1. Chapter 4: Chemical Composition
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2. Questions for Consideration
How can we describe the mass composition of elements in a compound?
How can we determine the number of atoms in a given mass of a material? The number of molecules? The number of formula units?
How can we use the masses of elements in a compound to determine its chemical formula?
How can we express the composition of a solution?

3. Chapter 4 Topics: Percent Composition Mole Quantities
Determining Empirical and Molecular Formulas
Chemical Composition of Solutions

4. Chapter 4 Math Toolbox: 4.1 Mole Quantities
This math toolbox gives a concise overview of the mole-quantity calculations that are introduced in more detail throughout this chapter.

5. 4.1 Percent Composition
We have used chemical formulas to express the composition of compounds, but where do these formulas come from?
We cannot directly determine a chemical formula. However, we can measure the mass of each element in a sample of a compound by using appropriate analytical techniques.
Because the mass of each element in a compound will vary from one sample of the compound to another, we must have some expression of composition that is the same for all samples.

6. Percent Composition
A convenient method for expressing composition is percent composition by mass.
For any element, E, in a compound, the percent composition by mass is given by the following equation:

7. Moles, Masses, and Particles
How can we describe the composition of a compound if we know the mass of the elements in the compound?
A 3.67-g sample of the mineral chalcopyrite was determined to contain 1.27 g Cu, 1.12 g Fe, and 1.28 g S.
What is the mass percent of each element in this compound?

8. Percent Composition
The percent composition for all the elements present in a compound must add up to 100%.
Figure 4.5

9. Composition of Chalcopyrite
Would this mass % differ for a different sample of this mineral?
If you had a 100-gram sample, what mass of copper would it contain?
Figure 4.5

10. Activity: Percent Composition
A sample of a copper compound weighs 1.63 g and contains 1.30 g of Cu. The other element in the compound is oxygen. What is the percent composition of this compound?
By difference, the sample contains 0.33 g O. Applying the formula for percent composition gives:

11. Activity: Percent Composition
What are the percent iron and the percent sulfur in an 8.33-g sample of chalcopyrite that contains 2.54 g Fe and 2.91 g S?

12. Activity Solution: Percent Composition
What are the percent iron and the percent sulfur in an 8.33-g sample of chalcopyrite that contains 2.54 g Fe and 2.91 g S?

13. Activity: Percent Composition
A 4.55-g sample of limestone (CaCO3) contains 1.82 g of calcium. What is the percent Ca in limestone?

14. Activity Solutions: Percent Composition
A 4.55-g sample of limestone (CaCO3) contains 1.82 g of calcium. What is the percent Ca in limestone?

15. 4.2 Mole Quantities
When working with amounts of a substance on a macroscopic scale, we cannot simply count atoms or molecules. There are too many. Instead, we use the moles of atoms, which are related by Avogadro’s number:
1 mole = × 1023 particles
1 mole C = × 1023 carbon atoms
1 mole H2S = × 1023 H2S molecules
1 mol Cu2O = × 1023 Cu2O formula units

16. The Mole
The mole unit acts as a bridge between the microscopic world and the macroscopic world.
One mole of substance contains as many basic particles (atoms, molecules, or formula units) as there are atoms in exactly 12 g of carbon-12.
One mole of a substance contains × 1023 particles (molecules, atoms, ions, formula units, etc.)
This number is called Avogadro’s number.

17. Moles of Various Elements and Compounds
Figure 4.8

18. Activity: Atoms in H2S How many sulfur atoms are in 1 mol of H2S?
How many hydrogen atoms are in 1 mol of H2S?
Figure 4.6

19. Molar Mass The Mass of 1 Mole
Avogadro’s number has been defined so that the mass of 1 mol of C-12 has a mass of exactly 12 grams.
This means that the average mass of an atom of any substance in amus is the same numerical value as the mass of 1 mole of that substance in grams (molar mass).
The molar mass of carbon is g/mol.
The molar mass of oxygen is g/mol
The molar mass of CO2 is:
(16.00) = g/mol
We’ll use
4 sig. figs

20. Activity: Molar Mass Complete the table. Element or Compound
Atomic Mass
Molar Mass H O Na
Ca(ClO3)2
H2O
NaCl

21. Activity Solutions: Molar Mass
Element or Compound
Atomic Mass
Molar Mass H
1.008 amu
1.008 g/mol O
16.00 amu
16.00 g/mol Na
22.99 amu
22.99 g/mol
Ca(ClO3)2
amu
g/mol
H2O
18.02 amu
18.02 g/mol
NaCl
58.44 amu
58.44 g/mol

22. Molar Mass 1 gram of copper or 1 gram of gold?
Which contains the greatest number of atoms?
1 mole of copper or 1 mole of gold?
1 gram of copper or 1 gram of gold?
Answer: Because 1 mole is a fixed quantity of particles (6.022 × 1023 ), 1 mole of copper contains the same number of atoms as 1 mole of gold.
Answer: Because each copper atom has less mass than each gold atom, 1 gram of copper contains more atoms than 1 gram of gold.

23. Converting Between Grams and Moles
Convert 10.0 grams of CO2 to moles.
Convert 0.50 mol CO2 to grams.
Figure from p. 132

24. Activity: Converting Between Grams and Moles
Convert 10.0 g O2 to moles.
Figure from p. 132

25. Grams  Moles  Particles
Once we know the number of moles of a substance, we can use Avogadro’s number (6.022×1023) to determine the number of particles in that sample of the substance.
1 mole = × 1023 particles
Figure from p. 134

26. Activity: Number of Particles
What mass of MgCl2 contains × 1023 Cl ions?
The molar mass of MgCl2 is g/mol.
Figure from p. 134

27. Extra Activity: Conversions with Molar Mass
How many moles of aspartame (C14H18N2O5) are found in 40.0 mg of aspartame? How many molecules of aspartame are found in this mass?

28. Extra Activity Solution: Conversions with Molar Mass
How many moles of aspartame (C14H18N2O5) are found in 40.0 mg of aspartame? How many molecules of aspartame are found in this mass?
We first need to convert from mg to g:
Next, we need to find the molar mass of C14H18N2O5:

29. Extra Activity Solution: Conversions with Molar Mass
How many moles of aspartame (C14H18N2O5) are found in 40. mg of aspartame? How many molecules of aspartame are found in this mass?
Next, convert from grams to moles:
Finally, we convert from moles to molecules:

30. Extra Activity: Conversions with Molar Mass
If one aspirin tablet contains g of acetylsalicylic acid (C9H8O4), then how many molecules of acetylsalicylic acid are in 2 aspirin tablets?

31. Extra Activity Solution: Conversions with Molar Mass
If one aspirin tablet contains g of acetylsalicylic acid (C9H8O4), then how many molecules of acetylsalicylic acid are in 2 aspirin tablets?

32. 4.3 Determining Empirical and Molecular Formulas
The formula for a substance also tells us about the composition of a compound:
A formula unit for an ionic compound tells us the ratio of ions of different elements in the compound. (MgCl2 has a 1:2 ratio of Mg2+ to Cl.)
A molecular formula tells the number of atoms of each element in a molecule and the atom ratio. (CO2 has a 1:2 ratio of C to O atoms.)

33. Empirical and Molecular Formulas
Empirical formula
Expresses the simplest ratios of atoms in a compound
Written with the smallest whole-number subscripts
Molecular formula
Expresses the actual number of atoms in a compound
Can have the same subscripts as the empirical formula or some multiple of them

34. Empirical Formulas What is the same about these two compounds?
Figure 4.13

35. Empirical Formulas
The empirical formula of a substance is the ratio of atoms of different elements, in terms of the smallest whole numbers.
What are the empirical formulas of these two compounds?
What are the empirical formulas of H2O2 and H2O? Are they different compounds?
Figure 4.13

36. What is the empirical formula for copper(II) oxide?
Figure 4.12

37. Empirical or Molecular Formulas

38. Activity: Empirical Formula
Determine the empirical formulas of these substances. For which substances is the empirical formula the same as the molecular formula?
Figure 4.14

39. Activity Solutions: Empirical Formula
Determine the empirical formulas of these substances. For which substances is the empirical formula the same as the molecular formula?
same
C2HCl
same
NO2
HCO2
CH2O
NH2
Figure 4.14
CH3
CH2
P2O3

40. Determining Empirical Formulas from % Composition
If we know the masses of the elements in a compound, or its percent composition, we can determine its mole ratio, and therefore the compound’s empirical formula.
Consider the compound commonly called chalcopyrite. Any sample will have the following % composition:
Figure 4.5

41. Determining Empirical Formulas
Since the percent composition does not change from sample to sample, assume any size sample. The most convenient is 100 grams so % value = mass value.
Convert grams to moles for each element using its molar mass as a conversion factor.
Without changing the relative amounts, change moles to whole numbers. Do this by dividing all by the same smallest value. If all do not convert to whole numbers, multiply to get whole numbers.

42. Determining Empirical Formulas
grams chalcopyrite contains:
30.5 g Fe
34.6 g Cu
34.9 g S
2. mol Fe = (30.5g Fe)(1mol/55.85 g) = mol Fe
mol Cu = (34.6g Cu)(1mol/63.55 g) = mol Cu
mol S = (34.9g S)(1mol/32.07 g) = mol S
3. ( mol Fe)/( ) = mol Fe
( mol Cu)/( ) = mol Cu
(1.088 mol S)/( ) = mol S
Figure 4.5
FeCuS2

43. Steps for Determining Empirical Formula
Figure from p. 139

44. Activity: Empirical Formula from Percent Composition
A compound was determined to have the following percent composition:
50.0% sulfur
50.0% oxygen
What is the empirical formula for the compound?
SO2

45. Activity: Determining Empirical Formulas
Determine the empirical formula for the mineral covellite, which has the percent composition 66.5% Cu and 33.5% S.

46. Activity Solution: Determining Empirical Formulas
Determine the empirical formula for the mineral covellite, which has the percent composition 66.5% Cu and 33.5% S.
First, reassign the percentages to units of grams: g Cu and 33.5 g S.
Then, convert to moles and divide both numbers by the lowest number.
The whole numbers then become our subscripts.
The empirical formula is therefore: CuS

47. Activity: Determining Empirical Formulas
Shattuckite is a fairly rare copper mineral. It has the composition 48.43% copper, 17.12% silicon, 34.14% oxygen, and 0.31% hydrogen. Calculate the empirical formula of shattuckite.

48. Activity Solutions: Determining Empirical Formulas
Shattuckite is a fairly rare copper mineral. It has the composition 48.43% copper, 17.12% silicon, 34.14% oxygen, and 0.31% hydrogen. Calculate the empirical formula of shattuckite.
Therefore, the empirical formula is: Cu5Si4O14H2

49. Molecular Formulas from Empirical Formulas
Benzene and acetylene have the same empirical formulas but different molecular formulas.
How much greater in mass is benzene than acetylene?
How much greater in mass is each of these than the empirical formula?
Figure 4.13

50. Empirical and Molecular Formulas
What are the ratios of masses for the molecular and empirical formulas?

51. Activity: Molecular Formulas from Empirical Formulas
A compound was determined to have an empirical formula of CH2. Its molar mass was determined to be g/mol. What is the molecular formula for this compound?
The mass of the empirical formula is g/mol, so the ratio of masses is The ratio of the formulas therefore is 3, giving a molecular formula of (CH2)3 or C3H6.

52. Molecular Formulas
To determine a molecular formula, the problem must give a piece of experimental data, such as a molar mass, MM.
To find the molecular formula:
Find the empirical formula first.
Divide the empirical formula’s molar mass into the experimental molar mass (which is given).
Figure 4.15

53. Activity: Molecular Formulas
Potassium persulfate is a strong bleaching agent. It has a percent composition of 28.93% potassium, 23.72% sulfur, and 47.35% oxygen. The experimental molar mass of g/mol. What are the empirical and molecular formulas of potassium persulfate?

54. Activity Solution: Molecular Formulas
Potassium persulfate is a strong bleaching agent. It has a percent composition of 28.93% potassium, 23.72% sulfur, and 47.35% oxygen. The experimental molar mass of g/mol. What are the empirical and molecular formulas of potassium persulfate?
First, find the empirical formula:
Thus, the empirical formula is KSO4.

55. Activity Solution: Molecular Formulas
Potassium persulfate is a strong bleaching agent. It has a percent composition of 28.93% potassium, 23.72% sulfur, and 47.35% oxygen. The experimental molar mass of g/mol. What are the empirical and molecular formulas of potassium persulfate?
To find the molecular formula:
Calculate the MM of KSO4 = (1 mol K × g/mol K) + (1 mol S × g/mol S) + (4 mol O × g/mol O) = g/mol KSO4
Thus, the molecular formula is K2S2O8.

56. Determining Percent Composition from Empirical or Molecular Formula
If you know the formula for a compound, you can determine the mass percent composition.
Assume you have 1 mole of compound, and convert moles of the element and compound to grams.
Mass Percent Composition
Mass Ratio
Empirical or Molecular Formula
Atom or Mole Ratio

57. Determining Percent Composition
Assume 1 mole of compound, and remember the mass of 1 mole is the molar mass.
Figure from p. 144

58. Activity: Determining Percent Composition
What is the percent sodium in Na2O?

59. Activity: Percent Composition
Which of these compounds has the greater percentage of Cu?
Cuprite,
Cu2O
Answer: cuprite
Chalcocite,
Cu2S
Figure 4.16

60. 4.4 Chemical Composition of Solutions
A solution is a homogeneous mixture.
This is a solution being prepared by adding the CuSO4 (solute) to water (solvent).
The composition of the solution depends on the relative amounts of the solute and solvent.
Figure 4.17

61. Concentration
Which aqueous CuSO4 solution has the greatest concentration (most concentrated)? Which is most dilute?
Figure 4.18

62. Percent by Mass of Solute
Solution concentration is often expressed as the mass percent of solute:

63. Activity: Percent by Mass of Solute
What is the mass percent of NaCl in a solution that is prepared by adding 10.0 g NaCl to 50.0 g water?

64. Molarity (M)
Another common way to express the concentration of a solution is in molarity units:

65. Activity: Preparing a CuSO4 Solution
6.25 grams ( mol) of CuSO45H2O is added to a 250-mL volumetric flask.
Water is added to the mark so that the total volume is mL.
What is the molarity of this solution?
Figure 4.19

66. Activity: Molarity
How many moles of NaCl are in 1.85 L of a 0.25 M NaCl solution?
moles NaCl = 1.85 L × 0.25 mol/L = 0.46 mol

67. Activity: Solution Concentration
Bluestone is copper(II) sulfate pentahydrate, CuSO4•5H2O, with a molar mass of g/mol. A sample of pond water was found to have a concentration of 6.2  105 M copper(II) sulfate. If the pond has a volume of 1.8  107 L, then what mass of bluestone did the farmer add to the pond as an algicide?

68. Activity Solution: Solution Concentration
Bluestone is copper(II) sulfate pentahydrate, CuSO4•5H2O, with a molar mass of g/mol. A sample of pond water was found to have a concentration of 6.2  105 M copper(II) sulfate. If the pond has a volume of 1.8 x 107 L, then what mass of bluestone did the farmer add to the pond as an algicide?

69. Dilution
Suppose you want to dilute a 0.25 M solution to a concentration of M. What are some ways to do this?

70. Dilution
Figure 4.21

71. Describe this process
Figure from p. 163

72. Dilution Molesinitial= Molesfinal MinitialVinitial = MfinalVfinal
Moles = Molarity × Volume
Moles = mol/L × L
MinitialVinitial = MfinalVfinal

73. Activity: Dilution MinitialVinitial = MfinalVfinal
What is the concentration of a solution prepared by adding water to 25.0 mL of 6.00 M NaOH to a total volume of mL?

74. Activity: Dilution
If 42.8 mL of 3.02 M H2SO4 solution is diluted to a final volume mL, what is the molarity of the diluted solution of H2SO4?

75. Activity Solution: Dilution
If 42.8 mL of 3.02 M H2SO4 solution is diluted to a final volume mL, what is the molarity of the diluted solution of H2SO4?
MconVcon = MdilVdil
3.02 M × 42.8 mL = Mdil× mL

76. Activity: Dilution
What is the concentration of a solution prepared by diluting 35.0 mL of M KBr to mL?

77. Activity Solution: Dilution
What is the concentration of a solution prepared by diluting 35.0 mL of M KBr to mL?
MconVcon = MdilVdil
0.150 M × 35.0 mL = Mdil × mL