1. Chemical Formula Relationships
Chapter 7
Chemical Formula Relationships
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

2. Section 7.1 The Number of Atoms in a Formula
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

3. Goal 1
Given the formula of a chemical compound (or a name from which the formula may be written), state the number of atoms of each element in the formula unit.
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

4. In writing the formula of a substance,
subscript numbers are used to indicate
the number of atoms or groups of atoms
of each element in the formula unit
However, subscript numbers
of 1 are omitted
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

5. ammonium carbonate: (NH4)2CO3
2 nitrogen atoms
8 hydrogen atoms
1 carbon atom
3 oxygen atoms
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6. Section 7.2 Molecular Mass and Formula Mass
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7. Goal 2
Distinguish among atomic mass, molecular mass, and formula mass.
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

8. Goal 3
Calculate the formula (molecular) mass of any compound whose formula is given (or known).
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

9. The average mass of all atoms of an element
Atomic mass:
The average mass of all atoms of an element
as they occur in nature. Based on 12C at a
mass of 12 amu, and weighted by
natural isotopic distribution.
(Section 5.5)
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

10. The sum of the atomic masses of the atoms that make up the molecule.
Molecular mass:
The sum of the atomic masses of the atoms
that make up the molecule.
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
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11. The mass of one formula unit of any compound.
Formula mass:
The mass of one formula unit of any compound.
Term must be used for ionic compounds;
may be used for any compound.
Formula mass =
Sum of the atomic masses in the formula unit
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

12. What is the formula mass of calcium phosphate?
Example:
What is the formula mass of calcium phosphate?
Phosphate ion: PO43–
Calcium ion: Ca2+
Calcium phosphate: Ca3(PO4)2
3 x amu = amu Ca
2 x amu = amu P O
8 x amu = amu
amu
Ca3(PO4)2
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

13. Section 7.3 The Mole Concept
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14. Goal 4
Define the term mole. Identify the number of objects that corresponds to one mole.
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

15. Goal 5
Given the number of moles (or units) in any sample, calculate the number of units (or moles) in the sample.
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

16. of atoms in exactly 12 grams of carbon-12.
Mole (mol):
One mole is the amount of any substance that contains the same number of units as the number
of atoms in exactly 12 grams of carbon-12.
The definition of a mole refers to a number of particles, but it doesn’t say what that number is.
Experiments must be conducted to find
the number of particles in a mole.
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

17. Avogadro’s Number (N): The experimentally-determined
number of atoms in a mole.
X 1023/mol
(1998 value)
Analogy:
12 items
dozen
6.02 X 1023 items
mole
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

18. How many carbon dioxide molecules are in 2.0 moles of carbon dioxide?
Example:
How many carbon dioxide molecules are
in 2.0 moles of carbon dioxide?
6.02 x 1023 molecules CO2
mol CO2 =
2.0 mol CO2 x
1.2 x 1024 molecules CO2
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19. Section Molar Mass
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20. Goal 6
Define molar mass or interpret statements in which the term molar mass is used.
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

21. Goal 7
Calculate the molar mass of any substance whose chemical formula is given (or known).
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22. The mass in grams of one mole of a substance. Units: g/mol
Molar mass:
The mass in grams of one mole of a substance.
Units: g/mol
Molar mass is a bridge between
a macroscopic quantity of matter, grams,
and a particulate-level quantity of matter, moles
Chemists count particles by weighing
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

23. 1 amu = 1/12 the mass of a carbon-12 atom
12 X
X 12
1 amu = 1/12 the mass of a carbon-12 atom
12 amu = the mass of a carbon-12 atom
1 mol = # atoms in 12 g carbon-12
12 g = the mass of 1 mol of carbon-12 atoms
12 amu/atom 12C
12 g/mol 12C
An atomic, molecular, or formula mass in amu
is numerically equal to
molar mass in g/mol
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

24. 1 mole of... C Cu Zn S8 Hg Br2 Ni Ca I2
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25. 1 mole of ... CuSO4 Hg2I2 ZnS CaBr2 NiCO3 CuBr2 CuS
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26. What is the mass of an ammonia molecule?
Example:
What is the mass of an ammonia molecule?
What is the mass of a mole of ammonia molecules?
Ammonia is NH3
14.01 amu N + 3(1.008 amu H) = amu NH3
An ammonia molecule has a mass of amu.
To change from molecular mass to molar mass,
change the units from amu to g/mol.
A mole of ammonia has a mass of g/mol.
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

27. Section 7.5 Conversion among Mass, Number of Moles, and Number of Units
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28. Goal 8
Given any one of the following for a substance whose formula is given (or known), calculate the other two: (a) mass, (b) number of atoms, (c) number of formula units, molecules, or atoms.
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

29. Number of atoms, 6.02 X 1023 particles molecules, mole formula units
Moles
# grams
mole
Grams
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

30. How many hydrogen atoms are in
Example:
How many hydrogen atoms are in
a 1.0-L sample of water?
1000 mL H2O
L H2O
1.00 g H2O
mL H2O x
1.0 L H2O x
1 mol H2O
18.02 g H2O x
6.02 x 1023 molecules H2O
mol H2O x
2 atoms H
molecule H2O x
= 6.7 x 1025 atoms H
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

31. Section 7.6 Mass Relationships among Elements in a Compound: Percentage Composition
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32. Goal 9
Calculate the percentage composition of any compound whose formula is given (or known).
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

33. Goal 10
Given the mass of a sample of any compound whose formula is given (or known), calculate the mass of any element in the sample; or, given the mass of any element in the sample, calculate the mass of the sample or the mass of any other element in the sample.
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

34. The percentage composition of a compound is the percentage by mass of
% of A = parts of A X 100
total parts
The percentage composition of a
compound is the percentage by mass of
each element in the compound.
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

35. Determine the percentage composition of sugar, C12H22O11
Example:
Determine the percentage composition of sugar, C12H22O11
In one mole of C12H22O11:
12(12.01 g C) + 22(1.008 g H) + 11(16.00 g O) =
g C12H22O11
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

36. (12.01 g C) X = % C
g C12H22O11
(1.008 g H) X = % H
g C12H22O11
(16.00 g O) X = % O
g C12H22O11
Check:
42.10% % % = %
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

37. In one mole of sugar: 12(12.01 g C) + 22(1.008 g H) + 11(16.00 g O) =
g C12H22O11
12(12.01 g C) PER g C12H22O11
12(12.01 g C) PER 22(1.008 g H)
and so on
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

38. How many grams of sugar are needed
Example:
How many grams of sugar are needed
to obtain 10.0 g carbon?
g C12H22O11
12(12.01 g C)
10.0 g C x
= 23.8 g C12H22O11
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

39. Section 7.7 Empirical Formula of a Compound
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40. Goal 11
Distinguish between an empirical formula and a molecular formula.
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

41. Goal 12
Given data from which the mass of each element in a sample of a compound can be determined, find the empirical formula of the compound.
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

42. The lowest whole-number ratio of atoms of the elements in a compound
Empirical formula:
The lowest whole-number ratio
of atoms of the elements in a compound
The empirical formula of C2H4 is CH2
Likewise, the empirical formula of C3H6 is CH2
All compounds with the general formula CnH2n
have the same empirical formula and the
same percentage composition
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

43. Write the empirical formula of benzene, C6H6, and of octane, C8H18.
Example:
Write the empirical formula
of benzene, C6H6, and of octane, C8H18.
C6H6: Both divisible by 6
EF: CH
C8H18: Both divisible by 2
EF: C4H9
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

44. How to Find an Empirical Formula
1. Find the masses of different elements
in a sample of the compound
2. Convert the masses into moles of
atoms of the different elements
3. Express the moles of atoms as the
smallest possible ratio of integers
4. Write the empirical formula, using the
number for each atom in the integer
ratio as the subscript in the formula
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

45. What is the empirical formula of a compound that analyzes as
Example:
What is the empirical formula
of a compound that analyzes as
79.95% carbon, 9.40% hydrogen,
and 10.65% oxygen?
Many empirical formula questions
can be solved with the following table:
Element
Grams
Moles
Mole
Ratio
Formula
Empirical
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46. Mole Ratio Formula Ratio Element Grams Moles C H O 79.95 9.40 10.65 10
Empirical
Formula
Element
Grams
Moles
79.95 g C
12.01 g/mol C
9.40 g H
1.008 g/mol H
10.65 g O
16.00 g O
6.657
0.6656
9.33 C H O
79.95
9.40
10.65 10 14 1
6.657
9.33
0.6656
10.00
14.0
1.000
C10H14O
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47. Section 7.8 Determination of a Molecular Formula
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48. Goal 13
Given the molar mass and empirical formula of a compound, or information from which they can be found, determine the molecular formula of the compound.
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

49. The molecular formula of a compound is found by determination of the
number of empirical formula units
in the molecule
For example, if the empirical formula
of a compound is CH2,
the molecular formula is
CH2 or C2H4 or C3H6 or C4H8, etc
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

50. Empirical formula units in one molecule: molar mass of compound
molar mass of empirical formula
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

51. How to Find the Molecular Formula of a Compound
1. Determine the empirical formula of the compound
2. Calculate the molar mass of
the empirical formula unit
3. Determine the molar mass of the compound
(which will be given at this time)
4. Divide the molar mass of the compound by the molar mass of the empirical formula unit to get n, the number of empirical formula units per molecule
5. Write the molecular formula
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

52. What is the molecular formula of a compound with the empirical formula
Example:
What is the molecular formula
of a compound with the empirical formula
C5H10O and a molar mass of 258 g/mol?
Molar mass of C5H10O:
5(12.01 g/mol C) + 10(1.008 g/mol H) g/mol O = g/mol C5H10O
258 g/mol molecule
86.13 g/mol ef unit
= 3
(C5H10O)3
C15H30O3
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.