Mole Problems
Solution Guide
How many water molecules are found in
18.0g water?




602000000000000000000000
6.02x1023 water molecules
If you had 6.02x1023 dollar bills you would need to
spend $238 trillion dollars per second for every day
of an 80 year life span to spend that many dollars, if
you shared it with each person on the planet they
would have to spend $34,000 dollars per second
every day of their 80 year life span to spend it all.
$6.02x1023
= $238,616,184,677,188.51
(80yrs ) (365 d) ( 24hr) ( 60min) (60s )
second
( 1yr ) ( 1 d) ( 1hr ) (1min)
How many water molecules are found in
18.0g water?




602000000000000000000000
6.02x1023 water molecules
If you had 6.02x1023 dollar bills you would need to
spend $238 trillion dollars per second for every day
of an 80 year life span to spend that many dollars, if
you shared it with each person on the planet they
would have to spend $34,000 dollars per second
every day of their 80 year life span to spend it all.
$6.02x1023
= $238,616,184,677,188.51
(80yrs ) (365 d) ( 24hr) ( 60min) (60s )
second
( 1yr ) ( 1 d) ( 1hr ) (1min)
How many water molecules are found in
18.0g water?




602000000000000000000000
6.02x1023 water molecules
If you had 6.02x1023 dollar bills you would need to
spend $238 trillion dollars per second for every day
of an 80 year life span to spend that many dollars, if
you shared it with each person on the planet they
would have to spend $34,000 dollars per second
every day of their 80 year life span to spend it all.
$6.02x1023
= $238,616,184,677,188.51
(80yrs ) (365 d) ( 24hr) ( 60min) (60s )
second
( 1yr ) ( 1 d) ( 1hr ) (1min)
How many water molecules are found in
18.0g water?




602000000000000000000000
6.02x1023 water molecules
If you had 6.02x1023 dollar bills you would need to
spend $238 trillion dollars per second for every day
of an 80 year life span to spend that many dollars, if
you shared it with each person on the planet they
would have to spend $34,000 dollars per second
every day of their 80 year life span to spend it all.
$6.02x1023
= $238,616,184,677,188.51
(80yrs ) (365 d) ( 24hr) ( 60min) (60s )
second
( 1yr ) ( 1 d) ( 1hr ) (1min)
How many water molecules are found in
18.0g water?




602000000000000000000000
6.02x1023 water molecules
If you had 6.02x1023 dollar bills you would need to
spend $238 trillion dollars per second for every day
of an 80 year life span to spend that many dollars, if
you shared it with each person on the planet they
would have to spend $34,000 dollars per second
every day of their 80 year life span to spend it all.
$6.02x1023
= $238,616,184,677,188.51
(80yrs ) (365 d) ( 24hr) ( 60min) (60s )
second
( 1yr ) ( 1 d) ( 1hr ) (1min)
How many water molecules are found in
18.0g water?




602000000000000000000000
6.02x1023 water molecules
If you had 6.02x1023 dollar bills you would need to
spend $238 trillion dollars per second for every day
of an 80 year life span to spend that many dollars, if
you shared it with each person on the planet they
would have to spend $34,000 dollars per second
every day of their 80 year life span to spend it all.
$6.02x1023
= $238,616,184,677,188.51
(80yrs ) (365 d) ( 24hr) ( 60min) (60s )
second
( 1yr ) ( 1 d) ( 1hr ) (1min)
In any measurable sample of matter there are too many particles to count directly!
Atoms are exceedingly small!!!



1
One H has a relative mass of ______ amu.
1


10

1
H have a relative mass of ______ amu.
1
12
One
C has a relative mass of
6
10
12
C have a relative mass of
6
_______ amu
______ amu

1


100




106
H have a relative mass of _____ amu.
1
1
H have a relative mass of ______ amu.
1
12
100
C have a relative mass of
______ amu
6
106
12
C have a relative mass of ______ amu
6



1
One H has a relative mass of __1___ amu.
1


10

1
H have a relative mass of ______ amu.
1
12
One
C has a relative mass of
6
10
12
C have a relative mass of
6
_______ amu
______ amu

1


100




106
H have a relative mass of _____ amu.
1
1
H have a relative mass of ______ amu.
1
12
100
C have a relative mass of
______ amu
6
106
12
C have a relative mass of ______ amu
6



1
One H has a relative mass of __1___ amu.
1


10

1
H have a relative mass of ______ amu.
1
12
One
C has a relative mass of
6
10
12
C have a relative mass of
6
___12__ amu
______ amu

1


100




106
H have a relative mass of _____ amu.
1
1
H have a relative mass of ______ amu.
1
100
106
12
C have a relative mass of
6
______ amu
12
C have a relative mass of ______ amu
6



1
One H has a relative mass of __1___ amu.
1


10

1
H have a relative mass of _10___ amu.
1
12
One
C has a relative mass of
6
10
12
C have a relative mass of
6
___12__ amu
______ amu

1


100




106
H have a relative mass of _____ amu.
1
1
H have a relative mass of ______ amu.
1
12
100
C have a relative mass of
______ amu
6
106
12
C have a relative mass of ______ amu
6



1
One H has a relative mass of __1___ amu.
1


10

1
H have a relative mass of _10___ amu.
1
12
One
C has a relative mass of
6
10
12
C have a relative mass of
6
100
12
C have a relative mass of
6
___12__ amu
120__ amu

1


100




106
H have a relative mass of ____ amu.
1
1
H have a relative mass of ______ amu.
1
106
______ amu
12
C have a relative mass of ______ amu
6



1
One H has a relative mass of __1___ amu.
1


10

1
H have a relative mass of _10___ amu.
1
12
One
C has a relative mass of
6
10
12
C have a relative mass of
6
___12__ amu
120__ amu

1


100




106
H have a relative mass of 100 amu.
1
1
H have a relative mass of ______ amu.
1
100_
106
12
C have a relative mass of
6
______ amu
12
C have a relative mass of ______ amu
6



1
One H has a relative mass of __1___ amu.
1


10

1
H have a relative mass of _10___ amu.
1
12
One
C has a relative mass of
6
10
12
C have a relative mass of
6
___12__ amu
120__ amu

1


100




106
H have a relative mass of 100 amu.
1
1
H have a relative mass of ______ amu.
1
100_
106
12
C have a relative mass of
6
1200_ amu
12
C have a relative mass of ______ amu
6



1
One H has a relative mass of __1___ amu.
1


10

1
H have a relative mass of _10___ amu.
1
12
One
C has a relative mass of
6
10
12
C have a relative mass of
6
___12__ amu
120__ amu

1


100




106
H have a relative mass of 100 amu.
1
100_
1
H have a relative mass of 1million amu. 106
1
12
C have a relative mass of
6
1200_ amu
12
C have a relative mass of ______ amu
6



1
One H has a relative mass of __1___ amu.
1


10

1
H have a relative mass of _10___ amu.
1
12
One
C has a relative mass of
6
10
12
C have a relative mass of
6
___12__ amu
120__ amu

1


100




106
H have a relative mass of 100 amu.
1
100_
1
H have a relative mass of 1million amu. 106
1
12
C have a relative mass of
6
1200_ amu
12
C have a relative mass of 12 million amu
6

1a) The mass of the sample of carbon is
always ______ more than an equivalent
number of hydrogens
Mole Unit Questions

1a) The mass of the sample of carbon is
always _12x__ more than an equivalent
number of hydrogens
Mole Unit Question 1

6.02x1023 atoms of hydrogen have a mass of
1.0 g that equals the atomic mass of
hydrogen found on the periodic table

6.02x1023 atoms of carbon have a mass of
12.0 g that equals the atomic mass of carbon
found on the periodic table
Mole Unit Question 1

6.02x1023 atoms of hydrogen have a mass of
1.0 g that equals the atomic mass of
hydrogen found on the periodic table

6.02x1023 atoms of carbon have a mass of
12.0 g that equals the atomic mass of carbon
found on the periodic table
Questions 2-8
2. What is the mass of 3.01 x 1023 carbon–12 atoms?
3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half
of 12 g = 6g
3. What is the mass of 1.51 x 1023 carbon–12 atoms?
1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one
fourth of 12 g = 3g
4. What is the mass of 12.04 x 1023 carbon–12 atoms?
12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g =
24 g
Questions 2-8
2. What is the mass of 3.01 x 1023 carbon–12 atoms?
3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half
of 12 g = 6g
3. What is the mass of 1.51 x 1023 carbon–12 atoms?
1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one
fourth of 12 g = 3g
4. What is the mass of 12.04 x 1023 carbon–12 atoms?
12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g =
24 g
Questions 2-8
2. What is the mass of 3.01 x 1023 carbon–12 atoms?
3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half
of 12 g = 6g
3. What is the mass of 1.51 x 1023 carbon–12 atoms?
1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one
fourth of 12 g = 3g
4. What is the mass of 12.04 x 1023 carbon–12 atoms?
12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g =
24 g
Questions 2-8
2. What is the mass of 3.01 x 1023 carbon–12 atoms?
3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half
of 12 g = 6g
3. What is the mass of 1.51 x 1023 carbon–12 atoms?
1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one
fourth of 12 g = 3g
4. What is the mass of 12.04 x 1023 carbon–12 atoms?
12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g =
24 g
Questions 2-8
2. What is the mass of 3.01 x 1023 carbon–12 atoms?
3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half
of 12 g = 6g
3. What is the mass of 1.51 x 1023 carbon–12 atoms?
1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one
fourth of 12 g = 3g
4. What is the mass of 12.04 x 1023 carbon–12 atoms?
12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g =
24 g
Questions 2-8
2. What is the mass of 3.01 x 1023 carbon–12 atoms?
3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half
of 12 g = 6g
3. What is the mass of 1.51 x 1023 carbon–12 atoms?
1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one
fourth of 12 g = 3g
4. What is the mass of 12.04 x 1023 carbon–12 atoms?
12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g =
24 g
Questions 2-8
2. What is the mass of 3.01 x 1023 carbon–12 atoms?
3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half
of 12 g = 6g
3. What is the mass of 1.51 x 1023 carbon–12 atoms?
1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one
fourth of 12 g = 3g
4. What is the mass of 12.04 x 1023 carbon–12 atoms?
12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g =
24 g
Questions 2-8
2. What is the mass of 3.01 x 1023 carbon–12 atoms?
3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half
of 12 g = 6g
3. What is the mass of 1.51 x 1023 carbon–12 atoms?
1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one
fourth of 12 g = 3g
4. What is the mass of 12.04 x 1023 carbon–12 atoms?
12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g =
24 g
Questions 2-8
2. What is the mass of 3.01 x 1023 carbon–12 atoms?
3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half
of 12 g = 6g
3. What is the mass of 1.51 x 1023 carbon–12 atoms?
1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one
fourth of 12 g = 3g
4. What is the mass of 12.04 x 1023 carbon–12 atoms?
12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g =
24 g
Questions 2-8
2. What is the mass of 3.01 x 1023 carbon–12 atoms?
3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half
of 12 g = 6g
3. What is the mass of 1.51 x 1023 carbon–12 atoms?
1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one
fourth of 12 g = 3g
4. What is the mass of 12.04 x 1023 carbon–12 atoms?
12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g =
24 g
Questions 2-8
5. What is the mass of 24.01 x 1023 carbon–12 atoms?
24.01 x 1023 is four times 6.02 x 1023 therefore the mass
will be four time 12 g = 48g
Questions 2-8
5. What is the mass of 24.01 x 1023 carbon–12 atoms?
24.01 x 1023 is four times 6.02 x 1023 therefore the mass
will be four time 12 g = 48g
Questions 2-8
5. What is the mass of 24.01 x 1023 carbon–12 atoms?
24.01 x 1023 is four times 6.02 x 1023 therefore the mass
will be four time 12 g = 48g
Questions 2-8
6. How many carbon atoms are in 36.0 g

36.0g is three times 12.0g therefore there are three
times 6.02x1023 C = 18.06x1023C = 1.806x1024 C
7. How many carbon atoms are in 4.0 g

4.0 g is one third of 12.0g therefore one third of
6.02x1023 C = 2.01x1023C
8. How many carbon atoms are in 1.0 g

1.0 g is one twelfth of 12.0g therefore one twelfth
6.02x1023 C = 5.01x1022 C
Questions 2-8
6. How many carbon atoms are in 36.0 g

36.0g is three times 12.0g therefore there are three
times 6.02x1023 C = 18.06x1023C = 1.806x1024 C
7. How many carbon atoms are in 4.0 g

4.0 g is one third of 12.0g therefore one third of
6.02x1023 C = 2.01x1023C
8. How many carbon atoms are in 1.0 g

1.0 g is one twelfth of 12.0g therefore one twelfth
6.02x1023 C = 5.01x1022 C
Questions 2-8
6. How many carbon atoms are in 36.0 g

36.0g is three times 12.0g therefore there are three
times 6.02x1023 C = 18.06x1023C = 1.806x1024 C
7. How many carbon atoms are in 4.0 g

4.0 g is one third of 12.0g therefore one third of
6.02x1023 C = 2.01x1023C
8. How many carbon atoms are in 1.0 g

1.0 g is one twelfth of 12.0g therefore one twelfth
6.02x1023 C = 5.01x1022 C
Questions 2-8
6. How many carbon atoms are in 36.0 g

36.0g is three times 12.0g therefore there are three
times 6.02x1023 C = 18.06x1023C = 1.806x1024 C
7. How many carbon atoms are in 4.0 g

4.0 g is one third of 12.0g therefore one third of
6.02x1023 C = 2.01x1023C
8. How many carbon atoms are in 1.0 g

1.0 g is one twelfth of 12.0g therefore one twelfth
6.02x1023 C = 5.01x1022 C
Questions 2-8
6. How many carbon atoms are in 36.0 g

36.0g is three times 12.0g therefore there are three
times 6.02x1023 C = 18.06x1023C = 1.806x1024 C
7. How many carbon atoms are in 4.0 g

4.0 g is one third of 12.0g therefore one third of
6.02x1023 C = 2.01x1023C
8. How many carbon atoms are in 1.0 g

1.0 g is one twelfth of 12.0g therefore one twelfth
6.02x1023 C = 5.01x1022 C
Questions 2-8
6. How many carbon atoms are in 36.0 g

36.0g is three times 12.0g therefore there are three
times 6.02x1023 C = 18.06x1023C = 1.806x1024 C
7. How many carbon atoms are in 4.0 g

4.0 g is one third of 12.0g therefore one third of
6.02x1023 C = 2.01x1023C
8. How many carbon atoms are in 1.0 g

1.0 g is one twelfth of 12.0g therefore one twelfth
6.02x1023 C = 5.01x1022 C
Questions 2-8
6. How many carbon atoms are in 36.0 g

36.0g is three times 12.0g therefore there are three
times 6.02x1023 C = 18.06x1023C = 1.806x1024 C
7. How many carbon atoms are in 4.0 g

4.0 g is one third of 12.0g therefore one third of
6.02x1023 C = 2.01x1023C
8. How many carbon atoms are in 1.0 g

1.0 g is one twelfth of 12.0g therefore one twelfth
6.02x1023 C = 5.01x1022 C
Questions 2-8
6. How many carbon atoms are in 36.0 g

36.0g is three times 12.0g therefore there are three
times 6.02x1023 C = 18.06x1023C = 1.806x1024 C
7. How many carbon atoms are in 4.0 g

4.0 g is one third of 12.0g therefore one third of
6.02x1023 C = 2.01x1023C
8. How many carbon atoms are in 1.0 g

1.0 g is one twelfth of 12.0g therefore one twelfth
6.02x1023 C = 5.01x1022 C
Questions 2-8
6. How many carbon atoms are in 36.0 g

36.0g is three times 12.0g therefore there are three
times 6.02x1023 C = 18.06x1023C = 1.806x1024 C
7. How many carbon atoms are in 4.0 g

4.0 g is one third of 12.0g therefore one third of
6.02x1023 C = 2.01x1023C
8. How many carbon atoms are in 1.0 g

1.0 g is one twelfth of 12.0g therefore one twelfth
6.02x1023 C = 5.01x1022 C
Bonus Question
6.02x1023 C = 12.0 g C
1.0 x106 C
6.02x1023 C
=
x
12.0 g C
Cross multiply
1.0 x106 C multiplied by 12.0 g then divided by
6.0x1023 C = 1.99x10-17g C
Bonus Question
6.02x1023 C = 12.0 g C
1.0 x106 C
6.02x1023 C
=
x
12.0 g C
Cross multiply
1.0 x106 C multiplied by 12.0 g then divided by
6.0x1023 C = 1.99x10-17g C
Bonus Question
6.02x1023 C = 12.0 g C
1.0 x106 C
6.02x1023 C
=
x
12.0 g C
Cross multiply
1.0 x106 C multiplied by 12.0 g then divided by
6.0x1023 C = 1.99x10-17g C
Bonus Question
6.02x1023 C = 12.0 g C
1.0 x106 C
6.02x1023 C
=
x
12.0 g C
Cross multiply
1.0 x106 C multiplied by 12.0 g then divided by
6.0x1023 C = 1.99x10-17g C
Bonus Question
6.02x1023 C = 12.0 g C
1.0 x106 C
6.02x1023 C
=
x
12.0 g C
Cross multiply
1.0 x106 C multiplied by 12.0 g then divided by
6.0x1023 C = 1.99x10-17g C
Bonus Question
6.02x1023 C = 12.0 g C
1.0 x106 C
6.02x1023 C
=
x
12.0 g C
Cross multiply
1.0 x106 C multiplied by 12.0 g then divided by
6.0x1023 C = 1.99x10-17g C
Bonus Question
6.02x1023 C = 12.0 g C
1.0 x106 C
6.02x1023 C
=
x
12.0 g C
Cross multiply
1.0 x106 C multiplied by 12.0 g then divided by
6.0x1023 C = 1.99x10-17g C
Bonus Question
Cross multiply
1.0 x106 C multiplied by 12.0 g then divided by
6.0x1023 C
1.0 x106 C ( 12.0 g C ) = 1.99x10-17g C
(6.02x1023 C )
Bonus Question
Cross multiply
1.0 x106 C multiplied by 12.0 g then divided by
6.0x1023 C
1.0 x106 C ( 12.0 g C ) = 1.99x10-17g C
(6.02x1023 C )
Bonus Question
Cross multiply
1.0 x106 C multiplied by 12.0 g then divided by
6.0x1023 C
1.0 x106 C ( 12.0 g C ) = 1.99x10-17g C
(6.02x1023 C )
Bonus Question
Cross multiply
1.0 x106 C multiplied by 12.0 g then divided by
6.0x1023 C
1.0 x106 C ( 12.0 g C ) = 1.99x10-17g C
(6.02x1023 C )
Bonus Question
Cross multiply
1.0 x106 C multiplied by 12.0 g then divided by
6.0x1023 C
1.0 x106 C ( 12.0 g C ) = 1.99x10-17g C
(6.02x1023 C )
6.02 x 1023 = P.T. grams


9. How many silver atoms are in 25.0 grams of
silver?
6.02 x 1023 = P.T. grams



9. How many silver atoms are in 25.0 grams of
silver?
25.0g Ag
6.02 x 1023 = P.T. grams



9. How many silver atoms are in 25.0 grams of
silver?
25.0g Ag 6.02 x 1023 Ag =
6.02 x 1023 = P.T. grams




9. How many silver atoms are in 25.0 grams of
silver?
25.0g Ag 6.02 x 1023 Ag =
107.9g
6.02 x 1023 = P.T. grams




9. How many silver atoms are in 25.0 grams of
silver?
25.0g Ag 6.02 x 1023 Ag =1.39x1023 Ag
107.9Ag
6.02 x 1023 = P.T. grams



10. What is the mass of 1.00 x 109 silver
atoms?
1.00 x 109 Ag
6.02 x 1023 = P.T. grams




10. What is the mass of 1.00 x 109 silver
atoms?
1.00 x 109 Ag ( 107.9 g Ag ) =
6.02 x 1023 = P.T. grams




10. What is the mass of 1.00 x 109 silver
atoms?
1.00 x 109 Ag ( 107.9 g Ag ) =
(6.02 x 1023 Ag )
6.02 x 1023 = P.T. grams




10. What is the mass of 1.00 x 109 silver
atoms?
1.00 x 109 Ag ( 107.9 g Ag ) = 1.79 x 10-13 g Ag
(6.02 x 1023 Ag )
6.02 x 1023 = P.T. grams








11. What is the mass of 2.00 x 1023 gold
atoms?
2.00 x 1023 Au ( 197.0 g Au
) = 65.4 g Au
(6.02 x 1023 Au )
12. How many gold atoms are in 100 grams of
gold?
100g Au 6.02 x 1023 Au = 3.06 X1023 Au
197.0 g Au
6.02 x 1023 = P.T. grams








11. What is the mass of 2.00 x 1023 gold
atoms?
2.00 x 1023 Au ( 197.0 g Au
) = 65.4 g Au
(6.02 x 1023 Au )
12. How many gold atoms are in 100 grams of
gold?
100g Au 6.02 x 1023 Au = 3.06 X1023 Au
197.0 g Au
6.02 x 1023 = P.T. grams








11. What is the mass of 2.00 x 1023 gold
atoms?
2.00 x 1023 Au ( 197.0 g Au
) = 65.4 g Au
(6.02 x 1023 Au )
12. How many gold atoms are in 100 grams of
gold?
100g Au 6.02 x 1023 Au = 3.06 X1023 Au
197.0 g Au
6.02 x 1023 = P.T. grams








11. What is the mass of 2.00 x 1023 gold
atoms?
2.00 x 1023 Au ( 197.0 g Au
) = 65.4 g Au
(6.02 x 1023 Au )
12. How many gold atoms are in 100 grams of
gold?
100g Au 6.02 x 1023 Au = 3.06 X1023 Au
197.0 g Au
6.02 x 1023 = P.T. grams








11. What is the mass of 2.00 x 1023 gold
atoms?
2.00 x 1023 Au ( 197.0 g Au
) = 65.4 g Au
(6.02 x 1023 Au )
12. How many gold atoms are in 100 grams of
gold?
100g Au 6.02 x 1023 Au = 3.06 X1023 Au
197.0 g Au
6.02 x 1023 = P.T. grams








11. What is the mass of 2.00 x 1023 gold
atoms?
2.00 x 1023 Au ( 197.0 g Au
) = 65.4 g Au
(6.02 x 1023 Au )
12. How many gold atoms are in 100 grams of
gold?
100g Au 6.02 x 1023 Au = 3.06 X1023 Au
197.0 g Au
6.02 x 1023 = P.T. grams








11. What is the mass of 2.00 x 1023 gold
atoms?
2.00 x 1023 Au ( 197.0 g Au
) = 65.4 g Au
(6.02 x 1023 Au )
12. How many gold atoms are in 100 grams of
gold?
100g Au 6.02 x 1023 Au = 3.06 X1023 Au
197.0 g Au
6.02 x 1023 = P.T. grams








11. What is the mass of 2.00 x 1023 gold
atoms?
2.00 x 1023 Au ( 197.0 g Au
) = 65.4 g Au
(6.02 x 1023 Au )
12. How many gold atoms are in 100 grams of
gold?
100g Au 6.02 x 1023 Au = 3.06 X1023 Au
197.0 g Au
6.02 x 1023 = P.T. grams

13. What is the mass 2.00 x 1024 platinum
atoms?

2.00 x 1024 Pt ( 195.1 g Pt
) = 648 g Pt
(6.02 x 1023 Pt )


14. What is the mass of 2.00 x 1022 silicon
atoms?

2.00 x 1022 Si ( 28.1g Si
) = .93 g Si
(6.02 x 1023 Si )

6.02 x 1023 = P.T. grams

13. What is the mass 2.00 x 1024 platinum
atoms?

2.00 x 1024 Pt ( 195.1 g Pt
) = 648 g Pt
(6.02 x 1023 Pt )


14. What is the mass of 2.00 x 1022 silicon
atoms?

2.00 x 1022 Si ( 28.1g Si
) = .93 g Si
(6.02 x 1023 Si )

6.02 x 1023 = P.T. grams

13. What is the mass 2.00 x 1024 platinum
atoms?

2.00 x 1024 Pt ( 195.1 g Pt
) = 648 g Pt
(6.02 x 1023 Pt )


14. What is the mass of 2.00 x 1022 silicon
atoms?

2.00 x 1022 Si ( 28.1g Si
) = .93 g Si
(6.02 x 1023 Si )

6.02 x 1023 = P.T. grams

13. What is the mass 2.00 x 1024 platinum
atoms?

2.00 x 1024 Pt ( 195.1 g Pt
) = 648 g Pt
(6.02 x 1023 Pt )


14. What is the mass of 2.00 x 1022 silicon
atoms?

2.00 x 1022 Si ( 28.1g Si
) = .93 g Si
(6.02 x 1023 Si )

6.02 x 1023 = P.T. grams

13. What is the mass 2.00 x 1024 platinum
atoms?

2.00 x 1024 Pt ( 195.1 g Pt
) = 648 g Pt
(6.02 x 1023 Pt )


14. What is the mass of 2.00 x 1022 silicon
atoms?

2.00 x 1022 Si ( 28.1g Si
) = .93 g Si
(6.02 x 1023 Si )

6.02 x 1023 = P.T. grams

13. What is the mass 2.00 x 1024 platinum
atoms?

2.00 x 1024 Pt ( 195.1 g Pt
) = 648 g Pt
(6.02 x 1023 Pt )


14. What is the mass of 2.00 x 1022 silicon
atoms?

2.00 x 1022 Si ( 28.1g Si
) = .93 g Si
(6.02 x 1023 Si )

6.02 x 1023 = P.T. grams

13. What is the mass 2.00 x 1024 platinum
atoms?

2.00 x 1024 Pt ( 195.1 g Pt
) = 648 g Pt
(6.02 x 1023 Pt )


14. What is the mass of 2.00 x 1022 silicon
atoms?

2.00 x 1022 Si ( 28.1g Si
) = .93 g Si
(6.02 x 1023 Si )

6.02 x 1023 = P.T. grams

13. What is the mass 2.00 x 1024 platinum
atoms?

2.00 x 1024 Pt ( 195.1 g Pt
) = 648 g Pt
(6.02 x 1023 Pt )


14. What is the mass of 2.00 x 1022 silicon
atoms?

2.00 x 1022 Si ( 28.1g Si
) = .93 g Si
(6.02 x 1023 Si )

6.02 x 1023 = P.T. grams

13. What is the mass 2.00 x 1024 platinum
atoms?

2.00 x 1024 Pt ( 195.1 g Pt
) = 648 g Pt
(6.02 x 1023 Pt )


14. What is the mass of 2.00 x 1022 silicon
atoms?

2.00 x 1022 Si ( 28.1g Si
) = .93 g Si
(6.02 x 1023 Si )

6.02 x 1023 = P.T. grams

15. How many silicon atoms are in 12.0 grams of
silicon atoms?

12g Si

16. How many lead atoms are in 12.0 grams of lead?

12.0g Pb

6.02 x 1023 Si = 2.57X1023 Si
28.1g Si
6.02 x 1023 Pb = 3.49 X1022 Pb
207.2 g Pb
6.02 x 1023 = P.T. grams

15. How many silicon atoms are in 12.0 grams of
silicon atoms?

12g Si

16. How many lead atoms are in 12.0 grams of lead?

12.0g Pb

6.02 x 1023 Si = 2.57X1023 Si
28.1g Si
6.02 x 1023 Pb = 3.49 X1022 Pb
207.2 g Pb
6.02 x 1023 = P.T. grams

15. How many silicon atoms are in 12.0 grams of
silicon atoms?

12g Si

16. How many lead atoms are in 12.0 grams of lead?

12.0g Pb

6.02 x 1023 Si = 2.57X1023 Si
28.1g Si
6.02 x 1023 Pb = 3.49 X1022 Pb
207.2 g Pb
6.02 x 1023 = P.T. grams

15. How many silicon atoms are in 12.0 grams of
silicon atoms?

12g Si

16. How many lead atoms are in 12.0 grams of lead?

12.0g Pb

6.02 x 1023 Si = 2.57X1023 Si
28.1g Si
6.02 x 1023 Pb = 3.49 X1022 Pb
207.2 g Pb
6.02 x 1023 = P.T. grams

15. How many silicon atoms are in 12.0 grams of
silicon atoms?

12g Si

16. How many lead atoms are in 12.0 grams of lead?

12.0g Pb

6.02 x 1023 Si = 2.57X1023 Si
28.1g Si
6.02 x 1023 Pb = 3.49 X1022 Pb
207.2 g Pb
6.02 x 1023 = P.T. grams

15. How many silicon atoms are in 12.0 grams of
silicon atoms?

12g Si

16. How many lead atoms are in 12.0 grams of lead?

12.0g Pb

6.02 x 1023 Si = 2.57X1023 Si
28.1g Si
6.02 x 1023 Pb = 3.49 X1022 Pb
207.2 g Pb
6.02 x 1023 = P.T. grams

15. How many silicon atoms are in 12.0 grams of
silicon atoms?

12g Si

16. How many lead atoms are in 12.0 grams of lead?

12.0g Pb

6.02 x 1023 Si = 2.57X1023 Si
28.1g Si
6.02 x 1023 Pb = 3.49 X1022 Pb
207.2 g Pb
6.02 x 1023 = P.T. grams

15. How many silicon atoms are in 12.0 grams of
silicon atoms?

12g Si

16. How many lead atoms are in 12.0 grams of lead?

12.0g Pb

6.02 x 1023 Si = 2.57X1023 Si
28.1g Si
6.02 x 1023 Pb = 3.49 X1022 Pb
207.2 g Pb
6.02 x 1023 = P.T. grams

15. How many silicon atoms are in 12.0 grams of
silicon atoms?

12g Si

16. How many lead atoms are in 12.0 grams of lead?

12.0g Pb

6.02 x 1023 Si = 2.57X1023 Si
28.1g Si
6.02 x 1023 Pb = 3.49 X1022 Pb
207.2 g Pb
6.02 x 1023 = P.T. grams

17. What is the mass of 2.0 x109 Pb atoms?

2.0 x109 Pb 207.2g Pb
=6.88x10-13g Pb
6.02 x 1023 Pb

18. What is the mass of 1.51 x1023 C6H12O6?
1.51 x1023 C6H12O6 (180 g C6H12O6 )= 45.0g C6H12O6
(6.02 x 1023 C6H12O6)

6.02 x 1023 = P.T. grams

17. What is the mass of 2.0 x109 Pb atoms?

2.0 x109 Pb 207.2g Pb
=6.88x10-13g Pb
6.02 x 1023 Pb

18. What is the mass of 1.51 x1023 C6H12O6?
1.51 x1023 C6H12O6 (180 g C6H12O6 )= 45.0g C6H12O6
(6.02 x 1023 C6H12O6)

6.02 x 1023 = P.T. grams

17. What is the mass of 2.0 x109 Pb atoms?

2.0 x109 Pb 207.2g Pb
=6.88x10-13g Pb
6.02 x 1023 Pb

18. What is the mass of 1.51 x1023 C6H12O6?
1.51 x1023 C6H12O6 (180 g C6H12O6 )= 45.0g C6H12O6
(6.02 x 1023 C6H12O6)

6.02 x 1023 = P.T. grams

17. What is the mass of 2.0 x109 Pb atoms?

2.0 x109 Pb 207.2g Pb
=6.88x10-13g Pb
6.02 x 1023 Pb

18. What is the mass of 1.51 x1023 C6H12O6?
1.51 x1023 C6H12O6 (180 g C6H12O6 )= 45.0g C6H12O6
(6.02 x 1023 C6H12O6)

6.02 x 1023 = P.T. grams

17. What is the mass of 2.0 x109 Pb atoms?

2.0 x109 Pb 207.2g Pb
=6.88x10-13g Pb
6.02 x 1023 Pb

18. What is the mass of 1.51 x1023 C6H12O6?
1.51 x1023 C6H12O6 (180 g C6H12O6 )= 45.0g C6H12O6
(6.02 x 1023 C6H12O6)

6.02 x 1023 = P.T. grams

17. What is the mass of 2.0 x109 Pb atoms?

2.0 x109 Pb 207.2g Pb
=6.88x10-13g Pb
6.02 x 1023 Pb

18. What is the mass of 1.51 x1023 C6H12O6?
1.51 x1023 C6H12O6 (180 g C6H12O6 )= 45.0g C6H12O6
(6.02 x 1023 C6H12O6)

6.02 x 1023 = P.T. grams

17. What is the mass of 2.0 x109 Pb atoms?

2.0 x109 Pb 207.2g Pb
=6.88x10-13g Pb
6.02 x 1023 Pb

18. What is the mass of 1.51 x1023 C6H12O6?
1.51 x1023 C6H12O6 (180 g C6H12O6 ) )= 45.0g C6H12O6
(6.02 x 1023 C6H12O6)

6.02 x 1023 = P.T. grams

17. What is the mass of 2.0 x109 Pb atoms?

2.0 x109 Pb 207.2g Pb
=6.88x10-13g Pb
6.02 x 1023 Pb

18. What is the mass of 1.51 x1023 C6H12O6?
1.51 x1023 C6H12O6 (180 g C6H12O6 )= 45.0g C6H12O6
(6.02 x 1023 C6H12O6)

6.02 x 1023 = P.T. grams

17. What is the mass of 2.0 x109 Pb atoms?

2.0 x109 Pb 207.2g Pb
=6.88x10-13g Pb
6.02 x 1023 Pb

18. What is the mass of 1.51 x1023 C6H12O6?
1.51 x1023 C6H12O6 (180 g C6H12O6 )= 45.0g C6H12O6
(6.02 x 1023 C6H12O6)

6.02 x 1023 = P.T. grams

19. How many copper (II) nitrate are in 454g Cu(NO3)2

454g Cu(NO3)2 6.02 x 1023 Cu(NO3)2 =1.46x1024 Cu(NO )
187.5g Cu(NO3)2
3 2
6.02 x 1023 = P.T. grams

19. How many copper (II) nitrate are in 454g Cu(NO3)2

454g Cu(NO3)2 6.02 x 1023 Cu(NO3)2 =1.46x1024 Cu(NO )
187.5g Cu(NO3)2
3 2
6.02 x 1023 = P.T. grams

19. How many copper (II) nitrate are in 454g Cu(NO3)2

454g Cu(NO3)2 6.02 x 1023 Cu(NO3)2 =1.46x1024 Cu(NO )
187.5g Cu(NO3)2
3 2
6.02 x 1023 = P.T. grams

19. How many copper (II) nitrate are in 454g Cu(NO3)2

454g Cu(NO3)2 6.02 x 1023 Cu(NO3)2 =1.46x1024 Cu(NO )
187.5g Cu(NO3)2
3 2
6.02 x 1023 = P.T. grams

19. How many copper (II) nitrate are in 454g Cu(NO3)2

454g Cu(NO3)2 6.02 x 1023 Cu(NO3)2 =1.46x1024 Cu(NO )
187.5g Cu(NO3)2
3 2
Mole Concept









6.02 x 1023 is known as _____________________ number.
6.02 x 1023 particles is known as __________of particles
just as 12 particles is known as a _______
The mass of 1 particle in atomic mass units (amu) is
equivalent to the mass of 6.02 x 1023 of those particles or
___________ of those particle.
__________ g = 1 mole of particles = 6.02 x 1023 particles
atomic mass / molecular mass / formula mass
12
___________ g C = 1.00 moles of C = 6.02 x 1023 C atoms
6
Mole Concept









6.02 x 1023 is known as ___Avogadro’s___ number.
6.02 x 1023 particles is known as __________of particles
just as 12 particles is known as a _______
The mass of 1 particle in atomic mass units (amu) is
equivalent to the mass of 6.02 x 1023 of those particles or
___________ of those particle.
__________ g = 1 mole of particles = 6.02 x 1023 particles
atomic mass / molecular mass / formula mass
12
___________ g C = 1.00 moles of C = 6.02 x 1023 C atoms
6
Mole Concept









6.02 x 1023 is known as ___Avogadro’s___ number.
6.02 x 1023 particles is known as _1mole__of particles
just as 12 particles is known as a _______
The mass of 1 particle in atomic mass units (amu) is
equivalent to the mass of 6.02 x 1023 of those particles or
___________ of those particle.
__________ g = 1 mole of particles = 6.02 x 1023 particles
atomic mass / molecular mass / formula mass
12
___________ g C = 1.00 moles of C = 6.02 x 1023 C atoms
6
Mole Concept









6.02 x 1023 is known as ___Avogadro’s___ number.
6.02 x 1023 particles is known as _1mole__of particles
just as 12 particles is known as a _dozen__
The mass of 1 particle in atomic mass units (amu) is
equivalent to the mass of 6.02 x 1023 of those particles or
___________ of those particle.
__________ g = 1 mole of particles = 6.02 x 1023 particles
atomic mass / molecular mass / formula mass
12
___________ g C = 1.00 moles of C = 6.02 x 1023 C atoms
6
Mole Concept









6.02 x 1023 is known as ___Avogadro’s___ number.
6.02 x 1023 particles is known as _1mole__of particles
just as 12 particles is known as a _dozen__
The mass of 1 particle in atomic mass units (amu) is
equivalent to the mass of 6.02 x 1023 of those particles or
__1mole__ of those particle.
__________ g = 1 mole of particles = 6.02 x 1023 particles
atomic mass / molecular mass / formula mass
12
___________ g C = 1.00 moles of C = 6.02 x 1023 C atoms
6
Mole Concept









6.02 x 1023 is known as ___Avogadro’s___ number.
6.02 x 1023 particles is known as _1mole__of particles
just as 12 particles is known as a _dozen__
The mass of 1 particle in atomic mass units (amu) is
equivalent to the mass of 6.02 x 1023 of those particles or
__1mole__ of those particle.
___P.T.____ g = 1 mole of particles = 6.02 x 1023 particles
atomic mass / molecular mass / formula mass
12
___________ g C = 1.00 moles of C = 6.02 x 1023 C atoms
6
Mole Concept









6.02 x 1023 is known as ___Avogadro’s___ number.
6.02 x 1023 particles is known as _1mole__of particles
just as 12 particles is known as a _dozen__
The mass of 1 particle in atomic mass units (amu) is
equivalent to the mass of 6.02 x 1023 of those particles or
__1mole__ of those particle.
___P.T.____ g = 1 mole of particles = 6.02 x 1023 particles
atomic mass / molecular mass / formula mass
12
__12.0____ g C = 1.00 moles of C = 6.02 x 1023 C atoms
6
Mole Concept II







6.02 x 1023 = P.T. grams = 1.0 mole
20. What is the mass of 2 moles C?
2 mol C ( 12.0g C ) = 24.0g C
( 1mol C )
21. What is the mass of 7.5x1020C?
7.5x1020C ( 12.0g C ) = .015 g C
( 6.02x1023C)
Mole Concept II







6.02 x 1023 = P.T. grams = 1.0 mole
20. What is the mass of 2 moles C?
2 mol C ( 12.0g C ) = 24.0g C
( 1mol C )
21. What is the mass of 7.5x1020C?
7.5x1020C ( 12.0g C ) = .015 g C
( 6.02x1023C)
Mole Concept II







6.02 x 1023 = P.T. grams = 1.0 mole
20. What is the mass of 2 moles C?
2 mol C ( 12.0g C ) = 24.0g C
( 1mol C )
21. What is the mass of 7.5x1020C?
7.5x1020C ( 12.0g C ) = .015 g C
( 6.02x1023C)
Mole Concept II







6.02 x 1023 = P.T. grams = 1.0 mole
20. What is the mass of 2 moles C?
2 mol C ( 12.0g C ) = 24.0g C
( 1mol C )
21. What is the mass of 7.5x1020C?
7.5x1020C ( 12.0g C ) = .015 g C
( 6.02x1023C)
Mole Concept II







6.02 x 1023 = P.T. grams = 1.0 mole
20. What is the mass of 2 moles C?
2 mol C ( 12.0g C ) = 24.0g C
( 1mol C )
21. What is the mass of 7.5x1020C?
7.5x1020C ( 12.0g C ) = .015 g C
( 6.02x1023C)
Mole Concept II







6.02 x 1023 = P.T. grams = 1.0 mole
20. What is the mass of 2 moles C?
2 mol C ( 12.0g C ) = 24.0g C
( 1mol C )
21. What is the mass of 7.5x1020C?
7.5x1020C ( 12.0g C ) = .015 g C
( 6.02x1023C)
Mole Concept II







6.02 x 1023 = P.T. grams = 1.0 mole
20. What is the mass of 2 moles C?
2 mol C ( 12.0g C ) = 24.0g C
( 1mol C )
21. What is the mass of 7.5x1020C?
7.5x1020C ( 12.0g C ) = .015 g C
( 6.02x1023C)
Mole Concept II







6.02 x 1023 = P.T. grams = 1.0 mole
22. How many moles of carbon are in
1.806x1025 C atom sample?
1.806x1025 C(1mole C ) = 30 moles C
(6.02x1023 C)
23. How many atoms of carbon are in 100g
C?
100gC ( 6.02x1023C) = 5.04x1024 C atoms
( 12.0g C )
Mole Concept II







6.02 x 1023 = P.T. grams = 1.0 mole
22. How many moles of carbon are in
1.806x1025 C atom sample?
1.806x1025 C(1mole C ) = 30 moles C
(6.02x1023 C)
23. How many atoms of carbon are in 100g
C?
100gC ( 6.02x1023C) = 5.04x1024 C atoms
( 12.0g C )
Mole Concept II







6.02 x 1023 = P.T. grams = 1.0 mole
22. How many moles of carbon are in
1.806x1025 C atom sample?
1.806x1025 C(1mole C ) = 30 moles C
(6.02x1023 C)
23. How many atoms of carbon are in 100g
C?
100gC ( 6.02x1023C) = 5.04x1024 C atoms
( 12.0g C )
Mole Concept II







6.02 x 1023 = P.T. grams = 1.0 mole
22. How many moles of carbon are in
1.806x1025 C atom sample?
1.806x1025 C(1mole C ) = 30 moles C
(6.02x1023 C)
23. How many atoms of carbon are in 100g
C?
100gC ( 6.02x1023C) = 5.04x1024 C atoms
( 12.0g C )
Mole Concept II







6.02 x 1023 = P.T. grams = 1.0 mole
22. How many moles of carbon are in
1.806x1025 C atom sample?
1.806x1025 C(1mole C ) = 30 moles C
(6.02x1023 C)
23. How many atoms of carbon are in 100g
C?
100gC ( 6.02x1023C) = 5.04x1024 C atoms
( 12.0g C )
Mole Concept II







6.02 x 1023 = P.T. grams = 1.0 mole
22. How many moles of carbon are in
1.806x1025 C atom sample?
1.806x1025 C(1mole C ) = 30 moles C
(6.02x1023 C)
23. How many atoms of carbon are in 100g
C?
100gC ( 6.02x1023C) = 5.04x1024 C atoms
( 12.0g C )
Mole Concept II







6.02 x 1023 = P.T. grams = 1.0 mole
22. How many moles of carbon are in
1.806x1025 C atom sample?
1.806x1025 C(1mole C ) = 30 moles C
(6.02x1023 C)
23. How many atoms of carbon are in 100g
C?
100gC ( 6.02x1023C) = 5.04x1024 C atoms
( 12.0g C )
Mole Concept II







6.02 x 1023 = P.T. grams = 1.0 mole
22. How many moles of carbon are in
1.806x1025 C atom sample?
1.806x1025 C(1mole C ) = 30 moles C
(6.02x1023 C)
23. How many atoms of carbon are in 100g
C?
100gC ( 6.02x1023C) = 5.04x1024 C atoms
( 12.0g C )
Mole Concept II




6.02 x 1023 = P.T. grams = 1.0 mole
25. How many atoms of carbon are in .750
mole sample?
.750 mole C ( 6.02 x1023 C) = 4.52 x1023 C
1mole C
Mole Concept II




6.02 x 1023 = P.T. grams = 1.0 mole
25. How many atoms of carbon are in .750
mole sample?
.750 mole C ( 6.02 x1023 C) = 4.52 x1023 C
1mole C
Mole Concept II




6.02 x 1023 = P.T. grams = 1.0 mole
25. How many atoms of carbon are in .750
mole sample?
.750 mole C ( 6.02 x1023 C) = 4.52 x1023 C
1mole C
Mole Concept II




6.02 x 1023 = P.T. grams = 1.0 mole
25. How many atoms of carbon are in .750
mole sample?
.750 mole C ( 6.02 x1023 C) = 4.52 x1023 C
1mole C
6.02 x 1023 = P.T. grams

26. What is the mass of 3.50 moles of C2H5OH?

3.50 mole C2H5OH
46 g C2H5OH=161 g C2H5OH
1mole C2H5OH




27. How many molecules are in .250 mole of
C2H5OH?
.250 mole C2H5OH 6.02 x 1023 C2H5OH= 1.51 X1023
1mole C2H5OH
6.02 x 1023 = P.T. grams

26. What is the mass of 3.50 moles of C2H5OH?

3.50 mole C2H5OH
46 g C2H5OH=161 g C2H5OH
1mole C2H5OH




27. How many molecules are in .250 mole of
C2H5OH?
.250 mole C2H5OH 6.02 x 1023 C2H5OH= 1.51 X1023
1mole C2H5OH
6.02 x 1023 = P.T. grams

26. What is the mass of 3.50 moles of C2H5OH?

3.50 mole C2H5OH
46 g C2H5OH=161 g C2H5OH
1mole C2H5OH




27. How many molecules are in .250 mole of
C2H5OH?
.250 mole C2H5OH 6.02 x 1023 C2H5OH= 1.51 X1023
1mole C2H5OH
6.02 x 1023 = P.T. grams

26. What is the mass of 3.50 moles of C2H5OH?

3.50 mole C2H5OH
46 g C2H5OH=161 g C2H5OH
1mole C2H5OH




27. How many molecules are in .250 mole of
C2H5OH?
.250 mole C2H5OH 6.02 x 1023 C2H5OH= 1.51 X1023
1mole C2H5OH
6.02 x 1023 = P.T. grams

26. What is the mass of 3.50 moles of C2H5OH?

3.50 mole C2H5OH
46 g C2H5OH=161 g C2H5OH
1mole C2H5OH




27. How many molecules are in .250 mole of
C2H5OH?
.250 mole C2H5OH 6.02 x 1023 C2H5OH= 1.51 X1023
1mole C2H5OH
6.02 x 1023 = P.T. grams

26. What is the mass of 3.50 moles of C2H5OH?

3.50 mole C2H5OH
46 g C2H5OH=161 g C2H5OH
1mole C2H5OH




27. How many molecules are in .250 mole of
C2H5OH?
.250 mole C2H5OH 6.02 x 1023 C2H5OH= 1.51 X1023
1mole C2H5OH
6.02 x 1023 = P.T. grams

26. What is the mass of 3.50 moles of C2H5OH?

3.50 mole C2H5OH
46 g C2H5OH=161 g C2H5OH
1mole C2H5OH




27. How many molecules are in .250 mole of
C2H5OH?
.250 mole C2H5OH 6.02 x 1023 C2H5OH= 1.51 X1023
1mole C2H5OH
6.02 x 1023 = P.T. grams = 1.0 mole

28. How many moles of C2H5OH are in a 1.0x10 3
molecule sample of C2H5OH?

1.00 x 103 C2H5OH 1mole C2H5OH = 1.67x10-21moles
6.02 x 1023 C2H5OH

29. How many moles of C2H5OH are in 12.0 grams of
C2H5OH?

12.0 gC2H5OH 1mole C2H5OH= .26 moles C2H5OH
46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole

28. How many moles of C2H5OH are in a 1.0x10 3
molecule sample of C2H5OH?

1.00 x 103 C2H5OH 1mole C2H5OH = 1.67x10-21moles
6.02 x 1023 C2H5OH

29. How many moles of C2H5OH are in 12.0 grams of
C2H5OH?

12.0 gC2H5OH 1mole C2H5OH= .26 moles C2H5OH
46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole

28. How many moles of C2H5OH are in a 1.0x10 3
molecule sample of C2H5OH?

1.00 x 103 C2H5OH 1mole C2H5OH = 1.67x10-21moles
6.02 x 1023 C2H5OH

29. How many moles of C2H5OH are in 12.0 grams of
C2H5OH?

12.0 gC2H5OH 1mole C2H5OH= .26 moles C2H5OH
46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole

28. How many moles of C2H5OH are in a 1.0x10 3
molecule sample of C2H5OH?

1.00 x 103 C2H5OH 1mole C2H5OH = 1.67x10-21moles
6.02 x 1023 C2H5OH

29. How many moles of C2H5OH are in 12.0 grams of
C2H5OH?

12.0 gC2H5OH 1mole C2H5OH= .26 moles C2H5OH
46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole

28. How many moles of C2H5OH are in a 1.0x10 3
molecule sample of C2H5OH?

1.00 x 103 C2H5OH 1mole C2H5OH = 1.67x10-21moles
6.02 x 1023 C2H5OH

29. How many moles of C2H5OH are in 12.0 grams of
C2H5OH?

12.0 gC2H5OH 1mole C2H5OH= .26 moles C2H5OH
46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole

28. How many moles of C2H5OH are in a 1.0x10 3
molecule sample of C2H5OH?

1.00 x 103 C2H5OH 1mole C2H5OH = 1.67x10-21moles
6.02 x 1023 C2H5OH

29. How many moles of C2H5OH are in 12.0 grams of
C2H5OH?

12.0 gC2H5OH 1mole C2H5OH= .26 moles C2H5OH
46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole

28. How many moles of C2H5OH are in a 1.0x10 3
molecule sample of C2H5OH?

1.00 x 103 C2H5OH 1mole C2H5OH = 1.67x10-21moles
6.02 x 1023 C2H5OH

29. How many moles of C2H5OH are in 12.0 grams of
C2H5OH?

12.0 gC2H5OH 1mole C2H5OH= .26 moles C2H5OH
46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole

30. What is the mass of 2.50 moles of C2H5OH?


2.50 mole C2H5OH 46.0g C2H5OH
1mole C2H5OH

31. How many molecules are found in 454 g C2H5OH

454 gC2H5OH 6.02 X 10 23 C2H5OH= 5.94X10 24
46.0 g C2H5OH
= 115 g C2H5OH
6.02 x 1023 = P.T. grams = 1.0 mole

30. What is the mass of 2.50 moles of C2H5OH?


2.50 mole C2H5OH 46.0g C2H5OH
1mole C2H5OH

31. How many molecules are found in 454 g C2H5OH

454 gC2H5OH 6.02 X 10 23 C2H5OH= 5.94X10 24
46.0 g C2H5OH
= 115 g C2H5OH
6.02 x 1023 = P.T. grams = 1.0 mole

30. What is the mass of 2.50 moles of C2H5OH?


2.50 mole C2H5OH 46.0g C2H5OH
1mole C2H5OH

31. How many molecules are found in 454 g C2H5OH

454 gC2H5OH 6.02 X 10 23 C2H5OH= 5.94X10 24
46.0 g C2H5OH
= 115 g C2H5OH
6.02 x 1023 = P.T. grams = 1.0 mole

30. What is the mass of 2.50 moles of C2H5OH?


2.50 mole C2H5OH 46.0g C2H5OH
1mole C2H5OH

31. How many molecules are found in 454 g C2H5OH

454 gC2H5OH 6.02 X 10 23 C2H5OH= 5.94X10 24
46.0 g C2H5OH
= 115 g C2H5OH
6.02 x 1023 = P.T. grams = 1.0 mole

30. What is the mass of 2.50 moles of C2H5OH?


2.50 mole C2H5OH 46.0g C2H5OH
1mole C2H5OH

31. How many molecules are found in 454 g C2H5OH

454 gC2H5OH 6.02 X 10 23 C2H5OH= 5.94X10 24
46.0 g C2H5OH
= 115 g C2H5OH
6.02 x 1023 = P.T. grams = 1.0 mole

30. What is the mass of 2.50 moles of C2H5OH?


2.50 mole C2H5OH 46.0g C2H5OH
1mole C2H5OH

31. How many molecules are found in 454 g C2H5OH

454 gC2H5OH 6.02 X 10 23 C2H5OH= 5.94X10 24
46.0 g C2H5OH
= 115 g C2H5OH
6.02 x 1023 = P.T. grams = 1.0 mole

30. What is the mass of 2.50 moles of C2H5OH?


2.50 mole C2H5OH 46.0g C2H5OH
1mole C2H5OH

31. How many molecules are found in 454 g C2H5OH

454 gC2H5OH 6.02 X 10 23 C2H5OH= 5.94X10 24
46.0 g C2H5OH
= 115 g C2H5OH
6.02 x 1023 = P.T. grams = 1.0 mole

32. How many moles are in 0.10g Ca(NO3)2 ?


0.10 g Ca(NO3)2 1mole Ca(NO3)2 =.000609 mol Ca(NO )
164.1g Ca(NO3)2

33. What is the mass of .125 moles Ca(NO3)2?

.125 moles Ca(NO3)2 164.1 g Ca(NO3)2 =20.5 g Ca(NO3)2
1 mole Ca(NO3)2
3 2
6.02 x 1023 = P.T. grams = 1.0 mole

32. How many moles are in 0.10g Ca(NO3)2 ?


0.10 g Ca(NO3)2 1mole Ca(NO3)2 =.000609 mol Ca(NO )
164.1g Ca(NO3)2

33. What is the mass of .125 moles Ca(NO3)2?

.125 moles Ca(NO3)2 164.1 g Ca(NO3)2 =20.5 g Ca(NO3)2
1 mole Ca(NO3)2
3 2
6.02 x 1023 = P.T. grams = 1.0 mole

32. How many moles are in 0.10g Ca(NO3)2 ?


0.10 g Ca(NO3)2 1mole Ca(NO3)2 =.000609 mol Ca(NO )
164.1g Ca(NO3)2

33. What is the mass of .125 moles Ca(NO3)2?

.125 moles Ca(NO3)2 164.1 g Ca(NO3)2 =20.5 g Ca(NO3)2
1 mole Ca(NO3)2
3 2
6.02 x 1023 = P.T. grams = 1.0 mole

32. How many moles are in 0.10g Ca(NO3)2 ?


0.10 g Ca(NO3)2 1mole Ca(NO3)2 =.000609 mol Ca(NO )
164.1g Ca(NO3)2

33. What is the mass of .125 moles Ca(NO3)2?

.125 moles Ca(NO3)2 164.1 g Ca(NO3)2 =20.5 g Ca(NO3)2
1 mole Ca(NO3)2
3 2
6.02 x 1023 = P.T. grams = 1.0 mole

32. How many moles are in 0.10g Ca(NO3)2 ?


0.10 g Ca(NO3)2 1mole Ca(NO3)2 =.000609 mol Ca(NO )
164.1g Ca(NO3)2

33. What is the mass of .125 moles Ca(NO3)2?

.125 moles Ca(NO3)2 164.1 g Ca(NO3)2 =20.5 g Ca(NO3)2
1 mole Ca(NO3)2
3 2
6.02 x 1023 = P.T. grams = 1.0 mole

32. How many moles are in 0.10g Ca(NO3)2 ?


0.10 g Ca(NO3)2 1mole Ca(NO3)2 =.000609 mol Ca(NO )
164.1g Ca(NO3)2

33. What is the mass of .125 moles Ca(NO3)2?

.125 moles Ca(NO3)2 164.1 g Ca(NO3)2 =20.5 g Ca(NO3)2
1 mole Ca(NO3)2
3 2
6.02 x 1023 = P.T. grams = 1.0 mole

32. How many moles are in 0.10g Ca(NO3)2 ?


0.10 g Ca(NO3)2 1mole Ca(NO3)2 =.000609 mol Ca(NO )
164.1g Ca(NO3)2

33. What is the mass of .125 moles Ca(NO3)2?

.125 moles Ca(NO3)2 164.1 g Ca(NO3)2 =20.5 g Ca(NO3)2
1 mole Ca(NO3)2
3 2
6.02 x 1023 = P.T. grams = 1.0 mole

34. How many units are in 454g Ca(NO3)2?


454g Ca(NO3)2 6.02 X 10 23 Ca(NO3)2
164.1g Ca(NO3)2

35. What is the mass of .50 moles Ca(NO3)2?

.50 moles Ca(NO3)2(164.1g Ca(NO3)2=82.1g Ca(NO3)2
(1mol Ca(NO3)2

= 1.67x1024 Ca(NO3)2
6.02 x 1023 = P.T. grams = 1.0 mole

34. How many units are in 454g Ca(NO3)2?


454g Ca(NO3)2 6.02 X 10 23 Ca(NO3)2
164.1g Ca(NO3)2

35. What is the mass of .50 moles Ca(NO3)2?

.50 moles Ca(NO3)2(164.1g Ca(NO3)2=82.1g Ca(NO3)2
(1mol Ca(NO3)2

= 1.67x1024 Ca(NO3)2
6.02 x 1023 = P.T. grams = 1.0 mole

34. How many units are in 454g Ca(NO3)2?


454g Ca(NO3)2 6.02 X 10 23 Ca(NO3)2
164.1g Ca(NO3)2

35. What is the mass of .50 moles Ca(NO3)2?

.50 moles Ca(NO3)2(164.1g Ca(NO3)2=82.1g Ca(NO3)2
(1mol Ca(NO3)2

= 1.67x1024 Ca(NO3)2
6.02 x 1023 = P.T. grams = 1.0 mole

34. How many units are in 454g Ca(NO3)2?


454g Ca(NO3)2 6.02 X 10 23 Ca(NO3)2
164.1g Ca(NO3)2

35. What is the mass of .50 moles Ca(NO3)2?

.50 moles Ca(NO3)2(164.1g Ca(NO3)2=82.1g Ca(NO3)2
(1mol Ca(NO3)2

= 1.67x1024 Ca(NO3)2
6.02 x 1023 = P.T. grams = 1.0 mole

34. How many units are in 454g Ca(NO3)2?


454g Ca(NO3)2 6.02 X 10 23 Ca(NO3)2
164.1g Ca(NO3)2

35. What is the mass of .50 moles Ca(NO3)2?

.50 moles Ca(NO3)2(164.1g Ca(NO3)2=82.1g Ca(NO3)2
(1mol Ca(NO3)2

= 1.67x1024 Ca(NO3)2
6.02 x 1023 = P.T. grams = 1.0 mole

34. How many units are in 454g Ca(NO3)2?


454g Ca(NO3)2 6.02 X 10 23 Ca(NO3)2
164.1g Ca(NO3)2

35. What is the mass of .50 moles Ca(NO3)2?

.50 moles Ca(NO3)2(164.1g Ca(NO3)2=82.1g Ca(NO3)2
(1mol Ca(NO3)2

= 1.67x1024 Ca(NO3)2
6.02 x 1023 = P.T. grams = 1.0 mole

34. How many units are in 454g Ca(NO3)2?


454g Ca(NO3)2 6.02 X 10 23 Ca(NO3)2
164.1g Ca(NO3)2

35. What is the mass of .50 moles Ca(NO3)2?

.50 moles Ca(NO3)2(164.1g Ca(NO3)2=82.1g Ca(NO3)2
(1mol Ca(NO3)2

= 1.67x1024 Ca(NO3)2
6.02 x 1023 = P.T. grams = 1.0 mole

36. How many units are in .250 mol Ca(NO3)2?


.250 mole Ca(NO3)2 6.02x1023 Ca(NO3)2 =1.51x1023 Ca(NO3)2
1mole Ca(NO3)2

37. How many moles are in 1.0x109 Ca(NO3)2?

1.0x109 Ca(NO3)2 ( 1mole Ca(NO3)2= 1.66x10-15 mol
6.02x1023 Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole

36. How many units are in .250 mol Ca(NO3)2?


.250 mole Ca(NO3)2 6.02x1023 Ca(NO3)2 =1.51x1023 Ca(NO3)2
1mole Ca(NO3)2

37. How many moles are in 1.0x109 Ca(NO3)2?

1.0x109 Ca(NO3)2 ( 1mole Ca(NO3)2= 1.66x10-15 mol
6.02x1023 Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole

36. How many units are in .250 mol Ca(NO3)2?


.250 mole Ca(NO3)2 6.02x1023 Ca(NO3)2 =1.51x1023 Ca(NO3)2
1mole Ca(NO3)2

37. How many moles are in 1.0x109 Ca(NO3)2?

1.0x109 Ca(NO3)2 ( 1mole Ca(NO3)2= 1.66x10-15 mol
6.02x1023 Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole

36. How many units are in .250 mol Ca(NO3)2?


.250 mole Ca(NO3)2 6.02x1023 Ca(NO3)2 =1.51x1023 Ca(NO3)2
1mole Ca(NO3)2

37. How many moles are in 1.0x109 Ca(NO3)2?

1.0x109 Ca(NO3)2 ( 1mole Ca(NO3)2= 1.66x10-15 mol
6.02x1023 Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole

36. How many units are in .250 mol Ca(NO3)2?


.250 mole Ca(NO3)2 6.02x1023 Ca(NO3)2 =1.51x1023 Ca(NO3)2
1mole Ca(NO3)2

37. How many moles are in 1.0x109 Ca(NO3)2?

1.0x109 Ca(NO3)2 ( 1mole Ca(NO3)2= 1.66x10-15 mol
6.02x1023 Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole

36. How many units are in .250 mol Ca(NO3)2?


.250 mole Ca(NO3)2 6.02x1023 Ca(NO3)2 =1.51x1023 Ca(NO3)2
1mole Ca(NO3)2

37. How many moles are in 1.0x109 Ca(NO3)2?

1.0x109 Ca(NO3)2 ( 1mole Ca(NO3)2= 1.66x10-15 mol
6.02x1023 Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole

36. How many units are in .250 mol Ca(NO3)2?


.250 mole Ca(NO3)2 6.02x1023 Ca(NO3)2 =1.51x1023 Ca(NO3)2
1mole Ca(NO3)2

37. How many moles are in 1.0x109 Ca(NO3)2?

1.0x109 Ca(NO3)2 ( 1mole Ca(NO3)2= 1.66x10-15 mol
6.02x1023 Ca(NO3)2

Empirical Formula Extra Credit

50 g S 50 g O
50g S ( 1mole S) = 1.56 mol S1.0
( 32 g S ) 1.56
50 g O ( 1 mol O ) = 3.13 mol O2.0
( 16 g O ) 1.56

SO2




Empirical Formula Extra Credit

50 g S 50 g O
50g S ( 1mole S) = 1.56 mol S1.0
( 32 g S ) 1.56
50 g O ( 1 mol O ) = 3.13 mol O2.0
( 16 g O ) 1.56

SO2




Empirical Formula Extra Credit

50 g S 50 g O
50g S ( 1mole S) = 1.56 mol S1.0
( 32 g S ) 1.56
50 g O ( 1 mol O ) = 3.13 mol O2.0
( 16 g O ) 1.56

SO2




Empirical Formula Extra Credit

50 g S 50 g O
50g S ( 1mole S) = 1.56 mol S1.0
( 32 g S ) 1.56
50 g O ( 1 mol O ) = 3.13 mol O2.0
( 16 g O ) 1.56

SO2




Empirical Formula Extra Credit

50 g S 50 g O
50g S ( 1mole S) = 1.56 mol S1.0
( 32 g S ) 1.56
50 g O ( 1 mol O ) = 3.13 mol O2.0
( 16 g O ) 1.56

SO2




Empirical Formula Extra Credit

50 g S 50 g O
50g S ( 1mole S) = 1.56 mol S1.0
( 32 g S ) 1.56
50 g O ( 1 mol O ) = 3.13 mol O2.0
( 16 g O ) 1.56

SO2




Empirical Formula Extra Credit

50 g S 50 g O
50g S ( 1mole S) = 1.56 mol S1.0
( 32 g S ) 1.56
50 g O ( 1 mol O ) = 3.13 mol O2.0
( 16 g O ) 1.56

SO2




Empirical formula



Count the number of particles of each
element by determining the number of moles
of each element
Divide by the smallest number of moles in
order to determine the simplest whole
number ratio that the elements combine in to
form the compound
If the ratios are not whole numbers multiply
each ratio by a common factor until they are
whole number ratios
Empirical formula



Count the number of particles of each
element by determining the number of moles
of each element
Divide by the smallest number of moles in
order to determine the simplest whole
number ratio that the elements combine in to
form the compound
If the ratios are not whole numbers multiply
each ratio by a common factor until they are
whole number ratios
Empirical formula



Count the number of particles of each
element by determining the number of moles
of each element
Divide by the smallest number of moles in
order to determine the simplest whole
number ratio that the elements combine in to
form the compound
If the ratios are not whole numbers multiply
each ratio by a common factor until they are
whole number ratios
Empirical formula



Count the number of particles of each
element by determining the number of moles
of each element
Divide by the smallest number of moles in
order to determine the simplest whole
number ratio that the elements combine in to
form the compound
If the ratios are not whole numbers multiply
each ratio by a common factor until they are
whole number ratios
Empirical Formula

A) 35.6g Co and 64.4 g Cl
35.6gCo(1mole Co)=.603 mol Co1.0
(59.0gCo ) .603
64.4gCl ( 1mole Cl)=1.81 mol Cl3.0
( 35.5 g Cl) .603

CoCl3




Empirical Formula

A) 35.6g Co and 64.4 g Cl
35.6gCo(1mole Co)=.603 mol Co1.0
(59.0gCo ) .603
64.4gCl ( 1mole Cl)=1.81 mol Cl3.0
( 35.5 g Cl) .603

CoCl3




Empirical Formula

A) 35.6g Co and 64.4 g Cl
35.6gCo(1mole Co)=.603 mol Co1.0
(59.0gCo ) .603
64.4gCl ( 1mole Cl)=1.81 mol Cl3.0
( 35.5 g Cl) .603

CoCl3




Empirical Formula

A) 35.6g Co and 64.4 g Cl
35.6gCo(1mole Co)=.603 mol Co1.0
(59.0gCo ) .603
64.4gCl ( 1mole Cl)=1.81 mol Cl3.0
( 35.5 g Cl) .603

CoCl3




Empirical Formula

A) 35.6g Co and 64.4 g Cl
35.6gCo(1mole Co)=.603 mol Co1.0
(59.0gCo ) .603
64.4gCl ( 1mole Cl)=1.81 mol Cl3.0
( 35.5 g Cl) .603

CoCl3




Empirical Formula

A) 35.6g Co and 64.4 g Cl
35.6gCo(1mole Co)=.603 mol Co1.0
(59.0gCo ) .603
64.4gCl ( 1mole Cl)=1.81 mol Cl3.0
( 35.5 g Cl) .603

CoCl3




Empirical Formula

A) 35.6g Co and 64.4 g Cl
35.6gCo(1mole Co)=.603 mol Co1.0
(59.0gCo ) .603
64.4gCl ( 1mole Cl)=1.81 mol Cl3.0
( 35.5 g Cl) .603

CoCl3




Empirical Formula

A) 35.6g Co and 64.4 g Cl
35.6gCo(1mole Co)=.603 mol Co1.0
(59.0gCo ) .603
64.4gCl ( 1mole Cl)=1.81 mol Cl3.0
( 35.5 g Cl) .603

CoCl3




Empirical Formula

A) 35.6g Co and 64.4 g Cl
35.6gCo(1mole Co)=.603 mol Co1.0
(59.0gCo ) .603
64.4gCl ( 1mole Cl)=1.81 mol Cl3.0
( 35.5 g Cl) .603

CoCl3




Empirical Formula

A) 35.6g Co and 64.4 g Cl
35.6gCo(1mole Co)=.603 mol Co1.0
(59.0gCo ) .603
64.4gCl ( 1mole Cl)=1.81 mol Cl3.0
( 35.5 g Cl) .603

CoCl3




Empirical Formula

B) CoCl2
C) 52.9gAl and 47.1gO
52.9gAl(1mole Al)= 1.96 mol Al1.0
(27.0g Al ) 1.96
47.1 gO ( 1mole O)=2.94 mol O1.5
( 16.0 g O) 1.96

Al1.0O1.5 x 2 = Al2O3





Empirical Formula

B) CoCl2
C) 52.9gAl and 47.1gO
52.9gAl(1mole Al)= 1.96 mol Al1.0
(27.0g Al ) 1.96
47.1 gO ( 1mole O)=2.94 mol O1.5
( 16.0 g O) 1.96

Al1.0O1.5 x 2 = Al2O3





Empirical Formula

B) CoCl2
C) 52.9gAl and 47.1gO
52.9gAl(1mole Al)= 1.96 mol Al1.0
(27.0g Al ) 1.96
47.1 gO ( 1mole O)=2.94 mol O1.5
( 16.0 g O) 1.96

Al1.0O1.5 x 2 = Al2O3





Empirical Formula

B) CoCl2
C) 52.9gAl and 47.1gO
52.9gAl(1mole Al)= 1.96 mol Al1.0
(27.0g Al ) 1.96
47.1 gO ( 1mole O)=2.94 mol O1.5
( 16.0 g O) 1.96

Al1.0O1.5 x 2 = Al2O3





Empirical Formula

B) CoCl2
C) 52.9gAl and 47.1gO
52.9gAl(1mole Al)= 1.96 mol Al1.0
(27.0g Al ) 1.96
47.1 gO ( 1mole O)=2.94 mol O1.5
( 16.0 g O) 1.96

Al1.0O1.5 x 2 = Al2O3





Empirical Formula

B) CoCl2
C) 52.9gAl and 47.1gO
52.9gAl(1mole Al)= 1.96 mol Al1.0
(27.0g Al ) 1.96
47.1 gO ( 1mole O)=2.94 mol O1.5
( 16.0 g O) 1.96

Al1.0O1.5 x 2 = Al2O3





Empirical Formula

B) CoCl2
C) 52.9gAl and 47.1gO
52.9gAl(1mole Al)= 1.96 mol Al1.0
(27.0g Al ) 1.96
47.1 gO ( 1mole O)=2.94 mol O1.5
( 16.0 g O) 1.96

Al1.0O1.5 x 2 = Al2O3





Empirical Formula

B) CoCl2
C) 52.9gAl and 47.1gO
52.9gAl(1mole Al)= 1.96 mol Al1.0
(27.0g Al ) 1.96
47.1 gO ( 1mole O)=2.94 mol O1.5
( 16.0 g O) 1.96

Al1.0O1.5 x 2 = Al2O3





Empirical Formula

B) CoCl2
C) 52.9gAl and 47.1gO
52.9gAl(1mole Al)= 1.96 mol Al1.0
(27.0g Al ) 1.96
47.1 gO ( 1mole O)=2.94 mol O1.5
( 16.0 g O) 1.96

Al1.0O1.5 x 2 = Al2O3





Empirical Formula

B) CoCl2
C) 52.9gAl and 47.1gO
52.9gAl(1mole Al)= 1.96 mol Al1.0
(27.0g Al ) 1.96
47.1 gO ( 1mole O)=2.94 mol O1.5
( 16.0 g O) 1.96

Al1.0O1.5 x 2 = Al2O3





Empirical Formula

B) CoCl2
C) 52.9gAl and 47.1gO
52.9gAl(1mole Al)= 1.96 mol Al1.0
(27.0g Al ) 1.96
47.1 gO ( 1mole O)=2.94 mol O1.5
( 16.0 g O) 1.96

Al1.0O1.5 x 2 = Al2O3





Empirical Formula

B) CoCl2
C) 52.9gAl and 47.1gO
52.9gAl(1mole Al)= 1.96 mol Al1.0
(27.0g Al ) 1.96
47.1 gO ( 1mole O)=2.94 mol O1.5
( 16.0 g O) 1.96

Al1.0O1.5 x 2 = Al2O3





Empirical Formula

B) CoCl2
C) 52.9gAl and 47.1gO
52.9gAl(1mole Al)= 1.96 mol Al1.0
(27.0g Al ) 1.96
47.1 gO ( 1mole O)=2.94 mol O1.5
( 16.0 g O) 1.96

Al1.0O1.5 x 2 = Al2O3





Empirical Formula

B) CoCl2
C) 52.9gAl and 47.1gO
52.9gAl(1mole Al)= 1.96 mol Al1.0
(27.0g Al ) 1.96
47.1 gO ( 1mole O)=2.94 mol O1.5
( 16.0 g O) 1.96

Al1.0O1.5 x 2 = Al2O3





Empirical Formula

B) CoCl2
C) 52.9gAl and 47.1gO
52.9gAl(1mole Al)= 1.96 mol Al1.0
(27.0g Al ) 1.96
47.1 gO ( 1mole O)=2.94 mol O1.5
( 16.0 g O) 1.96

Al1.0O1.5 x 2 = Al2O3





Empirical Formula

B) CoCl2
C) 52.9gAl and 47.1gO
52.9gAl(1mole Al)= 1.96 mol Al1.0
(27.0g Al ) 1.96
47.1 gO ( 1mole O)=2.94 mol O1.5
( 16.0 g O) 1.96

Al1.0O1.5 x 2 = Al2O3





Empirical Formula








D) MgO
E) 2.83gK, .435g C, 1.735 g O
2.83 g K ( 1 mole K ) = .0724 mole
( 39.1 g K )
K2
.0363
.435 g C ( 1 mole C ) = .0363 mole C
( 12.0 g C )
.0363
1.735 g O ( 1 mole O ) = .108 mole
( 16.0 g O )
.0363
O3
Empirical Formula







E) 2.83gK, .435g C, 1.735 g O
2.83 g K ( 1 mole K ) = .0724 mole
( 39.1 g K )
K2
.0363
.435 g C ( 1 mole C ) = .0363 mole C
( 12.0 g C )
.0363
1.735 g O ( 1 mole O ) = .108 mole
( 16.0 g O )
.0363
O3
Empirical Formula







E)
2.83 g K ( 1 mole K ) = .0724 mole
( 39.1 g K )
K2
.0363
.435 g C ( 1 mole C ) = .0363 mole C
( 12.0 g C )
.0363
1.735 g C ( 1 mole O ) = .108 mole
( 16.0 g O )
.0363
O3
Empirical Formula







E)
2.83 g K ( 1 mole K ) = .0724 mole
( 39.1 g K )
K2
.0363
.435 g C ( 1 mole C ) = .0363 mole C
( 12.0 g C )
.0363
1.735 g C ( 1 mole O ) = .108 mole
( 16.0 g O )
.0363
O3
Empirical Formula







E)
2.83 g K ( 1 mole K ) = .0724 mole
( 39.1 g K )
K2
.0363
.435 g C ( 1 mole C ) = .0363 mole C
( 12.0 g C )
.0363
1.735 g C ( 1 mole O ) = .108 mole
( 16.0 g O )
.0363
O3
Empirical Formula







E)
2.83 g K ( 1 mole K ) = .0724 mole
( 39.1 g K )
K2
.0363
.435 g C ( 1 mole C ) = .0363 mole C
( 12.0 g C )
.0363
1.735 g C ( 1 mole O ) = .108 mole
( 16.0 g O )
.0363
O3
Empirical Formula






F) ZnSO3
G) CaC2O4
H) NaCN
I) Al2S3O12  Al2(SO4)3
J) Fe2S3O12  Fe2(SO4)3
K) MgCl2O8  Mg(ClO4)2