1. Quantitative Composition Chapter 7
Larry Emme
Chemeketa Community College

2. Counting by Weighing

5. Please send an order of 1,200,000 nuts!

7. Suppose I know that the mass of 1 nut is 2.5 grams

9. How many nuts are in 1.5109 grams?

12. Suppose I know that the mass of 1 Walnut is 5.0 grams
Remember the mass of 1 Hazelnut is 2.5 grams
The mass ratio of 1 Walnut to 1 Hazelnut is:

13. Equal numbers of nuts would always ratio down to:
If I weigh out 10 Walnuts and also
10 Hazelnuts the mass ratio would be:
Equal numbers of nuts would always ratio down to:

14. Suppose I have two large piles of nuts and the weight of Walnuts is twice the weight of Hazelnuts. What’s true?
Since the mass ratio of Walnuts to Hazelnuts is 2 to 1, the same number of nuts must be in both piles!

15. Suppose I have 240,000 Hazelnuts that weigh 600 kg
Suppose I have 240,000 Hazelnuts that weigh 600 kg. How many Walnuts are needed to weigh 1200 kg ?
Since the mass ratio of Walnuts to Hazelnuts is to 600 or 2 to 1, you will need 240,000 Walnuts!

16. Counting by Weighing on the Periodic Table

17. The standard to which the masses of all other atoms are compared to was originally chosen to be the most abundant isotope of hydrogen.

18. A mass of exactly 1 atomic mass units (amu) was assigned to this isotope.

19. Electric charge and mass of electrons, protons, & neutrons
Particle
symbol
Actual charge (coulombs)
Relative charge
Actual mass (g)
Relative mass (amu)
Electron e–
–1.602  10–19 –1
9.11  10–28
Proton p
 10–19 +1
1.67  10–24 1
Neutron n
1.68  10–24

20. The standard to which the masses of all other atoms are compared is now defined to be the most abundant isotope of carbon.

21. A mass of exactly 12 atomic mass units (amu) is assigned to this isotope.

22. The most abundant isotope of magnesium is:

23. A mass of exactly 24 atomic mass units (amu) would be assigned to this isotope.

24. What is the mass ratio of one magnesium atom to one carbon atom?

25. Suppose I have 3. 011020 of 12C atoms that weigh 0. 006 g
Suppose I have 3.011020 of 12C atoms that weigh g. How many 24Mg atoms are needed to weigh g ?
Since the mass ratio of 24Mg to 12C is to 600 or 2 to 1, you will need 240,000 of 24Mg atoms!

26. Balances in the lab room weigh in metric ,i. e. , grams not in amu’s
Balances in the lab room weigh in metric ,i.e., grams not in amu’s. So how do we change the number on the periodic table from amu’s to grams?
We need to know how many atoms of each element we need to obtain the amu weight in grams.
This amount (a big pile) would convert amu’s to grams for each element.

27. Suppose we tare a beaker and continue to add carbon atoms until the mass reads 12.01g (to account for isotopes of carbon).
0.000
The amount of carbons would be a large pile, but we would eventually get there.
12.01

28. How many magnesium atoms would you need?
Suppose we tare another beaker and add magnesium atoms until the mass reads 24.3 g (to account for isotopes of magnesium).
0.000
How many magnesium atoms would you need?
24.3
To get the 2 to 1 mass ratio of Mg to C, you would need exactly the same number!

29. Other units for counting:
This big pile of atoms would convert any amu value on the periodic table to grams.
Let’s define this number to be a “pile”. We can get fancy and use a Greek name that means the same thing, “Mole”
Other units for counting:
1 dozen = 12 things
1 score = 20 things
1 gross = 144 things
1 ream = 500 things
1 mole = ? things

30. Amadeo (Amedeo) Avogadro

31. 1 pile = 1 mole (in Greek)
1 mole = 6.021023 objects

32. 6.021023
is a very
LARGE
number

33. 6.021023 is
Avogadro’s Number
number

35. 1 mole of any element contains 6.021023 particles of that substance.

36. The atomic weight in grams of any element23 contains 1 mole of atoms
and is defined to be the mole weight

37. You might ask why Avogadro’s Number is approximate and not an exact counting number as we have for 1 dozen or 1 ream. It is approximate for two good reasons: First, we cannot actually see the atoms to count them. Second, even if we could somehow see them, counting up to this huge number would take longer than we might guess – see the next slide:

38. Stacking 6.02 × 1023 papers on top of each other would take you to the sun and back 1,851,083,193 times. 1 mole of marshmallows would be enough marshmallows to make a 19 km (12 mi) thick layer of marshmallows covering the entire face of the Earth. 1 mole of popcorn kernels could be spread uniformly over the USA if the thickness of the layer was about 14 km (9 mi). 1 mole of donut holes would cover the earth and be 8 km (5 mi) deep. 1 mole of blood cells would be more than the total number of blood cells found in every human on earth. 1 mole of pennies could be distributed to all the currently-living people of the world so that they could spend a million dollars per hour every hour (day and night) for the rest of their lives.

39. A Mole Poem
Say you had a mole of paper and stacked it towards the sky, Paper’s really thin, but that pile would get so high, I’d reach up into outer space, in fact I’d think you find, I’d go up to the moon and back, eighty billion times.

40. Examples

41. Species Quantity Number of H atoms
1 dozen 12

42. Species Quantity Number of H atoms
1 mole
6.021023

43. Species Quantity Number of H2 molecules
1 gross
144

44. Species Quantity Number of H2 molecules
6.021023

45. Species Quantity Number of Na atoms
1 ream
500

46. Species Quantity Number of Na atoms
1 mole
6.021023

47. Species Quantity Number of Fe atoms
1 mole
6.021023

48. Species Quantity Number of C6H6 molecules
6.021023

49. 1 mole of atoms =
6.021023 atoms
1 mole of molecules =
6.021023 molecules
1 mole of ions =
6.021023 ions

50. Summary The mole weight of an element is its atomic weight in grams.
It contains 6.021023 atoms (Avogadro’s number) of the element.

51. Element
Atomic mass
Mole weight
Number of atoms H
1.008 amu
1.008 g
6.021023 Mg
24.31 amu
24.31 g Na
22.99 amu
22.99 g

52. Problems

54. Weight (mass)
Numbers
Moles 
Periodic Table
Avogadro’s Number
6.021023

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

56. How many iron atoms are contained in 40.0 grams of iron?
Atomic weight iron = 55.85
Conversion sequence is:
grams Fe → moles Fe → atoms Fe

57. What is the mass of 3.01 x 1023 atoms of sodium (Na)?
Mole weight Na = g
Conversion sequence:
atoms Na → moles Na → grams Na

58. How many oxygen atoms are present in 2.00 moles of oxygen molecules?
Conversion sequence:
moles O2 → molecules O2 → atoms O

59. Mole Weight of Compounds

60. The mole weight of a compound can be determined by adding the mole weights of all of the atoms in its formula.

61. Calculate the mole weight 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

62. Calculate the mole weight of lithium perchlorate.
LiClO4
1 Li = 1(6.94 g) = g
1 Cl = 1(35.45 g) = g
4 O = 4(16.00 g) = g g

63. Calculate the mole weight of ammonium phosphate.
(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

64. Solids that contain water as part of their crystalline structure are known as hydrates.
Water in a hydrate is known as water of hydration or water of crystallization.

65. Formulas of hydrates are written by first writing the formula for the anhydrous compound and then adding a dot followed by the number of water molecules present.
6H2O
CoCl2

66. Sodium acetate trihydrate
Calculate the mole weight of NaC2H3O2 · 3 H2O
Sodium acetate trihydrate
1 Na = 1(22.99 g) = g
2 C = 2(12.01 g) = g
9 H = 9(1.01 g) = g
5 O = 5(16.00 g) = g g

67. Avogadro’s Number of Particles
6.02 x 1023 Particles
1 MOLE
Mole Weight

68. Avogadro’s Number of Ca atoms
6.02 x 1023 Ca atoms
1 MOLE Ca
g Ca

69. Avogadro’s Number of H2O molecules
6.02 x 1023 H2O molecules
1 MOLE H2O
18.02 g H2O

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

71. Calculate the mole weight of 1 mole of H atoms.
1 H = 1(1.01 g) = 1.01 g
Calculate the mole weight of 1 mole of H2 molecules.
2 H = 2(1.01 g) = 2.02 g

72. Problems

73. How many moles of benzene, C6H6, are present in 390.0 grams of benzene?
The mole weight of C6H6 is g.
Conversion sequence: grams C6H6 → moles C6H6

74. How many grams of (NH4)3PO4 are contained in 2.52 moles of (NH4)3PO4?
The mole weight of (NH4)3PO4 is g.
Conversion sequence:
moles (NH4)3PO4 → grams (NH4)3PO4

75. Conversion sequence: g N2 → moles N2 → molecules N2
56.04 g of N2 contains how many N2 molecules?
The mole weight of N2 is g.
Conversion sequence: g N2 → moles N2 → molecules N2
Use the conversion factors

76. Percent Composition of Compounds

77. 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

78. Percent Composition From Formula

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

81. Step 1 Calculate the mole weight of H2S.
Calculate the percent composition of hydrosulfuric acid H2S(aq) .
Step 1 Calculate the mole weight of H2S.
2 H = 2 (1.01 g) = g
1 S = 1 (32.07 g) = g
34.09 g

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

83. Percent Composition From Experimental Data

84. 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.

85. 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

86. 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%

87. Empirical Formula versus Molecular Formula

88. 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.

89. 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.

90. Examples

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

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

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

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

97. Calculating Empirical Formulas

98. The analysis of a compound shows that it contains 27% carbon and 73% oxygen. Calculate the empirical formula for this substance.
C___O___

99. Step 1 Assume a definite starting quantity (usually 100
Step 1 Assume a definite starting quantity (usually g) of the compound, if 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 mole weight.

100. 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.

101. The analysis of a compound shows that it contains 27% carbon and 73% oxygen. Calculate the empirical formula for this substance.
C___O___
27g
73g
12.0 g/mole
16.0 g/mole
C___O___
2.3moles
4.6moles
2.3moles
2.3moles

102. The analysis of a compound shows that it contains 27% carbon and 73% oxygen. Calculate the empirical formula for this substance.
C___O___ 1 2

103. 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2O1.52
Fe2O3

104. 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.

105. Problems

106. 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___C___O___
56.58g
8.68g
34.73g

107. 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 2 Convert the grams of each element to moles.
K___C___O___
56.58g
8.68g
34.73g
39.10 g/mole
12.01 g/mole
16.00 g/mole

108. Empirical formula so far:
K1.447C0.723O2.171

109. 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 3 Divide each number of moles by the smallest value.
K___C___O___
1.447
0.723
2.171
0.723
0.723
0.723

110. 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 3 Divide each number of moles by the smallest value.
K___C___O___ 2 1 3
The simplest ratio of K:C:O is 2:1:3
Empirical formula K2CO3

111. 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___O___
25.94g
74.06g

112. 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.
N___O___
25.94g
74.06g
14.01 g/mole
16.00 g/mole

113. Empirical formula so far:
N1.852O4.629

114. 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.
N___O___
1.852
4.629
1.852
1.852

115. 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.
N___O___
1.000
2.500
This is not a ratio of whole numbers.

116. 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

117. Calculating the Molecular Formula from the Empirical Formula

118. The molecular formula can be calculated from the empirical formula if the mole weight 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.

119. What is the molecular formula of a compound which has an empirical formula of CH2 and a mole weight 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

120. A compound made of 30. 4% N and 69
A compound made of 30.4% N and 69.6% O has a mole weight of 138 grams/mole. Find the molecular formula.
Old Way:

121. A compound made of 30. 4% N and 69
A compound made of 30.4% N and 69.6% O has a mole weight of 138 grams/mole. Find the molecular formula.
New Way:
N3O6

122. The End