1. Chapter 7 Chemical Quantities
7.1 The Mole
7.2 Molar Mass
7.3 Calculations Using Molar Mass
7.4 Percent Composition and Empirical Formulas

2. 7.1 A Mole
A mole contains 6.02 x 1023 particles (atoms, ions, molecules, formula unit)
The number 6.02 x 1023 is known as Avogadro’s number.
One mole of any element contains Avogadro’s number of atoms.
1 mole Na = x 1023 Na atoms
1 mole Au = x 1023 Au atoms

3. A Mole of Molecules
One mole of a covalent compound contains Avogadro’s number of molecules.
1 mole CO2 = x 1023 CO2 molecules
1 mole H2O = x 1023 H2O molecules
One mole of an ionic compound contains Avogadro’s number of formula units.
1 mole NaCl = 6.02 x 1023 NaCl formula units

4. Samples of One Mole Quantities

5. Avogadro’s Number Avogadro’s number is written as conversion factors.
6.02 x 1023 particles and mole
1 mole x 1023 particles
The number of molecules in 0.50 mole of CO2 molecules is calculated as
0.50 mole CO2 molecules x 6.02 x 1023 CO2 molecules
1 mole CO2 molecules
= 3.0 x 1023 CO2 molecules

6. Learning Check A. Calculate the number of atoms in 2.0 moles of Al.
B. Calculate the number of moles of S in 1.8 x 1024 S.

7. Solution
A. Calculate the number of atoms in 2.0 moles of Al moles Al x x 1023 Al atoms
1 mole Al
=1.2 x 1024 Al atoms
B. Calculate the number of moles of S in 1.8 x 1024 S.
1.8 x 1024 S atoms x mole S
6.02 x 1023 S atoms
= 3.0 mole S atoms

8. 7.2 Molar Mass The mass of one mole is called molar mass (g/mole).
The molar mass of an element is the atomic mass expressed in grams.

9. Learning Check Give the molar mass to the nearest 0.1 g.
A. 1 mole of K atoms = ________
B. 1 mole of Sn atoms = ________

10. Solution Give the molar mass to the nearest 0.1 g.
A. 1 mole of K atoms = 39.1 g
B. 1 mole of Sn atoms = g

11. Molar Mass of CaCl2 111.1amu = 111.1g/mol
For a compound, the molar mass is the sum of the molar masses of the elements in the formula. We calculate the molar mass of CaCl2 to the nearest 0.1 g as follows.
Formula mass of CaCl2 = [ (35.45)] = 111.1amu
Formula mass = molar mass, so
111.1amu = 111.1g/mol

12. Molar Mass of K3PO4 Molar Mass = [3(39.1) + 31.0 + 4(16)] = 212.3g/mol
Determine the molar mass of K3PO4 to 0.1 g.
Molar Mass = [3(39.1) (16)]
= 212.3g/mol

13. One-Mole Quantities
32.1 g g g g g

14. Learning Check A. 1 mole of K2O = ______g
B. 1 mole of antacid Al(OH)3 = ______g

15. Solution A. 1 mole of K2O 2 (39.1) + 1 (16.0) = 94.2 g/mol
1mole K2O x 94.2g/mol = 94.2g
B. 1 mole of antacid Al(OH)3
1 (27.0) + 3 (16.0) + 3 (1.0) = g/mol
1mole Al(OH)3 x 78.0 g/mol = 78.0g

16. Learning Check
Prozac, C17H18F3NO, is an antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac?

17. Solution
Prozac, C17H18F3NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac?
17C (12.0) + 18H (1.0) + 3F (19.0) + 1N (14.0) + 1 O (16.0) =
= 309 g/mole

18. Molar Mass Factors
Methane CH4 known as natural gas is used in gas cook tops and gas heaters.
1 mole CH4 = g
The molar mass of methane can be written as conversion factors.
16.0 g CH and mole CH4
1 mole CH g CH4

19. Learning Check
Acetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass conversion factors for acetic acid.

20. Solution
Acetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass factors for acetic acid.
2(12.0) + 4(1.0) + 2(16) = 60.0g/mol
1 mole of acetic acid = g acetic acid
1 mole acetic acid and g acetic acid
60.0 g acetic acid mole acetic acid

21. 7.3 Calculations with Molar Mass
Mole factors are used to convert between the grams of a substance and the number of moles.
Mole factor
Grams
Moles

22. Calculating Grams from Moles
Aluminum is often used for the structure of
lightweight bicycle frames. How many grams
of Al are in 3.00 moles of Al?
3.00 moles Al x g Al = g Al
1 mole Al
mole factor for Al

23. Learning Check
The artificial sweetener aspartame (Nutri-Sweet) C14H18N2O5 is used to sweeten diet foods, coffee and soft drinks. How many moles of aspartame are present in 225 g of aspartame?

24. Solution Calculate the molar mass of C14H18N2O5.
14 (12.0) + 18 (1.0) + 2 (14.0) + 5(16.0) = 294 g/mole
Set up the calculation using a mole factor.
225 g aspartame x 1 mole aspartame
294 g aspartame
mole factor(inverted)
= mole aspartame

25. 7.4 Percent Composition
In a compound, the percent composition is the percent by mass of each element in the formula.
In one mole of CO2 there are 12.0 g of C and 32.0 g of O (molar mass 44.0 g/mol),
12.0 g C x = % C
44.0 g CO2
32.0 g O x = % O
44.0 g CO %

26. Learning Check
What is the percent carbon in C5H8NNaO4 (MSG monosodium glutamate), a compound used to flavor foods and tenderize meats?
1) %C
2) %C
3) %C

27. Solution 2) 35.5 %C Molar mass = 169.1 g % = total g C x 100
total g MSG
= g C x = % C
169.1 g MSG

28. Types of Formulas
The molecular formula is the true or actual number of the atoms in a molecule.
The empirical formula is the simplest whole number ratio of the atoms.
The empirical formula is calculated by dividing the subscripts in the molecular formula by a whole number to give the lowest ratio. C5H10O5  = C1H2O = CH2O molecular empirical formula formula

29. Some Molecular and Empirical Formulas

30. Learning Check A. What is the empirical formula for C4H8?
1) C2H4 2) CH ) CH
B. What is the empirical formula for C8H14?
1) C4H7 2) C6H ) C8H14
C. Which is a possible molecular formula for
CH2O?
1) C4H4O4 2) C2H4O ) C3H6O3

31. Solution A. What is the empirical formula for C4H8? 2) CH2 C4H8  4
B. What is the empirical formula for C8H14?
1) C4H7 C8H14  2
C. Which is a possible molecular formula for
CH2O?
2) C2H4O ) C3H6O3

32. Learning Check
If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain.
1) SN
2) SN4
3) S4N4

33. Solution
If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain.
3) S4N4
If the molecular formula has 4 atoms of N, and S and N are related 1:1, then there must also be 4 atoms of S.

34. Relating Empirical and Molecular Formulas
A molecular formula is equal to or a multiple of the empirical formula.
Thus, the molar mass is equal to or a multiple of the empirical mass.
molar mass = a whole number empirical mass
Multiply the empirical formula by the whole number to determine the molecular formula.

35. Finding the Molecular Formula
Determine the molecular formula of a compound
that has a molar mass of 78.0 and an empirical
formula of CH.
1. Empirical mass of CH = g/mol
2. Divide the molar mass by the empirical mass.
g/mol = 6.00
13.0 g/mol
4. Multiply the subscripts in CH by 6.
5. Molecular formula = (CH)6 = C6H6

36. Learning Check
A compound has a formula mass of and an empirical formula of C3H4O3. What is the molecular formula?

37. Solution
A compound has a formula mass of and an empirical formula of C3H4O3. What is the molecular formula?
C3H4O3 = 3(12.0) + 4(1.0) + 3(16.0) = 88.0 g/mol
176.0 g/mol (molar mass) =
88.0 g/mol (empirical mass)
Molecular formula = 2 (empirical formula)
(C3H4O3 )2 = C6H8O6

38. Finding the Molecular Formula
A compound is C %, H 4.07%, and Cl 71.65%. The molar mass is known to be 99.0 g. What are the empirical and molecular formulas?
1. Write the mass percents as the grams in a
g sample of the compound.
C g H g Cl g

39. Finding the Molecular Formula Continued
2. Calculate the number of moles of each element.
24.27 g C x 1 mole C = moles C
12.0 g C
4.07 g H x 1 mole H = moles H
1.01 g H
71.65 g Cl x 1 mole Cl = moles Cl
35.5 g Cl

40. Finding the Molecular Formula (continued)
3. Divide each by the smallest moles C = 1 mole C moles H = 2 moles H moles Cl = 1 mole Cl
Empirical formula = C1H2Cl1 = CH2Cl

41. Finding the Molecular Formula (continued)
4. Calculate empirical mass (EM) empirical mass CH2Cl = g/mol
5. Divide molar mass by empirical mass Molar mass = g/mol = 2 Empirical mass g/mol
6. Determine Molecular formula (CH2Cl)2 = C2H4Cl2

42. Learning Check
Aspirin is 60.0% C, 4.5 % H and 35.5 % O. Calculate its empirical (simplest) formula.

43. Solution (continued)
In 100 g of aspirin, there are g C, 4.5 g H, and g O.
60.0 g C x 1 mole C = moles C
12.0 g C
4.5 g H x 1 mole H = moles H
1.01 g H
35.5 g O x 1mole O = moles O
16.0 g O

44. Solution (continued) Divide by the smallest number of moles.
5.00 moles C = moles C
2.22
4.5 moles H = 2.0 moles H
2.22 moles O = mole O
Note that the results are not all whole numbers. To obtain whole numbers, multiply by a factor to give whole numbers.

45. Solution (continued) Multiply each number of moles by 4
C: moles C x 4 = 9 moles C
H: 2.0 moles H x 4 = 8 moles H
O: mole O x 4 = 4 moles O
Use the whole numbers as subscripts to
obtain the simplest formula
C9H8O4

46. Learning Check
A compound is 27.4% S, 12.0% N and 60.6 % Cl. If the compound has a molar mass of 351 g, what is the molecular formula?

47. Solution In 100.0 g, there are 27.4 g S, 12.0 g N and 60.6 g Cl.
27.4 g S x 1 mole S = mole S
32.1 g S
12.0 g N x 1 mole N = moles N
14.0 g N
60.6 g Cl x 1mole Cl = moles Cl
35.5 g Cl

48. Solution (continued) Dividing by the smallest number of moles
0.854 mole S / = mole S
0.857 mole N/ = mole N
1.71 moles Cl/ = moles Cl
Empirical formula = SNCl2 = g/mol
Molar Mass/ Empirical mass =
Molecular formula = (SNCl2)3 = S3N3Cl6