1. Quantitative Composition of Compounds Chapter 7
Hein and Arena
Eugene Passer Chemistry Department Bronx Community College
© John Wiley and Sons, Inc
Version 2.0
12th Edition

2. Chapter Outline 7.1 The Mole
7.4 Empirical Formula versus Molecular Formula
7.2 Molar Mass of Compounds
7.5 Calculating Empirical Formulas
7.3 Percent Composition of Compounds
7.6 Calculating the Molecular Formula from the Empirical Formula

3. 7.1 The Mole

4. The mass of a single atom is too small to measure on a balance.
mass of hydrogen atom = x g

5. This is an
infinitesimal
mass
1.673 x g

6. MOLE
Chemists require a unit for counting which can express large numbers of atoms using simple numbers.
Chemists have chosen a unit for counting atoms.
That unit is the

7. 1 mole = x 1023 objects

8. 6.022 x 1023
is a very
LARGE
number

9. 6.022 x 1023 is
Avogadro’s Number
number

10. 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.

11. 1 mole of any element contains 6.022 x 1023
particles of that substance.

12. The atomic mass in grams
of any element23
contains 1 mole of atoms.

13. This is the same number of particles 6
This is the same number of particles x as there are in exactly 12 grams of

14. Examples

15. Species Quantity Number of H atoms
1 mole
6.022 x 1023

16. Species Quantity Number of H2 molecules
6.022 x 1023

17. Species Quantity Number of Na atoms
1 mole
6.022 x 1023

18. Species Quantity Number of Fe atoms
1 mole
6.022 x 1023

19. Species Quantity Number of C6H6 molecules
6.022 x 1023

20. 1 mol of atoms =
6.022 x 1023 atoms
1 mol of molecules =
6.022 x 1023 molecules
1 mol of ions =
6.022 x 1023 ions

21. The molar mass of an element is its atomic mass in grams.
It contains x 1023 atoms (Avogadro’s number) of the element.

22. Element
Atomic mass
Molar mass
Number of atoms H
1.008 amu
1.008 g
6.022 x 1023 Mg
24.31 amu
24.31 g Na
22.99 amu
22.99 g

23. Problems

24. How many moles of iron does 25.0 g of iron represent?
Atomic mass iron = 55.85
Conversion sequence: grams Fe → moles Fe
Set up the calculation using a conversion factor between moles and grams.

25. How many iron atoms are contained in 25.0 grams of iron?
Atomic mass iron = 55.85
Conversion sequence: grams Fe → atoms Fe
Set up the calculation using a conversion factor between atoms and grams.

26. What is the mass of 3.01 x 1023 atoms of sodium (Na)?
Molar mass Na = g
Conversion sequence: atoms Na → grams Na
Set up the calculation using a conversion factor between grams and atoms.

27. What is the mass of 0.365 moles of tin?
Atomic mass tin = 118.7
Conversion sequence: moles Sn → grams Sn
Set up the calculation using a conversion factor between grams and atoms.

28. How many oxygen atoms are present in 2.00 mol of oxygen molecules?
Two conversion factors are needed:
Conversion sequence: moles O2 → molecules O → atoms O

29. 7.2 Molar Mass of Compounds

30. The molar mass of a compound can be determined by adding the molar masses of all of the atoms in its formula.

31. Calculate the molar mass of C2H6O.
2 C = 2(12.01 g) = g
6 H = 6(1.01 g) = g
1 O = 1(16.00 g) = g
46.08 g

32. Calculate the molar mass of LiClO4.
1 Li = 1(6.94 g) = 6.94 g
1 Cl = 1(35.45 g) = g
4 O = 4(16.00 g) = g g

33. Calculate the molar mass of (NH4)3PO4 .
3 N = 3(14.01 g) = g
12 H = 12(1.01 g) = g
1 P = 1(30.97 g) = g
4 O = 4(16.00 g) = g g

34. Avogadro’s Number of Particles
6 x 1023 Particles
1 MOLE
Molar Mass

35. Avogadro’s Number of Ca atoms
6 x 1023 Ca atoms
1 MOLE Ca
g Ca

36. Avogadro’s Number of H2O molecules
6 x 1023 H2O molecules
1 MOLE H2O
18.02 g H2O

37. 1 molar mass HCl molecules
These relationships are present when hydrogen combines with chlorine. H Cl
HCl
6.022 x 1023 H atoms
6.022 x 1023 Cl atoms
6.022 x 1023 HCl molecules
1 mol H atoms
1 mol Cl atoms
1 mol HCl molecules
1.008 g H
35.45 g Cl
36.46 g HCl
1 molar mass H atoms
1 molar mass Cl atoms
1 molar mass HCl molecules

38. In dealing with diatomic elements (H2, O2, N2, F2, Cl2, Br2, and I2), distinguish between one mole of atoms and one mole of molecules.

39. Calculate the molar mass of 1 mole of H atoms.
1 H = 1(1.01 g) = 1.01 g
Calculate the molar mass of 1 mole of H2 molecules.
2 H = 2(1.01 g) = 2.02 g

40. Problems

41. How many moles of benzene, C6H6, are present in 390.0 grams of benzene?
The molar mass of C6H6 is g.
Conversion sequence: grams C6H6 → moles C6H6

42. How many grams of (NH4)3PO4 are contained in 2.52 moles of (NH4)3PO4?
The molar mass of (NH4)3PO4 is g.
Conversion sequence: moles (NH4)3PO → grams (NH4)3PO4

43. Conversion sequence: g N2 → moles N2 → molecules N2
56.04 g of N2 contains how many N2 molecules?
The molar mass of N2 is g.
Conversion sequence: g N2 → moles N2 → molecules N2
Use the conversion factors

44. 56.04 g of N2 contains how many nitrogen atoms?
The molar mass of N2 is g.
Conversion sequence: g N2 → moles N2 → molecules N2
→ atoms N
Use the conversion factor

45. 7.3 Percent Composition of Compounds

46. Percent composition of a compound is the mass percent of each element in the compound.
H2O
11.19% H by mass
88.79% O by mass

47. Percent Composition From Formula

48. If the formula of a compound is known, a two-step process is needed to calculate the percent composition.
Step 1 Calculate the molar mass of the formula.
Step 2 Divide the total mass of each element in the formula by the molar mass and multiply by 100.

50. Step 1 Calculate the molar mass of H2S.
Calculate the percent composition of hydrosulfuric acid H2S.
Step 1 Calculate the molar mass of H2S.
2 H = 2 x 1.01g = g
1 S = 1 x g = g
34.09 g

51. Calculate the percent composition of hydrosulfuric acid H2S.
Step 2 Divide the mass of each element by the molar mass and multiply by 100.
H 5.93%
S 94.07%

52. Percent Composition From Experimental Data

53. Percent composition can be calculated from experimental data without knowing the composition of the compound.
Step 1 Calculate the mass of the compound formed.
Step 2 Divide the mass of each element by the total mass of the compound and multiply by 100.

54. Step 1 Calculate the total mass of the compound
A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition.
Step 1 Calculate the total mass of the compound
1.52 g N
3.47 g O
4.99 g
= total mass of product

55. A compound containing nitrogen and oxygen is found to contain 1
A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition.
Step 2 Divide the mass of each element by the total mass of the compound formed.
N 30.5%
O 69.5%

56. 7.4 Empirical Formula versus Molecular Formula

57. The empirical formula or simplest formula gives the smallest whole-number ratio of the atoms present in a compound.
The empirical formula gives the relative number of atoms of each element present in the compound.

58. The molecular formula is the true formula of a compound.
The molecular formula represents the total number of atoms of each element present in one molecule of a compound.

59. Examples

60. Smallest Whole Number Ratio
Molecular Formula
C2H4
CH2
Empirical Formula
C:H 1:2
Smallest Whole Number Ratio

61. Smallest Whole Number Ratio
Molecular Formula
C6H6
Empirical Formula CH
C:H 1:1
Smallest Whole Number Ratio

62. Smallest Whole Number Ratio
Molecular Formula
H2O2
Empirical Formula HO
Smallest Whole Number Ratio
H:O 1:1

64. Two compounds can have identical empirical formulas and different molecular formulas.

66. 7.5 Calculating Empirical Formulas

67. Step 1 Assume a definite starting quantity (usually 100
Step 1 Assume a definite starting quantity (usually g) of the compound, if the actual amount is not given, and express the mass of each element in grams.
Step 2 Convert the grams of each element into moles of each element using each element’s molar mass.

68. Step 3 Divide the moles of atoms of each element by the moles of atoms of the element that had the smallest value.
– If the numbers obtained are whole numbers, use them as subscripts and write the empirical formula.
– If the numbers obtained are not whole numbers, go on to step 4.

69. Step 4 Multiply the values obtained in step 3 by the smallest numbers that will convert them to whole numbers
Use these whole numbers as the subscripts in the empirical formula.
FeO1.5
Fe1 x 2O1.5 x 2
Fe2O3

70. The results of calculations may differ from a whole number.
If they differ ±0.1, round off to the next nearest whole number.
2.9
 3
Deviations greater than 0.1 unit from a whole number usually mean that the calculated ratios have to be multiplied by a whole number.

71. Some Common Fractions and Their Decimal Equivalents
Resulting Whole Number 1 2 3
Common Fraction
0.25
0.333…
Multiply the decimal equivalent by the number in the denominator of the fraction to get a whole number.
0.666…
0.5
0.75

72. Problems

73. The analysis of a salt shows that it contains 56. 58% potassium (K); 8
The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance.
Step 1 Express each element in grams. Assume 100 grams of compound.
K = g
C = g
O = g

74. C has the smallest number of moles
The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance.
Step 2 Convert the grams of each element to moles.
C has the smallest number of moles

75. C has the smallest number of moles
The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance.
Step 3 Divide each number of moles by the smallest value.
C has the smallest number of moles
The simplest ratio of K:C:O is 2:1:3
Empirical formula K2CO3

76. The percent composition of a compound is 25. 94% nitrogen (N), and 74
The percent composition of a compound is 25.94% nitrogen (N), and % oxygen (O). Calculate the empirical formula for this substance.
The percent composition of a compound is 25.94% nitrogen (N), and % oxygen (O). Calculate the empirical formula for this substance.
Step 1 Express each element in grams. Assume 100 grams of compound.
N = g
O = g

77. The percent composition of a compound is 25. 94% nitrogen (N), and 74
The percent composition of a compound is 25.94% nitrogen (N), and % oxygen (O). Calculate the empirical formula for this substance.
Step 2 Convert the grams of each element to moles.

78. The percent composition of a compound is 25. 94% nitrogen (N), and 74
The percent composition of a compound is 25.94% nitrogen (N), and % oxygen (O). Calculate the empirical formula for this substance.
Step 3 Divide each number of moles by the smallest value.
This is not a ratio of whole numbers.

79. The percent composition of a compound is 25. 94% nitrogen (N), and 74
The percent composition of a compound is 25.94% nitrogen (N), and % oxygen (O). Calculate the empirical formula for this substance.
Step 4 Multiply each of the values by 2.
N: (1.000)2 = 2.000
O: (2.500)2 = 5.000
Empirical formula N2O5

80. 7.6 Calculating the Molecular Formula from the Empirical Formula

81. The molecular formula can be calculated from the empirical formula if the molar mass is known.
The molecular formula will be equal to the empirical formula or some multiple, n, of it.
To determine the molecular formula evaluate n.
n is the number of units of the empirical formula contained in the molecular formula.

82. What is the molecular formula of a compound which has an empirical formula of CH2 and a molar mass of g?
Let n = the number of formula units of CH2.
Calculate the mass of each CH2 unit
1 C = 1(12.01 g) = 12.01g
2 H = 2(1.01 g) = g
14.03g
The molecular formula is (CH2)9 = C9H18

83. The End