Percentage Composition
• If you have a box containing 100 golf balls and 100 ping
pong balls, which type of ball contributes the most to
the mass of the box?
• The same principle applies to finding the % composition
of a compound. Different elements have different
masses and this must be taken into consideration.
1
Agenda
• Day 52 - Percent Composition
• Lesson: PPT- Percent Composition
• Handouts: 1. Percent Composition Handout; 2.
Percent Composition Worksheet
• Text: 1. P. 284 - 288- Percent Composition
• HW: 1. Finish all the worksheets; 2. P. 288 #
4
Percent Composition (by mass)
• Percent Composition (by mass): Identifies the elements
present in a compound as a mass percent of the total
compound mass.
• The mass percent is obtained by dividing the mass of each
element by the total mass of a compound and converting
to percentage.
%compotion =
total
How to find the percent composition
of a compound:
1. Write a correct formula for the compound
2. Find the molar mass of the compound
3. Divide the total atomic mass of EACH ELEMENT by
the molar mass
4. Multiply by 100 to convert your results to a percent
5. Since you have no significant figures to go by,
express your answer to TWO decimal places with
the % sign.
What is the percent composition of each element in
NH4OH?
N:
14.01g
 100% 
35.05 g
39.97% N
5.04 g
 100% 
35.05 g
14.38% H
16.00 g
O:
 100% 
35.05 g
45.65% O
H:
N : 1  14.01g = 14.01g
H : 5  1.0078 g = 5.04 g
O : 1  16.00 g = 16.00 g
Molar mass = 35.05 g
Practice:
1. Find the percentage composition by mass of
aluminum thiocyanate.
2. A student prepares a compound of tungsten
chloride from 3.946 g of tungsten and 3.806 g of
chlorine. Assuming the reaction goes to
completion, calculate the percent composition.
3. How many grams of sodium will combine with
567.0 g of sulfur to form Na2S?
Agenda
• Day 53 - Simplest and True Formula
• Lesson: PPT- Simplest and True Formula
• Handouts: 1. Percent Composition Handout; 2.
Empirical Formula Calculations; 3. Molecular
Formula Calculations
• Text: 1. P. 289-300- Simplest and True Formula
• HW: 1. Finish all the worksheets; 2. P. 293 # 2
– 10; P. 300 # 2 pract. # 6,7
•
Empirical Formulas
• The simplest ratio of elements in a compound
• It uses the smallest possible whole number ratio of
atoms present in a formula unit of a compound
• If the percent composition is known, an empirical
formula can be calculated
Compound
Formula
Hydrogen peroxide H2O2
Benzene
C6H6
Ethylene
C2H4
Propane
C3H8
Empirical Formula
OH
CH
CH2
C3H8
Simplest and molecular formulae
Consider NaCl (ionic)
vs.
H2O2 (covalent)
Na Cl Na Cl
Cl Na Cl Na
• Chemical formulas are either “simplest” (a.k.a.
“empirical”) or “molecular”. Ionic compounds are always
expressed as simplest formulas.
• Covalent compounds can either be molecular formulas
(I.e. H2O2) or simplest (e.g. HO)
Q - Write simplest formulas for propene (C3H6),
C2H2, glucose (C6H12O6), octane (C8H14)
Q - Identify these as simplest formula, molecular
formula, or both H2O, C4H10, CH, NaCl
Answers
Q - Write simplest formulas for propene (C3H6),
C2H2, glucose (C6H12O6), octane (C8H14)
Q - Identify these as simplest formula, molecular
formula, or both H2O, C4H10, CH, NaCl
A - CH2
CH
CH2O
C4H7
A - H2O is both simplest and molecular
C4H10 is molecular (C2H5 would be simplest)
CH is simplest (not molecular since CH cannot form a
molecule - recall Lewis diagrams)
NaCl is simplest (It is ionic, thus it does not form
molecules; it has no molecular formula)
Empirical Formulas
Calculations to find the simplest formula incorporate
this rhyme:
• % to mass
• Mass to mole
• Divide by small
• Multiply till whole
A chart form may help to organize work
A compound contained 29.08 % Na, 40.58 % S, and 30.34 % O.
Find the empirical formulae for this compound.
Species % or
MM n=m/MM
mass( g) g/mol
mol
Na
29.08/22.99
29.08
22.99
Smallest
ratio
1.266
S
40.58
32.06 40.58/32.06
= 1.266
1.266
30.34
16.00 30.34/16.00
= 1.896
1.896
= 1.266
O
The formula is Na2S2O3
Whole
no.
/1.266 = 1 1 x 2 = 2
/1.266 = 1 1 x 2 = 2
/1.266 = 1.5
or Sodium thiosulfate
1.5 x 2 =3
A 5.72 g sample of washing soda(Na2CO3 . xH2O) is
heated to give 2.12 g of anhydrous Na2CO3. What is the
simplest formula of the hydrated salt.
Species % or
MM n=m/MM
mass( g) g/mol
mol
Na2CO3
2.12/106
2.12
106
Smallest
ratio
0.02
Whole
no.
= 0.02
/ 0.02 = 1
1
H2O
3.60/18
= 0.2
0.2
3.60
18
/0.02 = 10
10
Hydrated salt = Anhydrous salt + H2O
5.72g =
2.12g
+ 3.60g
The formula is Na2CO3 . 10H2O
or Sodium carbonate
decahydrate
A compound contained 40.0g C, 6.71g H, and 53.3g O.
Find its empirical formula and the empirical formula
mass.
40.0 g C 
6.71 g H 
mol C
 3.33 mol C
12.0 g C
mol H
 6.66 mol H
1.008 g H
53.3 g O 
mol O
 3.33 mol O
16.00 g O
The ratio of C to H to O is 1 to 2 to 1
3.33
 1.00
3.33
6.66
H:
 2.00
3.33
C:
O:
3.33
 1.00
3.33
Empirical
formula is
CH2O
Empirical formula mass = 12.01 + 2 (1.008) + 16.00 = 30.03 g/mol
Mole ratios and simplest formula
Given the following mole ratios for the hypothetical
compound AxBy, what would x and y be if the mole ratio of
A and B were:
AB3
A = 1 mol, B = 2.98 mol
A = 1.337 mol, B = 1 mol A4B3
A = 2.34 mol, B = 1 mol A7B3
A = 1 mol, B = 1.48 mol A2B3
If any result from Step 3 is a mixed number, you must
multiply ALL values by some number to make it a whole
number. Ex: 1.33 x 3, 2.25 x 4, 2.50 x 2, etc.
Formulas for Compounds
Empirical Formula
• Smallest possible set of subscript numbers
• Smallest whole number ratio
• All ionic compounds are given as empirical formulas
Molecular Formulas
• The actual formulas of molecules
• It shows all of the atoms present in a molecule
• It may be the same as the EF or a whole- number
multiple of its EF
Molecular formula = n х Empirical formula
Relating Empirical and
Molecular Formulas
n represents a whole number multiplier from 1 to as
large as necessary
molar mass ( g / mol )
n
empirical formula mass ( g / mol )
Calculate the empirical formula and the mass of the
empirical formula
Divide the given molecular mass by the calculated
empirical mass
Answer is a whole number multiplier
Example: Lactic acid has a molar mass of 90.08 g and
has this percent composition:40.0% C, 6.71% H, 53.3% O
What is the empirical and molecular formula of lactic
acid? Assume a 100.0 g sample size
1. Use Chart to find the Empirical Formula - CH2O
2. Obtain the mass of the Empirical Formula - 30.03 g/ mol
3. Obtain the value of n (whole number multiplier)
4. Multiply the empirical formula by the multiplier
molar mass ( g / mol )
n
empirical formula mass ( g / mol )
90.08 g / mol
3
30.03 g / mol
Molecular formula = n х empirical formula
Molecular formula = 3 (CH2O)
ANS: C3H6O3
1. What information must be known to determine
a) the empirical formula of a substance?
b) the molecular formula of a substance?
2. Determine the molecular formula for each
compound below from the information listed.
substance
simplest formula molar mass(g/mol)
a) octane
b) ethanol
c) naphthalene
d) melamine
C 4H 9
C 2H 6O
C 5H 4
CH2N2
114
46
128
126
Question 1
• For the empirical formula we need to know the
moles of each element in the compound
(which can be derived from grams or %).
For the molecular formula we need the above
information & the molar mass of the compound
Question 2
2. a) C8H18 (C4H9 = 57 g/mol, 114/57 = 2)
b) C2H6O (C2H6O = 46 g/mol, 46/46 = 1)
c) C10H8 (C5H4 = 64 g/mol, 128/64 = 2)
d) C3H6N6 (CH2N2 = 54 g/mol, 126/42 = 3)