1. Chapter 6
Chemical
Composition

2. Why Is Knowledge of Chemical 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 sugar in a can of pop.
The amount of hydrogen in water for hydrogen fuel.
The amount of chlorine in freon to estimate ozone depletion.

3. Counting Nails by the Pound
People usually buy thousands of nails for construction
Hardware stores sell nails in the pound.
How do I know how many nails I am buying when I buy a pound of nails?

4. Counting Nails by the Pound, Continued
A hardware store customer buys 2.60 pounds of nails. A dozen nails has a mass of pounds. How many nails did the customer buy?
1 dozen nails = lbs.
12 nails = 1 dozen nails
Solution map:

5. Counting Nails by the Pound, Continued
The customer bought 2.60 lbs of nails and received 208 nails. He counted the nails by weighing them!

6. Counting Nails by the Pound, Continued
What if he bought a different size nail?
Would the mass of a dozen be lbs?
Would there still be 12 nails in a dozen?
Would there be 208 nails in 2.60 lbs?
How would this effect the conversion factors?

7. Counting Atoms by Moles
Just as we can use the dozen to count nails, we can use the mole to count atoms in a sample.
1 mole = x 1023 things.
1 mole of atoms = x 1023 atoms
1 mole of apple = x 1023 apples
Note that just like the “dozen” the mole is a constant number and this number is called the Avogadro’s number.

8. Using Moles as Conversion Factors
Since 1 mole atoms = x1023 atoms OR

9. Tro's "Introductory Chemistry", Chapter 6
Example 6.1—A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?
Given:
Find:
1.1 x 1022 atoms Ag
moles Ag
Solution Map:
Relationships:
1 mol = x 1023 atoms
atoms Ag
mol Ag
Solution:
Check:
Since the number of atoms given is less than Avogadro’s number, the answer makes sense.
Tro's "Introductory Chemistry", Chapter 6

10. Example 6.1a—Calculate the number of atoms in 2.45 mol of copper.
Given:
Find:
2.45 mol Cu
atoms Cu
Solution Map:
Relationships:
1 mol = x 1023 atoms
mol Cu
atoms Cu
Solution:
Tro's "Introductory Chemistry", Chapter 6

11. Practice—Calculate the number of atoms in 1.45 mol of Sodium.
Answer: 8.73 x 1023 Na atoms

12. Relationship Between Moles and Mass
The mass of one mole of a substance is called the molar mass.
The molar mass of an element, in grams, is numerically equal to the element’s atomic mass, in amu
E.g., 1 Carbon atom weighs amu, therefore the molar mass of Carbon is g
The lighter the atom, the less a mole weighs.
Carbon (C) is lighter than Sodium (Na); 1 mol C weighs g and 1 mol Na weighs g.

13. Mole and Mass Relationships
carbon
12.01 g
1 mole
sulfur
32.06 g

14. Example 6.2a—Calculate the moles of sulfur in 57.8 g of sulfur.
Given:
Find:
57.8 g S
mol S
Solution Map:
Relationships:
1 mol S = g
g S
mol S
Solution:

15. Example 6.2b—Calculate the moles of carbon in 0.0265 g of pencil lead.
Given:
Find:
g C
mol C
Solution Map:
Relationships:
1 mol C = g
g C
mol C
Solution:

16. Example 6.3—How Many Aluminum Atoms Are in a Can Weighing 16.2 g?
Given:
Find:
16.2 g Al
atoms Al
Solution Map:
Relationships:
1 mol Al = g, 1 mol = x 1023
g Al
mol Al
atoms Al
Solution:

17. Tro's "Introductory Chemistry", Chapter 6
Practice 1) Calculate the grams of carbon in 2.21 x 10-3 mol of lead pencil. 2) How many copper atoms are in a penny weighing 3.10 g?
Tro's "Introductory Chemistry", Chapter 6

18. Practice 1) Calculate the grams of carbon in 2
Practice 1) Calculate the grams of carbon in 2.21 x 10-3 mol of lead pencil.
Answer: g Carbon

19. Practice 2) How many copper atoms are in a penny weighing 3.10 g?
Answer: 2.94 x Copper atoms

20. Molar Mass of Compounds
The molar mass of a compound can be calculated by adding up the molar mass of all the atoms that the compound is compose of.
Example: Calculate the molar mass of water (H2O)
Since 1 mole of H2O contains 2 moles of H and 1 mole of O.
Molar mass = 2(1.01 g H) g O = g.

21. Example 6.4—Calculate the mass of 1.75 mol of H2O.
Given:
Find:
1.75 mol H2O
g H2O
Solution Map:
Relationships:
1 mol H2O = g
mol H2O
g H2O
Solution:
Check:
Since the given amount is more than 1 mol, the mass being > 18 g makes sense.

22. Example 6. 4a—How many moles are in 50. 0 g of PbO2. (Pb = 207
Example 6.4a—How many moles are in 50.0 g of PbO2? (Pb = 207.2, O = 16.00)
Given:
Find:
50.0 g mol PbO2
moles PbO2
Solution Map:
Relationships:
1 mol PbO2 = g
g PbO2
mol PbO2
Solution:
Check:
Since the given amount is less than g, the moles being < 1 makes sense.

23. Practice—What is the mass of 4.78 x 1024 NO2 molecules?
Answer: 365 g

24. Example 6.5—How many formula units are in 50.0 g of PbO2? (PbO2 = 239.2)
Given:
Find:
50.0 g PbO2
formula units PbO2
Solution Map:
Relationships:
239.2 g = x 1023
g PbO2
units PbO2
Solution:

25. Mole Relationships in Chemical Formulas
Examples
Moles of compound
Moles of constituents
1 mol NaCl
1 mol Na, 1 mol Cl
1 mol H2O
2 mol H, 1 mol O
1 mol CaCO3
1 mol Ca, 1 mol C, 3 mol O
1 mol C6H12O6
6 mol C, 12 mol H, 6 mol O
Fact check: How many moles of H atoms are in
1 mole of CH3OCH2CH3
Tro's "Introductory Chemistry", Chapter 6

26. Example 6.6—Calculate the moles of oxygen in 1.7 moles of CaCO3.
Given:
Find:
1.7 mol CaCO3
mol O
Solution Map:
Relationships:
1 mol CaCO3 = 3 mol O
mol CaCO3
mol O
Solution:
Tro's "Introductory Chemistry", Chapter 6

27. Example 6.7—Find the mass of carbon in 55.4 g C10H14O.
Given:
Find:
55.4 g C10H14O
g C
Solution Map:
Relationships:
1 mol C10H14O = g, 1 mol C = g,
10 mol C : 1 mol C10H14O
g C10H14O
mol C10H14O
mol C
g C
Solution:

28. Practice—Find the Mass of Sodium in 6. 2 g of NaCl (MM: Na = 22
Practice—Find the Mass of Sodium in 6.2 g of NaCl (MM: Na = 22.99, Cl = 35.45)
Answer: 2.4 g

29. Percent Composition of Compounds
Percentage of each element in a compound by mass can be determined from:
The formula of the compound.
The experimental mass analysis of the compound.
Tro's "Introductory Chemistry", Chapter 6

30. Example 6.9—Find the mass percent of Cl in C2Cl4F2.
Given:
Find:
C2Cl4F2
% Cl by mass
Solution Map:
Relationships:
Solution:

31. Practice—Determine the mass percent composition of Ca and Cl in CaCl2:
Answer: % Ca and % Cl

32. Mass Percent as a Conversion Factor
The mass percent tells you the mass of a constituent element in 100 g of the compound.
The fact that NaCl is 39% Na by mass means that 100 g of NaCl contains 39 g Na.
This can be used as a conversion factor.
100 g NaCl = 39 g Na
Tro's "Introductory Chemistry", Chapter 6

33. Example. : The FDA recommends that you consume 2. 4 g of sodium a day
**Example**: The FDA recommends that you consume 2.4 g of sodium a day. How many grams of NaCl must you consume a day in order to keep up with your sodium intake. NaCl is 39 % sodium by mass
Solve this in class
Answer: 6.2 g NaCl

34. Empirical Formulas
The simplest, whole-number ratio of atoms in a molecule is called the empirical formula.
Can be determined from percent composition or combining masses.
The molecular formula is a multiple of the empirical formula.

35. Empirical Formulas, Continued
Hydrogen Peroxide
Molecular formula = H2O2
Empirical formula = HO
Benzene
Molecular formula = C6H6
Empirical formula = CH
Glucose
Molecular formula = C6H12O6
Empirical formula = CH2O

36. Example—Determine the Empirical Formula of Benzopyrene, C20H12,
Find the greatest common factor (GCF) of the subscripts.
20 and 12
GCF = 4
Divide each subscript by the GCF to get the empirical formula.
C20/4H12/4 = C5H3
Empirical formula = C5H3

37. Finding an Empirical Formula From % Mass
1. Convert the percentages to grams.
a. Skip if already grams.
2. Convert grams to moles.
a. Use molar mass of each element.
3. Write a pseudoformula using moles as subscripts.
4. Divide all by smallest number of moles.
5. Multiply all mole ratios by number to make all whole numbers, if necessary.
a. -If ratio is 0.5, multiply all by 2
-if ratio is 0.33 or 0.67, multiply all by 3
-if ratio is multiply by 4
b. Skip if already whole numbers after Step 4.

38. Tro's "Introductory Chemistry", Chapter 6
Example 6.11—Finding an empirical formula from experimental data
A laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula.
C = 60.00%
H = 4.48%
O = 35.53%
Tro's "Introductory Chemistry", Chapter 6

39. Example: Find the empirical formula of aspirin with the given mass percent composition.
Step 1: Convert percentages to grams
Given: C = 60.00%
H = 4.48%
O = 35.53%
Therefore, in 100 g of aspirin there are g C, g H, and g O.

40. Step 2: Convert masses to moles:
Example: Find the empirical formula of aspirin with the given mass percent composition.
Information:
Given: g C, 4.48 g H, g O
Find: empirical formula, CxHyOz
Conversion Factors:
1 mol C = g
1 mol H = 1.01 g
1 mol O = g
Step 2: Convert masses to moles:

41. Example: Find the empirical formula of aspirin with the given mass percent composition.
Information:
Given: mol C, 4.44 mol H,
2.221 mol O
Write the empirical formula, CxHyOz
Step 3: Write a pseudoformula in the form CxHyOz .
C4.996H4.44O2.221

42. Find: empirical formula, CxHyOz Solution Map: g C,H,O  mol C,H,O 
Example: Find the empirical formula of aspirin with the given mass percent composition.
Information:
Given: C4.996H4.44O2.221
Find: empirical formula, CxHyOz
Solution Map:
g C,H,O  mol C,H,O 
mol ratio  empirical formula
Step 4: Find the mole ratio by dividing by the smallest number of moles in C4.996H4.44O2.221

43. Step 5: Multiply subscripts by factor to give whole number.
Example: Find the empirical formula of aspirin with the given mass percent composition.
Information:
Given: C2.25H2O1
Find: empirical formula, CxHyOz
Solution Map:
g C,H,O  mol C,H,O 
mol ratio  empirical formula
Step 5: Multiply subscripts by factor to give whole number.
{ } x 4
C9H8O4

44. Example 6.12—Finding an Empirical Formula from Experimental Data
A 3.24-g sample of titanium reacts with oxygen to form 5.40 g of the metal oxide. What is the formula of the oxide?

45. Example 6.12 Solution: Step 1: Convert percentages to grams
Skip, since masses are given
Given: Ti = 3.24 g compound = 5.40 g
Therefore mass O = 5.40 g – 3.24 g = 2.16 g
Step 2: Convert masses to moles

46. Example 6.12 Cont’d Step 3: Write a pseudoformula in the form TixOy
Step 4: Find the mole ratio by dividing by the smallest number of moles in Ti0.0677O0.135
Ti0.0677O0.135

47. Practice 1—Determine the Empirical Formula of Tin Fluoride, which Contains 75.7% Sn and the Rest Fluorine.
Practice 2—Determine the Empirical Formula of a compound which Contains 69.94% Fe and the Rest Oxygen.

48. Practice 1—Determine the Empirical Formula of Stannous Fluoride, which Contains 75.7% Sn (118.70) and the Rest Fluorine (19.00).
Answer: SnF2

49. Tro's "Introductory Chemistry", Chapter 6
Answer: SnF2

50. Practice 2—Determine the Empirical Formula of a compound which Contains % Fe (55.85) and the Rest Oxygen (16.00).
Answer: Fe2O3

51. Answer: Fe2O3

52. Empirical Formula and Molecular Formula.
Name
Empirical
Formula
Molecular
Molar
Mass, g
Glyceraldehyde
CH2O
C3H6O3 90
Erythrose
C4H8O4
120
Arabinose
C5H10O5
150
Glucose
C6H12O6
180

53. Molecular Formulas
The molecular formula is a multiple of the empirical formula.
To determine the molecular formula, you need to know the empirical formula and the molar mass of the compound.
Molar massreal formula
Molar massempirical formula
= Factor used to multiply subscripts

54. Example 6.13—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8.
Determine the empirical formula.
May need to calculate it as previous.
C5H8
Determine the molar mass of the empirical formula.
5 C = 60.05, 8 H = 8.064
C5H8 = g/mol

55. Tro's "Introductory Chemistry", Chapter 6
Example 6.13—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8, Continued.
Divide the given molar mass of the compound by the molar mass of the empirical formula.
Round to the nearest whole number.
Tro's "Introductory Chemistry", Chapter 6

56. Tro's "Introductory Chemistry", Chapter 6
Example 6.13—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8, Continued.
Multiply the empirical formula by the factor above to give the molecular formula.
(C5H8)3 = C15H24
Tro's "Introductory Chemistry", Chapter 6

57. Practice—Benzopyrene has a Molar Mass of 252 g and an Empirical Formula of C5H3. What is its Molecular Formula? (C = 12.01, H=1.01)
Answer: C20H12

58. Tro's "Introductory Chemistry", Chapter 6
Example 6.14—Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g and is 74.0% C, 8.7% H, and the Rest N. (C=12.01, H=1.01, N=14.01)
Tro's "Introductory Chemistry", Chapter 6

59. Example 6.14—Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g and is 74.0% C, 8.7% H, and the Rest N, Continued
Given: 74.0% C, 8.7% H, {100 – ( )} = 17.3% N 
in 100 g nicotine there are 74.0 g C, 8.7 g H, and 17.3 g N.
Find: CxHyNz
Conversion Factors:
1 mol C = g; 1 mol H = 1.01 g; 1 mol N = g
Solution Map:
g C
mol C
whole
number
ratio
mole
ratio
pseudo-
formula
empirical
formula
g H
mol H
g N
mol N

60. Example 6.14—Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g and is 74.0% C, 8.7% H, and the Rest N, Continued.
Apply solution map:
C5 = 5(12.01 g) = g
N1 = 1(14.01 g) = g
H7 = 7(1.01 g) = g
C5H7N = g
C6.16H8.6N1.23
{C5H7N} x 2 = C10H14N2

61. Recommended Study Problems Chapter 6
NB: Study problems are used to check the student’s understanding of the lecture material. Students are EXPECTED TO BE ABLE TO SOLVE ALL THE SUGGESTED STUDY PROBLEMS.
If you encounter any problems, please talk to your professor or seek help at the HACC-Gettysburg learning center.
Questions from text book Chapter 6, p 189
3, 7, 11, 17, 19, 25, 27, 37, 51, 57, 59, 69, 71, 73, 79, 85, 87, 89, 97, 102, 117, 116
ANSWERS
-The answers to the odd-numbered study problems are found at the back of your textbook