 x100
Al= 15.8%
342.3
S =96.3/342.3 x100
amu
S= 28.1%
O =192/342.3 x100
O= 56.1%
Percent Composition
3.
Formula
Determine the % composition of all the
elements in Sodium Hydroxide.
Total
mass
Math
%
Na = 23.0/40.0 x100 Na= 57.5%
NaOH
40.0
amu
H = 1/40.0 x100
H= 2.5%
O = 16/40.0 x100
O= 40.0%
Percent Composition
4. Determine the % composition of
C in Carbon Dioxide.
Formula
CO2
Total
mass
Math
%
44.0
amu
C =12.0/44.0 x100
C= 27.3%
O =32.0/44.0 x100
O= 72.7%
Percent Composition
5. In a sample of Pitchblende, Uranium ore,
it is determined that there are 2.5 g of
Uranium in a
500.0 g sample. What
% of the ore is Uranium?
Little different problem but still % comp.
Total Ore = 500.0 g
Uranium = 2.5 g
% = U / Ore x 100  2.5 g/ 500.0 g x100
U = 0.50% of the Ore
% Composition Practice Problem
What is the % composition of all the
elements in Ammonium Phosphite.
Step 1  Formula (NH4)3PO3
Step 2  Molar Mass
3(14.0) + 12(1.0) + 31.0 + 3 (16.0) = 133.0 g
Step 3  % Comp
N = 42/133 x 100 = 31.6 %
H = 12/133 x 100 = 9.0 %
P = 31/133 x 100 = 23.3 %
O = 48/133 x100 = %36.1 %
Empirical Formula
What if you were given the % and
asked to find the formula.
A compound contains
32.38 % Na
22.65 % S
44.99 % O
What is the formula of this compound?
Empirical Formula
Step 1  Calculate the number of moles for all
elements in compound.
a) If given as mass convert using molar mass.
b) If given as a %, treat as if it is a 100 g
Step 2  Determine the smallest mole value.
Step 3  Determine mole ratio of elements by
dividing by lowest mole value.
Empirical Formula
Step 4  If a mole ratio develops that is not a whole #
or easily rounded off, multiply all ratios
by a small integer (2, 3 or 4) to convert
the fraction into a whole #.
Step 5  Apply mole ratio to determine the Empirical
formula. (The whole # ratio become the subscripts!)
Bonus Step  If asked to find the Molecular formula,
divide the empirical formula mass into the molecular
mass to get an integer. Multiply that integer through the
empirical formulas subscripts. The result will be the
molecular formula.
Empirical Formula
Step 1 Convert all values to moles
Start with a 100 g sample if in %
32.38 % Na  32.38 g of Na / 23.0 g Na
22.65 % S  22.65 g of S / 32.1 g S
44.99 % O  44.99 g of O / 16.0 g O
Empirical Formula
Step 2:: Determine the smallest value.
32.38 g of Na = 1.41 mol Na
22.65 g of S = 0.706 mol S
44.99 g of O = 2.81 mol O
 smallest
Empirical Formula
Step 3:: Divide through to get mole ratio.
Na  1.41 mol Na /0.706  1.997 mol  2
0.706 mol
S  0.706 mol S /0.706  1 mol  1
O  2.81 mol O/0.706
 3.98 mol  4
Empirical Formula
Step 4:: Ratio are whole #  Skip step
Step 5 :: whole # ratios convert to subscripts
Na  2
S1
Na2SO4
O4
Sodium Sulfate
Empirical Formula Homework
From your books::
Go to page 246.
Review sample problems L, M and N
Complete practice problems 1-3  Pg. 247
Complete practice problems 1-2  Pg. 249