1. Chapter 6
Chemical Composition
2006, Prentice Hall

2. CHAPTER OUTLINE The Mole Concept Molecular & Molar Mass
Calculations Using the Mole
Percent Composition
Chemical Formulas
Calculating Empirical Formulas
Molecular Formulas

3. Why is Knowledge of Composition Important?
everything in nature is either chemically or physically combined with other substances
to know the amount of a material in a sample, you need to know what fraction of the sample it is
Some Applications:
the amount of sodium in sodium chloride for diet
the amount of iron in iron ore for steel production
the amount of hydrogen in water for hydrogen fuel
the amount of chlorine in freon to estimate ozone depletion
Tro's Introductory Chemistry, Chapter 6

4. THE MOLE CONCEPT
Chemists find it more convenient to use mass relationships in the laboratory, while chemical reactions depend on the number of atoms present.
In order to relate the mass and number of atoms, chemists use the SI unit mole (abbreviated mol).

5. Avogadro’s number (NA)
THE MOLE CONCEPT
The number of particles in a mole is called Avogadro’s number and is 6.02x1023.
1 mole
6.02 x 1023
equals to
Avogadro’s number (NA)

6. A mole is a very large quantity
THE MOLE CONCEPT
A mole is a very large quantity
6.02x1023
If 10,000 people started to count Avogadro’s number and counted at the rate of 100 numbers per minute each minute of the day, it would take over 1 trillion years to count the total number.

7. THE MOLE CONCEPT 1 mole H atoms = 6.02x1023 H atoms
1 mole H2 molecules
= 6.02x1023 H2 molecules
= 2 x (6.02x1023) H atoms
1 mole H2O molecules
= 6.02x1023 H2O molecules
= 2 x (6.02x1023) H atoms
= 6.02x1023 O atoms
1 mole Na+ ions
= 6.02x1023 Na+ ions

8. Mass of 1 mol H atoms = 1.008 grams
THE MOLE CONCEPT
The atomic mass of one atom expressed in amu is numerically the same as the mass of 1 mole of atoms of the element expressed in grams.
Mass of 1 Mg atom = amu
Mass of 1 H atom = amu
Mass of 1 Cl atom = amu
Mass of 1 mol Cl atoms = g
Mass of 1 mol Mg atoms = g
Mass of 1 mol H atoms = grams

9. MOLE DAY
Chemists and chemistry students celebrate two days in the year in honor of the Mole and call them Mole Days.
October 23rd
6:02 a.m.
Jun 2nd
10:23 a.m.

10. Mass of one molecule of H2O
MOLECULAR MASS
The sum of atomic masses of all the atoms in one molecule of a substance is called molecular mass, and is measured in amu.
Mass of one molecule of H2O
2 H atoms =
2 (1.008 amu) = amu
1 O atom =
1 (16.00 amu) = amu
Molecular mass
18.02 amu

11. Relationship Between Moles and Mass
The mass of one mole of atoms is called the molar mass
The molar mass of an element, in grams, is numerically equal to the element’s atomic mass, in amu

12. MOLAR MASS
The mass of one mole of a substance is called molar mass, and is measured in grams.
Mass of one mole of H2O
2 mol H atoms =
2 (1.008 g) = g
1 mol O atom =
1 (16.00 g) = g
Molar mass
18.02 g

13. CALCULATIONS USING THE MOLE
Conversions between mass, mole and particles can be done using molar mass and Avogadro’s number.
Avogadro’s number
Molar mass
Mass of a substance
Moles of a substance
Particles of a substance MM NA

14. Example 1: What is the mass of 5.00 mol of water? 18.02 90.1 1
3 significant figures
Molar mass

15. Example 2: Avogadro’s number Molar mass
How many Mg atoms are present in 5.00 g of Mg?
mass  mol  atoms 1
6.02 x 1023
24.3 1
1.24x1023 atoms Mg
Avogadro’s number
Molar mass
3 significant figures

16. Example 3: How many molecules of HCl are present in 25.0 g of HCl?
mass  mol  molecules 1
6.02 x 1023
36.45 1
4.13 x 1023 molecules HCl
3 significant figures

17. PERCENT COMPOSITION
The percent composition of a compound is the mass percent of each element in the compound.
Total mass of element
Total mass of compound

18. Example 1:
Calculate the percent composition of sodium chloride (NaCl).
Step 1:
determine molar mass of NaCl
1 mol Na atoms =
1 (22.99 g) = g
1 (35.45 g) = g
1 mol Cl atom =
Molar mass
58.44 g/mol

19. Example 1: Step 2: calculate the mass % of each element Sum = 100%
% Na =
% Cl =

20. Example 2: mass of sample =
1.63 g of zinc combines with 0.40 g of oxygen to form zinc oxide. Determine the % composition of the compound formed.
Step 1:
determine total mass of sample
mass of sample =
1.63 g g = 2.03 g

21. Example 2: Step 2: calculate the mass % of each element Sum = 100%
% Zn =
% O =

22. Example 3:
Calculate the percent composition of sodium hydroxide (NaOH).
Step 1:
determine molar mass of NaOH
1 mol Na atoms =
1 (23.0 g) = 23.0 g
1 (16.0 g) = 16.0 g
1 mol O atom =
1 (1.01 g) = 1.01 g
1 mol H atoms =
Molar mass
40.0 g/mol

23. Example 3:
Calculate the percent composition of sodium hydroxide (NaOH).
Step 1:
determine molar mass of NaOH
1 mol Na atoms =
1 (23.0 g) = 23.0 g
1 (16.0 g) = 16.0 g
1 mol O atom =
1 (1.01 g) = 1.01 g
1 mol H atoms =
Molar mass
40.0 g/mol

24. CHEMICAL FORMULAS Molecular formula Empirical formula
Shows the actual number of atoms in a compound
Can be written for molecular compounds only
Shows the simplest ratio of atoms in a compound
Can be written for molecular and ionic compounds

25. CHEMICAL FORMULAS Molecular Empirical Multiplier H2O2 H2O C6H12O6 C6H6

26. CHEMICAL FORMULAS
Several compounds may possess the same percent composition and empirical formula, but different molecular formulas.
Formula
Composition
% C % H
Molar mass
(g/mol) CH
92.3
7.7
C2H2
C6H6
13.02
26.04 (2x13.02)
78.12 (6x13.02)

27. Finding an Empirical Formula
convert the percentages to grams
skip if already grams
convert grams to moles
use molar mass of each element
write a pseudoformula using moles as subscripts
divide all by smallest number of moles
multiply all mole ratios by number to make all whole numbers
if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3, etc.
skip if already whole numbers

28. If, after dividing by the smallest number of moles, the subscripts are not whole numbers, multiply all the subscripts by a small whole number to arrive at whole-number subscripts.

29. CALCULATING EMPIRICAL FORMULAS
Arsenic (As) reacts with oxygen (O) to form a compound that is 75.7% As and 24.3% oxygen by mass. What is the empirical formula for this compound?
Assume 100 g
Step 1:
Percent to mass
75.7% As
75.7 g As
24.3% O
24.3 g O

30. CALCULATING EMPIRICAL FORMULAS
Step 2:
Mass to mole
75.7 g As
24.3 g O
Use atomic mass of oxygen

31. CALCULATING EMPIRICAL FORMULAS
Step 3:
Divide by small
As =
O =

32. CALCULATING EMPIRICAL FORMULAS
Step 4:
Multiply till Whole
As2O3
As1.00O1.50
x 2 = 2
x 2 = 3

33. Example 1:
Determine the empirical formula for a compound containing 11.2% H and 88.8% O.
mol H =
mol O =
H2O

34. Example 2:
Determine the empirical formula for a compound with the following percent composition: % C, 13.12% H, 34.73% O.
C2H6O
mol C =
mol H =
mol O =

35. Molecular formula = (empirical formula) n
MOLECULAR FORMULAS
Molecular formula can be calculated from empirical formula if molar mass is known.
Molecular formula = (empirical formula) n

36. Example 1:
A compound of N and O with a molar mass of 92.0 g, has the empirical formula of NO2. What is its molecular formula?
Mass of empirical formula =
(16.0) = 46.0
Molecular formula =
2 x (NO2) = N2O4

37. Example 2:
Calculate the empirical and molecular formulas of a compound that contains 80.0% C and 20.0% H, and has a molar mass of g.
mol C =
mol H =
Empirical formula
CH3

38. Example 2: Empirical formula = CH3 Mass of empirical formula =
(1.01) = 15.0
Molecular formula =
2 x (CH3) = C2H6

39. THE END