Chapter 10: Chemical
Quantities
10.1 Moles
I. Numerical Abbreviations
• A. Dozen = 12
• B. Pair = 2
• C. Gross = 144
• D. Ream = 500
II. Moles
• A. Representative particle: type of particle in
substance (Atoms, molecules, ions, formula units)
• B. Mole: abbreviation for
number of particles in a
substance
• C. Avogadro’s number:
6.02x1023 particles in a mole
III. Determining Rep. Part.
• A. Depends on what you’re dealing with
• B. There are 6.02x1023 atoms in a mole of Carbon
• C. There are 6.02x1023 molecules in a mole of H2O
A mole of water
is only 18
milliliters.
IV. Mass of a Mole
• A. Molar mass: mass of one mole of a substance, use
atomic mass in grams
• B. For the mass of a molecule add atomic masses
together
• C. Example: H2O = 2 (mass of
Hydrogen) + 1(mass of Oxygen) =
2(1.01)+1(16.00) =
18.02 grams
V. Chemical Conversions
• A. 12 moles Carbon  atoms Carbon
• B. 4 moles of H2  grams H2
10.2: Mole-Volume
Relationships
• A. Avogadro’s Hypothesis: equal gas
volumes at same pressure and
temperature contain same # of particles
• B. Gas volumes usually measured at
Standard Temperature and Pressure
(STP)
• C. Temp= 0° C, Pressure = 1 Atmosphere
• D. Volume of a mole of gas at STP =
22.4 Liters
I. Molar
Volume
II. Sample Conversions
• A. 5.0 moles of O2  Liters of O2 @ STP
• B. 45.2 grams H2  atoms of H
10.3: Percent Composition
and Chemical Formulas
H
C
N
O
I. Percent Composition
• A. Percent, by mass, of each element in a compound
• B. % of element = grams element x 100
grams compound
• C. Ex. H2O: % H = (2g/18g) x 100 = 11.1%
% O = (16g/18g) x 100 = 88.9%
II. Chemical Formulas
• A. Empirical formula: smallest whole number ratio
of elements in molecular compound
• B. Molecular formula: normal compound ratio of
elements
• C. Ex. C2H4 (mol.), CH2 (emp.)
Ex. C6H12O6 (mol), CH2O (emp.)
Ex. H2O (mol), H2O (emp.)
III. Calculating Empirical Formula
•
A. What is emp. form. of a compound with 25.9%
of Nitrogen and 74.1% of Oxygen?
1. Change percents to mass. (25.9 g N, 74.1 g O)
2. Convert mass to moles for each.
25.9 g N
74.1 g O
1 moles N
14
gN
1
moles O
16 g O
=
1.85 moles N
= 4.63 moles O
3. Divide each mole # by smallest value.
1.85 N/ 1.85 = 1 N
4.63 O/ 1.85 = 2.5 O
N1O2.5
4. Change “mole ratios” to smallest whole numbers
(in this case multiply by 2) 2(N1O2.5) = N2O5
Ex. Emp. Formula
•
Calc. emp. form. of compound with 42.9% C
and 57.1% O.
IV. Calculating Molecular Formula
• A. Calc. the mol. form. of a compound with an emp.
form. of CH2 and a molecular mass of 56 grams.
• 1. Set up the ratio:
Mol. Form =
Mol. Mass
Emp. Form
Emp. Mass
• 2. Plug in values:
__X__
=
56 g
CH2
14 g
• 3. Solve: X = 4(CH2) = C4H8