1. Percentage Composition
Way to specify relative masses of each element in a compound
List of percentage by mass of each element
Percentage by Mass
Ex. Na2CO3 is
43.38% Na
11.33% C
45.29% O
What is sum of % by mass?
100.00%

2. Ex. Percent Composition
Determine percentage composition based on chemical analysis of substance
Ex. A sample of a liquid with a mass of g was decomposed into its elements and gave g of carbon, g of hydrogen, and g of oxygen. What is the percentage composition of this compound?
Analysis:
Total mass = g
Mass of each element

3. Ex. % Composition of Compound
For C:
For H:
For O:
% composition tells us mass of each element in g of substance
In g of our liquid
60.26 g C, g H & g O
= 60.26% C
= 11.11% H
= 28.62% O
Sum of percentages: 99.99%

4. Your Turn!
A sample was analyzed and found to contain g nitrogen and g oxygen. What is the percentage composition of this compound?
1. Calculate total mass of sample
Total sample mass = g g = g
2. Calculate % Composition of N
3. Calculate % Composition of O
= 25.94% N
Percent composition
25.94% N & 74.06% O
= 74.06% O

5. Calculating mass percentage from a chemical formula
The mass percentage composition allows you to determine the fraction of the total mass each element contributes to the compound.
Example
Ammonium nitrate (NH4NO3) is an important compound in the fertiliser industry. What is the mass % composition of ammonium nitrate?
Step 1: Calculate the molar mass of ammonium nitrate
Molar mass NH4NO3 = gmol-1
Mass fraction of N = g x100
80.05g
= 34.99% ≈ 35%
Mass fraction of H = g x100
80.05g
= 5.04% ≈ 5%
Oxygen: 48.00g O in one mol of ammonium nitrate
As above, the mass % composition of O is found to be 60%

6. Your Turn
If a sample containing only phosphorous & oxygen has percent composition 56.34% P & 43.66% O, is this P4O10?
Yes No
4 mol P  1 mol P4O10
10 mol O  1 mol P4O10
4 mol P = 4  g/mol P = g P
10 mol O = 10 16.00 g/mol O = g O
1 mol P4O10 = g P4O10
= % P
= % O

7. Question What is the mass % composition of C12H22O11?
Answer: % Carbon: 42.1%
% Hydrogen: 6.5%
% Oxygen: 51.4%

8. Determining Empirical & Molecular Formulas
When making or isolating new compounds one must characterize them to determine structure &
Empirical Formula
Simplest ratio of atoms of each element in compound
Obtained from experimental analysis of compound
Molecular Formula
Multiple of EF
4.3 | Determining Empirical and Molecular Formulas
Empirical formula CH2O
glucose
Molecular formula C6H12O6

9. 2 Ways to Calculate Empirical Formulas
From Masses of Elements
Ex g sample of which g is Fe and g is O.
From Percentage Composition
Ex % P and % O.

10. Strategy for Determining Empirical Formulas
Determine mass in g of each element
Convert mass in g to moles
Divide all quantities by smallest number of moles
Convert any non-integers into integer numbers.
If number ends in decimal equivalent of fraction, multiply all quantities by least common denominator
Otherwise, round numbers to nearest integers if within 0.10

11. 1. Empirical Formula from Mass Data
When a g sample of a compound was analyzed, it was found to contain g of C, g of H, and g of N. Calculate the empirical formula of this compound.
Step 1: Calculate moles of each substance
3.722  103 mol C
1.860  102 mol H
3.723  103 mol N

12. 1. Empirical Formula from Mass Data
Step 2: Select the smallest # of moles.
Lowest is x 10–3 mole
C =
H =
Step 3: Divide all # of moles by the smallest one
Mole ratio
Integer ratio
1.000
= 1
4.997
= 5
N =
1.000
= 1
Empirical formula = CH5N

13. Empirical Formula from Mass Composition #2
One of the compounds of iron and oxygen, “black iron oxide,” occurs naturally in the mineral magnetite. When a g sample was analyzed it was found to have g of Fe and g of O. Calculate the empirical formula of this compound. Assembling the tools: 1 mol Fe = g Fe 1 mol O = g O 1. Calculate moles of each substance
mol Fe
mol O

14. 1. Empirical Formula from Mass Data
2. Divide both by smallest #mol to get smallest whole # ratio.
=1.000 Fe
× 3 = Fe
=1.33 O
× 3 = 3.99 O Or
Empirical Formula = Fe3O4

15. Determining Molecular Formulas
Empirical formula
Accepted formula unit for ionic compounds
Molecular formula
Preferred for molecular compounds
In some cases molecular & empirical formulas are the same
When they are different, & the subscripts of molecular formula are integer multiples of those in empirical formula
If empirical formula is AxBy
Molecular formula will be An×xBn×y

16. Determining Molecular Formula
Need molecular mass & empirical formula
Calculate ratio of molecular mass to mass predicted by empirical formula & round to nearest integer
Ex. Glucose
Molecular mass is g/mol
Empirical formula = CH2O
Empirical formula mass = g/mol
If molecular mass & empirical formula mass are same, then molecular for
Molecular formula = C6H12O6

17. Learning Check
The empirical formula of a compound containing phosphorous and oxygen was found to be P2O5. If the molar mass is determined to be g/mol, what is the molecular formula?
Step 1: Calculate empirical mass
Step 2: Calculate ratio of molecular to empirical mass
= 2
Molecular formula = P4O10

18. Your Turn!
The empirical formula of hydrazine is NH2, and its molecular mass is What is its molecular formula?
NH2
N2H4
N3H6
N4H8
N1.5H3
Molar mass of NH2 =
(1×14.01)g + (2×1.008)g = g
n = (32.0/16.02) = 2 and (B)