# Tong. ch_6.0910 - NordoniaHonorsChemistry ```Chapter six
DAY 1: 6.1-6.3
•H O W M U C H S O D I U M ?
•C O U N T I N G N A I L S B Y T H E P O U N D
•C O U N T I N G A T O M S B Y T H E G R A M
Why Is Knowledge of Composition
Important?
2
 Some Applications:
 The amount of sodium in sodium chloride for
diet.
 The amount of iron in iron ore for steel
production.
 The amount of hydrogen in water for hydrogen
fuel.
 The amount of chlorine in freon to estimate
ozone depletion.
Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Counting Nails by the Pound
3
A hardware store customer buys 2.60 pounds
of nails. A dozen nails has a mass of 0.150
pounds. How many nails did the customer
1 dozen nails = 0.150 lbs.
12 nails = 1 dozen nails
Solution map:
Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Counting Nails by the Pound, Continued
4
1 doz. nails 12 nails
2.60 lbs. 

 208 nails
0.150 lbs. 1 doz.
 The customer bought 2.60 lbs of nails and
received 208 nails. He counted the nails by
weighing them!
Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Counting Atoms by Moles
5
 If we can find the mass of a particular number of
atoms, we can convert the mass of an element sample
to the number of atoms in the sample.
 The conversion factor we will use is the MOLE

1 mole = 6.022 x 1023 things.


Like 1 dozen = 12 things.
Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Moles- as a conversion factor
6
 Mole = Number of things equal to the number of
atoms in 12 g of C-12.


1 atom of C-12 weighs exactly 12 amu.
1 mole of C-12 weighs exactly 12 g.
 In 12 g of C-12 there are 6.022 x1023 C-12 atoms.
6.022  10 atoms
1 mole
23
Tro's &quot;Introductory Chemistry&quot;, Chapter 6
1 mole
6.022  10 23 atoms
Conversions
 When converting between moles and number of
atoms

Use the conversion factor:

1 mole = 6.02 x1023 atoms
 Practice: A silver ring contains 1.1 x 1022 silver atoms.
How many moles of silver are in the ring?
Conversions continued
 When converting from grams to moles of an element
 Use the conversion factor:

Molar mass (atomic mass in grams) = 1 mole
 Practice: Calculate the number of moles of sulfur in
57.8 g of sulfur.
Conversions, continued
 When converting between grams and number of
atoms

Use two conversion factors

molar mass= 1 mole and then 1 mole = 6.022 x1023 atoms
 Practice: How many aluminum atoms are in an
aluminum can with a mass of 16.2 g?
Day 2: 6.4-6.5
•C O U N T I N G M O L E C U L E S B Y T H E G R A M
•C H E M I C A L F O R M U L A S A S C O N V E R S I O N
FACTORS
Molar Mass of Compounds
12
 The relative weights of molecules can be calculated
from atomic weights.
Formula mass = 1 molecule of H2O
= 2(1.01 amu H) + 16.00 amu O = 18.02 amu.
 Since 1 mole of H2O contains 2 moles of H and 1 mole
of O.
Molar mass = 1 mole H2O
= 2(1.01 g H) + 16.00 g O = 18.02 g.
Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Conversions
 When converting from grams to moles of a
compound

Use the conversion factor:

Formula mass (in grams) = 1 mole
 Practice: Calculate the number of moles of Hydrogen
in 57.8 g of water.
Conversions, continued
 When converting between grams and number of
molecules

Use two conversion factors

formula mass= 1 mole and then 1 mole = 6.022 x1023 molecules
 Practice: How many carbon dioxide molecules are in
45.3g of dry ice (carbon dioxide)?
Chemical Formulas as Conversion Factors
15
 1 spider  8 legs.
 1 chair  4 legs.
 1 H2O molecule  2 H atoms  1 O atom.
Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Counting Parts
16
 We can use the total number to determine the
number of each part:

When all the desks in the room have 4 legs, if there are
30 desks in the room, there will be 120 legs (4 x 30).

Since every H2O molecule has 2 H atoms, in 100 H2O
molecules, there are 200 H atoms.

In 1 mole of H2O molecules, there are 2 moles of H
atoms.
Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Mole Relationships in Chemical Formulas
17
 Since we count atoms and molecules in mole
units, we can find the number of moles of a
constituent element if we know the number of
moles of the compound.
Moles of compound
1 mol NaCl
1 mol H2O
1 mol CaCO3
Moles of constituents
1 mol Na, 1 mol Cl
2 mol H, 1 mol O
1 mol Ca, 1 mol C, 3 mol O
1 mol C6H12O6
6 mol C, 12 mol H, 6 mol O
Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Converting from moles of compound to moles of
(constituent) part
 Calculate the Moles of Oxygen in 1.7 Moles of
CaCO3


1.7 moles of CaCO3 x 3 mole O
= 5.1 mol O
1 mole CaCO3
Practice:
 How many grams of Cl are in CF3Cl?
 How many grams of sodium are in 24g of sodium
phosphate?
Day 3:
QUIZ 6.1-6.5
SECTION: 6.6 &amp; 6.7
Mass Percent
 This is the % of the compound that is composed of
one of its elements.
 Mass % =
mass of one element
x 100
mass of compound sample
Mass percent is a conversion factor too!
100g of the compound/ mass % of element
Practice
 What is the mass percent of calcium in calcium
nitrate?


The atomic mass of calcium is 40.08g
The formula mass of calcium nitrate is 164g

40.08g x 1oo = 24.4%
164 g
Using %mass as a conversion factor

Silver Chloride, used in silver plating, contains 75.27%
Ag. Calculate the mass of silver chloride in grams,
required to make 4.8 g of silver plating.

4.8 g Ag x 100 g AgCl = 6.38g AgCl
75.27 g Ag
Mass % in a compound
 Mass of single element/ formula mass of compound
Practice:
What is the mass % of Cl in: C2Cl4F2
Day 4:
BAGGIE LAB
&amp;
WRITE UP
Day 5
6.8-6.9
EMPIRICAL FORMULAS
Empirical Formulas
28
 The simplest, whole-number ratio of atoms in a
molecule is called the empirical formula.

Can be determined from percent composition or combining
masses.
 The molecular formula is a multiple of the empirical
formula.
%A
100g
mass A (g)
Molar MassA
moles A
moles A
moles B
%B
100g
mass B (g)
Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Molar MassB
moles B
Empirical Formulas, Continued
29
Hydrogen Peroxide
Molecular formula = H2O2
Empirical formula = HO
Glucose
Molecular formula = C6H12O6
Empirical formula = CH2O
Benzene
Molecular formula = C6H6
Empirical formula = CH
Finding an Empirical Formula
30
1.
Convert the percentages to grams.
a.
2. Convert grams to moles.
a.
Use molar mass of each element.
3. Write a pseudoformula using moles as
subscripts.
4. Divide all by smallest number of moles.
5. Multiply all mole ratios by number to make all
whole numbers, if necessary.
a.
If ratio ?.5, multiply all by 2; if ratio ?.33 or
?.67, multiply all by 3, etc.
b.
Skip if already whole numbers after Step 4.
Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Empirical Formula from Data
 Empirical formula is a ratio of moles not mass
 Convert any data in grams into moles (use molar mass)
 Look at the number of moles and divide the number
of moles by the smallest in the formula

If the number you are left with is a decimal, multiply by a
whole number (see previous slide) to determine subscript.
Find the empirical formula of
aspirin with the given mass
percent composition.
32
A laboratory analysis of aspirin determined
the following mass percent composition:
Carbon= 60.00%, hydrogen= 4.48%, and
oxygen = 35.53%
(Remember % mass= g of element/100g
compound)
gC
gH
gO
Information:
Given:60.00 g C, 4.48 g H,
35.53 g O
Find: empirical formula,
CxHyOz
Conversion Factors:
1 mol C = 12.01 g; 1 mol H
= 1.01 g; 1 mol O = 16.00 g
mol C
mol H
mol O
Tro's &quot;Introductory Chemistry&quot;, Chapter 6
pseudoformula
mole
ratio
whole
number
ratio
empirical
formula
Example:
Find the empirical
formula of aspirin
with the given mass
percent composition.
33
Information:
Given: 60.00 g C, 4.48 g H, 35.53
gO
Find: empirical formula, CxHyOz
Conversion Factors:
1 mol C = 12.01 g;
1 mol H = 1.01 g; 1 mol O = 16.00
g
Solution Map: g C,H,O  mol
C,H,O 
mol ratio  empirical formula
 Apply the solution map:

Calculate the moles of each element.
1 mol C
 4.996 mol C
12.01 g C
1 mol H
4.48 g H 
 4.44 mol H
1.01 g H
1 mol O
35.53 g O 
 2.221 mol O
16.00 g O
60.00 g C 
Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Practice—Determine the Empirical Formula of Stannous Fluoride,
which Contains 75.7% Sn (118.70) and the Rest Fluorine (19.00)
34
Given: 75.7% Sn, (100 – 75.3) = 24.3% F 
in 100 g stannous fluoride there are 75.7 g Sn and 24.3 g F.
Find:
SnxFy
Conversion Factors:
1 mol Sn = 118.70 g; 1 mol F = 19.00 g
Solution Map:
g Sn
gF
mol Sn
mol F
Tro's &quot;Introductory Chemistry&quot;, Chapter 6
pseudoformula
whole
mole number
empirical
ratio ratio
formula
Calculating the molecular formula
 If you are given the empirical formula you can figure
out the molecular formula.
 Empirical formula x n = molecular formula
Molar mass of compound
. =n
Formula mass of empirical formula
Practice
 The empirical formula for Butane is C2H5, and has a
molar mass of 58.12 g/mol. What is the molecular
formula?
58.12 g = 2
29.07 g
C2H5 x 2 = C4H10
Day 6
EMPIRICAL LAB
Day 7:Lab write up
Day 8: Review
Day 9: Chapter 5/6 Test
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