1. Avogadro’s Constant and
Chemical Formulas
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2. Avogadro and the Mole
 First page: Title.
 Watch me do the first example(s).
 Copy content at the end of the slide.
 Use my examples to work through your
practice questions.
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3.  How do you write 6022 in scientific notation?
 C x 10x where 1 ≤ C < 10
 6.022 x 103
 Write 6022 and 200 in scientific notation,
and then multiply them, giving the answer in
scientific notation.
= 12.044 x 105
Scientific Notation Review
(Front = page 1)
3
6.022 x 103 2 x 102x
= 1 x 106

4.  Write 0.006022 and 0.0201 in scientific
notation, and then divide them, giving the
answer in scientific notation.
Scientific Notation Review
(page 1, continued)
4
6.022 x 10-3
= 2.996 x 10-1
2.01 x 10-2
6.022
2.01
10-3
10-2
x=
= 0.300

5. Conversion Review/Practice
(Page 2)
 What is the formula for conversions?
Original x
 How do you convert 2 days to minutes?
1 day = 24 hours; 1hour = 60 minutes
Therefore, 2 days
= 2880 minutes
Units To
Units From
x 24 hours
1 day
5
x 60 minutes
1 hour

6. Classwork Practice
(Page 3)
1. Convert 6022 km to centimeters.
2. How many hours are in 3 years?
15 minutes.
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7. Rules for Avogadro’s
Number (Page 4)
1. What is Avogadro’s number?
 6.022 x 1023
2. What unit is represented by Avogadro’s
number?
 A mole
3. What does the mole measure in Chemistry?
 Particles (atoms, molecules, ions, electrons,...)
4. How many particles are in 1 mole?
 6.022 x 1023
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8. Mole Particle Conversions
(Page 5)
1. How many atoms of iron are in 3 moles of Fe?
1 mole = 6.022 x 1023 atoms
Therefore, 3 moles
= 18.066 x 1023 atoms
2. How many moles are 5.66 x 1023 ions of Na+?
1 mole = 6.022 x 1023 ions
5.66 x 1023 ions
= 0.940 ions
8
x 6.022 x 1023 atoms
1 mole
= 1.8066 x 1024 atoms
x 1 mole
6.022 x 1023 ions

9. Conversion Practice 1
(Page 5)
1. How many atoms of Xe are in 0.187 moles
of Xe ?
2. How many moles of C atoms in 3.161 x
1021 molecules of CO2?
3. How many moles of O atoms in 3.161 x
1021 molecules of CO2?
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10. Rules for Avogadro’s
Number (Page 4)
5. If I have one mole of an element, how
much will it weigh?
 It’s average atomic mass in grams
6. The molar mass of an element is
measured in grams per mole = g/mol.
 For example, the molar mass of carbon is
 12.011 g/mol
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11. Mole-Mass Conversions
(Page 6)
1. What is the mass of 4 moles of Cu?
1 mole of Cu = 63.546 grams
Therefore, 4 moles
= 254.18 grams
2. What is the mass of 2.11 x 1024 atoms of Cu?
2.11 x 1024 atoms
= 223 grams 11
x 63.546 grams
1 mole
x 1 mole
6.022 x 1023 atoms
x 63.546 grams
1 mole

12. Molar Mass Conversions
(Rules - Page 4)
 If a formula tells you the ratio (ionic)
or number (molecular) of atoms in a compound,
how can we calculate the mass of a compound?
CH4 or CO2 or OH-
7. The molar mass of any molecule, formula unit,
or ion is the sum of the average atomic masses
of all atoms represented in its formula.
C + (H x 4) or C + (O x 2) or (O + H)
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13. Molar Mass Conversions
(Page 7)
3. What is the mass of 2 moles of CH4?
1 mole of CH4 = 1 mol of C atoms
+ 4 mol of H atoms
Molar mass of CH4 = 12.011 g/mol
+ 4 (1.008 g/mol)
= 16.043 g/mol
Mass of 2 moles of CH4 = 2 moles
= 32.086 grams 13
x 16.043 grams
1 mole

14. Classwork Practice
(Page 8)
1. Find the number of moles in 54g of water.
2. Determine the number of water molecules
from the previous question.
15 minutes.
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15. Rules for Avogadro’s
Number (Page 3 – final rule)
 How do I know I have one mole of a gas in
a container?
8. At STP, 1 mol of any gas occupies 22.4L
Practice:
What is the volume, in L, of 2 moles of CH4 gas?
1mol = 22.4L
2mol
= 44.8 Liters 15
x 22.4 Liters
1 mole

16. Chemical Formulas
 Front page:
Title and your name.
 Watch me do the first example(s).
 Copy content at the end of the slide.
 Use my examples to work through your
practice questions.
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17. Percentage Composition
(Page 1)
1. How do you calculate Percentage?
= x 100
2. If 4 eggs in a dozen are brown and the rest are
white, what % of the eggs are white? Brown?
White eggs = 12 - 4 = 8; x 100 = 66.67%
Brown? x 100 = 33.33%
3. If each brown egg weighs 10 grams and each
white egg weighs 9 grams:
What is the total weight of a dozen eggs?
(4 x 10g) + (8 x 9g) = 112g.
Part
Whole
8
12
4
12
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18. Percentage Composition
(Page 2)
Determine the percentage composition of water.
Formula of water = H2O
Formula mass of H2O = 2(1.008 amu) + 15.999 amu
= 18.159 amu
What percent is hydrogen? ( x 100) =
What percent is oxygen? ( x 100) =
Percentage composition:
The mass of each element in a compound expressed as a
percentage of the total mass of the compound.
2
18
18
11.11% H
16
18
88.89% O

19. Percentage Composition
Practice (Page 3)
1a) Percentage composition of lead (II) chloride.
1b) Percentage composition of barium nitrate.
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20. Percentage Composition
Practice Answers (Page 3)
1a) Percentage composition of PbCl2.
Formula mass of PbCl2= 207.2 + 2(35.45) amu = 278.1 amu
%Pb = x 100 = 74.51%
%Cl = x 100 = 25.49%
1b) Percentage composition of Ba(NO3)2.
Formula mass of Ba(NO3)2 =137.33 + 2(14.007) + 6(15.999)= 261.33amu
%Ba = x 100 = ?%
%N = x 100 = ?%
%O = x 100 = ?%
207.2 amu
278.1 amu
70.9 amu
278.1 amu
137.33 amu
261.33 amu
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21. Empirical Formula
(Page 4)
 Empirical Formula:
The symbols of each element in a compound with
subscripts showing the simplest whole number
ratio of each element in the compound.
 For example, hydrogen peroxide consists of 2
hydrogen atoms bonded to 2 oxygen atoms:
 But is H2O2 the simplest ratio (empirical formula)?
 No: the empirical formula is HO.
HH OO
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22. Empirical Formula
(Page 4 continued)
 We can use percentage composition as a ratio to
determine the empirical formula.
 For example, water consists of 11.11% H and 88.89% O
(earlier example).
1. Use 100g sample of water = 11.11g H; 88.89g O
2. Convert to a mole ratio = mol H mol O
(divide by element’s molar mass)
3. Simplify the ratio
(divide by the smaller mole value) =
Empirical formula = H2O
11.11g
1g/mol
88.89g
16g/mol
= 11.11mol = 5.55mol
5.55mol5.55mol
2 : 1
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23. Empirical Formula Practice
(Page 5)
 Determine the empirical formula for a compound
that consists of 32.38% sodium, 22.65% sulfur,
44.99% oxygen.
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24. Empirical Formula from Mass
(Page 6)
 We can use actual mass to determine the empirical
formula.
 Example: A 10.15g sample consists of 5.72g oxygen and
the rest phosphorus.
1. State mass of each element = (10.15-5.72)g P; 5.72g O
2. Convert to a mole ratio = mol P mol O
(divide by element’s molar mass)
3. Simplify the ratio
(divide by the smaller mole value) =
Empirical formula = P2O5
4.43g
31g/mol
5.72g
16g/mol
= 0.14 mol = 0.36mol
0.14mol0.14mol
1 : 2.57
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25. Molecular Formula
(Page 7)
 Molecular Formula:
A chemical formula with subscripts showing the
actual whole number ratio of each element in the
compound.
 The molecular formula is therefore a multiple of
the empirical formula:
Molecular formula = x (empirical formula)
Where x is a whole number.
Example : H2O2 = 2 (HO)
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26. Molecular Formula from
Empirical Formula (Page 7)
 Example: A compound with empirical formula CH has a
molar mass of 78 g/mol. What is the molecular formula?
1. Determine formula mass = 12amu + 1amu = 13 amu
(of empirical formula)
2. Formula mass of molecule (# amu) = molar mass (78 g/mol)
3. Find out formula ratio (a.k.a. the X factor) =
(empirical to molecular)
4. Multiply all the empirical subscripts by the X factor
empirical formula = CH
Molecular formula = C6H6
78 amu = 6
13 amu
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27. Molecular Formula Practice
(Back of foldable)
1. A brown gas has an empirical formula NO2 and a
molar mass of 46g/mol. What is the molecular
formula?
2. A compound has the empirical formula CH2O. Its
experimental molar mass is 90.0 g/mol. What is
its molecular formula?
Homework: Juno Quiz “Chemical Formulas”
due next class
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