### Chapter 10 PPT

```Chemical Quantities, the Mole,
and Conversions
 Measuring
Matter
-The amount of something is usually
determined one of three ways; by
counting, by mass, or by volume.
-Groups are made to make counting easier.
(Remember the bean lab?)
A bunch, a dozen, a ream, a can
 6.022
x 1023 of anything!
 Representative Particles are the smallest
division of a substance that maintains the
characteristics of that substance
-For molecular compounds R.Ps are
molecules
-For ionic compounds R.Ps are formula
units
-For elements R.Ps are atoms
1
mole of any element is equal to the
atomic mass on the periodic table
 AKA molar mass
 For example:
18g of H2O = 6.022 x 1023 molecules
 Number
of particles to moles…
RP x
1 mole
=Moles
6.02 x 1023 RPs
 How
many moles are 2.80 x 1024 atoms of
silicon?
 4.65
moles
 Moles
to number of particles…
 moles x 6.02 x 1023 RPs.
= RPs
1 mole
 How
many atoms are in 1.14 mol Co?
 6.86 x 1023 atoms Co
 The
atomic mass is the mass of one atom
of an element (in AMU)
 Based on the mass of one atom of
Carbon-12
 AKA – relative mass since each element
is compared to C-12
 Not very practical when working with
chemicals in real life
 Molar
Mass is the atomic mass of an
element expressed in grams or, the mass
of one mole of atoms of a particular
element expressed in grams
 Numerically
equal to the atomic mass in
amu’s because the Mole is a constant
number!



Formula Mass is the sum of the average atomic
masses of all the elements represented in the
formula (in amu’s)
Example: Use the atomic masses in the periodic
table to calculate the formula mass of SrCl2
FM = 87.62 + 2(35.453) = 158.526 amu
Importance – the formula mass of a compound
gives us useful information about the quantities
of reactants needed to make new products (must
follow the Law of Conservation of Mass)
 Molecular
Mass is the same as formula
mass, but this term in used specifically
for molecular compounds
 Not
used for ionic compounds
 Example
– water: H2O = 2(1.00797) +
15.9994 = 18.01534 amu
 Molar
Mass is the mass in GRAMS of one
mole (6.02 x 10 23 particles) numerically
equal to the formula mass of a compound
(units = g/mol)
 Mole-Mass
Relationship
Use the molar mass of an element or
compound to convert between the mass of a
substance and the moles of a substance
Moles to Mass (grams): Multiply by the
Molar Mass
Example – Find the mass, in grams, of 4.52 x
10-3 mol C20H42
 Step
1: Find the Molar Mass:
20 (12.01115) + 42 (1.00797)
= 282.55774 g/mol
 Step 2: Convert:
4.52 x 10-3 mol x
 1.28 g C20H42
282.55774 g C20H42 =
1 mol C20H42

Grams to moles: Divide by the Molar Mass
Example – Calculate the number of moles in 75.0 g of
Dinitrogen trioxide.
Step 1: Write the correct chemical formula: N2O3
Step 2: Find the Molar Mass:
2( 14.0067) + 3(15.9994)
= 76.0116 g/mol
Step 3: Convert:
75.0 g N2O3 x 1 mole N2O3
76.0116 g N2O3
0.987 mole N2O3
=
# of
RPs
#
# of Moles
x molar mass
Mass
(g)
Hypothesis states equal
volumes of gases at the same
temperature and pressure contain equal
numbers of molecules
 Standard Temperature
 (STP)
and Pressure
= 0 ˚C (273 ˚K) and 101.325 kPa (1
atm, 760 mmHg, 760 torr)
 Molar Volume
= the 22.4 L of space
occupied by 6.02 x 1023 representative
particles of any gas at STP
 Example – What is the
10-3 mol CO2 at STP?
 3.20
volume of 3.20 x
x 10-3 mol CO2 x 22.4 L = ?
1 mole
 Example: –
How many moles of CO2 are
in 57.0 L of CO2?
 57.0
L CO2 x
2.54 mol CO2
1 mole =
22.4 L
 Calculating
Molar Mass from Density

Background – the density of gases is
measured in g/L at a specific
temperature, and, the density of a gas at
STP is a constant value (characteristic
property)

To find Molar Mass of a gas from
density – multiply the density at STP by
the molar volume at STP
 Example
– A gaseous compound
composed of sulfur and oxygen, which is
linked to the formation of acid rain, has a
density of 3.58 g/L at STP. What is the
molar mass of this gas?
 Molar Mass = 3.58 g of gas x 22.4 L =
1L
1 mol
80.2 g/mol
 Definition
– relative amounts of
substances in a compound expressed as
percents by mass
 Percentage
Composition from Mass Data
 Equation:
Mass of Element x 100 = %
Mass Compound
 It
has been measured in lab that 1.000g
of water contains 0.112g Hydrogen. Find
the mass percentage of hydrogen
 Mass
%H = 0.112g x 100 = 11.2 %H
1.000g
 Percentage
Composition from the
Chemical Formula
(Atomic mass of elem.) x (# atoms of elem.) x 100
Molar mass of the compound
 What
is the percent composition by mass
of hydrogen in water?
 Mass
%H = (1.00797g) x 2 x 100 =
18.01534g
11.19013 %H
 Percentage
Factor
Composition as a Conversion
What is the mass of hydrogen in 30.0g of
water?
Mass H = (Mass %  100) x mass sample =
(11.2%  100) x 30.0g =
3.36gH
 Hydrates
(compounds containing water):
You must include the water of hydration
in the formula mass
 What
are the mass percentages of the
elements in Na2CO310 H2O?
 MM
= 2(22.98977) + 12.01115 +
3(15.9994) + 10(18.01534) = 286.14229g
 %Na
= 2(22.98977) x 100 = 16.06877 % Na
286.14229
 %C
= 12.01115 x 100 = 4.197614 % C
286.14229
 %O
= 3(15.9994) x 100 = 16.77424 % O
286.14229
 %H2O
= 10(18.01534) x 100 = 62.95938 % H2O
286.14229
A
chemical formula showing the smallest
(simplest) whole-number ratio of atoms
in a compound
 Steps for determining Empirical
Formula from % composition
Step 1: Determine the masses of the
elements in a 100g (since its based on a
percent by mass) sample
Step 2: Convert these masses in grams to
moles by dividing by the atomic mass
Step 3: Determine the simplest mole ratio
between the elements by dividing each
by the smallest number of moles
Step 4: Use these mole ratios to write the
subscripts in the formula


Calcium Fluoride occurs as the mineral Fluorite. It contains
51.3% by mass of Ca and 48.7% by mass of F. Determine the
Empirical formula.
51.3% of 100g = 51.3 g Ca
48.7% of 100g = 48.7 g F
Step 2: Convert to Moles
51.3 g Ca x 1 mole Ca = 1.28 mole Ca
40.08 g Ca

48.7 g F x

1 mole F
= 2.56 mole F
18.998403 g F
Step 3: Ratio mole F
= 2.56 mole F
= 2 - ratio is 1:2
mole Ca
1.28 mole Ca
 CaF2
is the empirical formula
 Another
Example!
Substance X and Y. Suppose you found the
number of moles of X and Y in 100.0g of a
compound to be 1.96 mole X and 2.94
mole Y. What is the empirical formula?
X2Y3
 C2H2
C8H8
 The
actual formula of a molecular
compound which is often a multiple of
the empirical formula (E.g. C2H6 instead
of CH3)
 The
Molecular Mass must be determined
by experimental analysis or be given in
the problem
 Divide
the Molecular Mass by the
Formula Mass (Molar Mass) and adjust
the subscripts in the Empirical Formula
by multiplying by this resulting number
Water – the simplest formula is H2O, but could
the molecular formula be H4O2 or H8O4?
 Molecular
Mass = 18.01 g by analysis
 Formula
Mass (Molar Mass) =
2(1.00797) + 15.9994 = 18.01534g
of MM/FM = 18.01  18.0534 or 1:1
so this is also the molecular formula
 Ratio
Hydrogen Peroxide – the simplest formula is
HO
 Molecular
(given)
Mass = 34.0g by analysis
 Formula
Mass (Molar Mass) = 1.00797 g
+ 15.9994 g = 17.00737 g
of MM/FM = 34.0  17.00737 = 2:1 so
the formula is doubled to H2O2
 Ratio
A compound is found by analysis to consist of 40.1% S
and 59.9% O. The Molecular Mass is 80.1g. Determine
the Empirical Formula, the Molecular Formula, and
name the compound


Step 1: in a 100.0g sample
40.1% of 100.0 g = 40.1 g S
59.9% of 100.0 g = 59.0 g O
Step 2: convert to moles
40.1 g S x 1 mole S = 1.25 mole S
32.06g S
59.9 g O x 1 mole O
15.9994 g O
= 3.74 mole O
Step 3: ratio
3.74 mole O  1.25 mole S =


2.992:1 or 3:1
Step 4: empirical formula
SO3
Step 5: mass ratios
FM* = 32.06 g S + 3(15.9994 g)O
= 80.0582 g SO3

MM = 80.1g (given in problem)
1:1 so the molecular formula = the empirical
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