1. The Mathematics of Chemical Equations Chapter 11
STOICHIOMETRY
aka USING THE REACTION EQUATION
LIKE A RECIPE
SAVE PAPER AND INK!!! When you print out the notes on PowerPoint, print "Handouts" instead of "Slides" in the print setup. Also, turn off the backgrounds (Tools>Options>Print>UNcheck "Background Printing")!

2. USING EQUATIONS
Nearly everything we use is manufactured from chemicals.
Soaps, shampoos, conditioners, cd’s, cosmetics, medications, clothes, etc.
For a manufacturer to make a profit the cost of making any of these items can’t be more than the money paid for them.
Chemical processes carried out in industry must be economical, this is where balanced equations help.

3. USING EQUATIONS Equations are a chemist’s recipe.
Equations tell chemists what amounts of reactants to mix and what amounts of products to expect.
When you know the quantity of one substance in a rxn, you can calculate the quantity of any other substance consumed or created in the rxn.
Quantity meaning the amount of a substance in grams, liters, molecules, or moles.

4. N2 + 3H2  2NH3 1 mol N2 + 3 mol N2  2 mol NH3 28 g N2 + 3 (2 g H2) 
2 (17 g NH3)
34 g reactants 
34 g products + 
22.4 L
22.4 L
22.4 L
22.4 L N2
67.2 L H2 
44.8 L NH3

5. USING EQUATIONS
The calculation of quantities in chemical reactions is called Stoichiometry.
When you bake cookies you probably use a recipe.
A cookie recipe tells you the amounts of ingredients to mix together to make a certain number of cookies.
If you need more cookies than the yield of the recipe, the amounts of ingredients can be doubled or tripled.

6. Chocolate Chip Cookies!!
1 cup butter
1/2 cup white sugar
1 cup packed brown sugar
1 teaspoon vanilla extract
2 eggs
2 1/2 cups all-purpose flour
1 teaspoon baking soda
1 teaspoon salt
2 cups semisweet chocolate chips
Makes 3 dozen
How many eggs are needed to make 3 dozen cookies?
How much butter is needed for the amount of chocolate chips used?
How many eggs would we need to make 9 dozen cookies?
How much brown sugar would I need if I had 1 ½ cups white sugar?

7. Cookies and Chemistry…Huh!?!?
Just like chocolate chip cookies have recipes, chemists have recipes as well
Instead of calling them recipes, we call them chemical equations
Furthermore, instead of using cups and teaspoons, we use moles
Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds & elements as ingredients

9. Chemistry Recipes
An equation tells us how much of something you need to react with something else to get a product (like the cookie recipe)
Be sure you have a balanced reaction before you start!
Example: 2 Na + Cl2  2 NaCl
This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride
What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

10. Conversion Factors from a Chemical Equation
2 C6H6 (l) O2 (g)  12 CO2 (g) + 6 H2O (g)
In this equation there are 2 molecules of benzene reacting with 15 molecules of oxygen to produce 12 molecules of carbon dioxide and 6 molecules of water.
This equation could also be read as 2 moles of benzene reacts with 15 moles of oxygen to produce 12 moles of carbon dioxide and 6 moles of water.

11. Conversion Factors from a Chemical Equation
Since 6.02 x 1023 is the common factor between the relationship between the actual number of molecules and the number of moles present we can use the ratio of the # moles of one substance in the equation to find the # of moles another substance in the equation.
This ratio, # moles of unknown
# moles of given
is known as the MOLE RATIO

12. Conversion Factors from a Chemical Equation
2 C6H6 (l) O2 (g)  12 CO2 (g) + 6 H2O (g)
The MOLE RATIO, from the balanced equation, for oxygen and carbon dioxide is 15 : 12. It can be written as:
12 moles CO or as moles of O2
15 moles O moles of CO2
NOTE: The MOLE RATIO is used for converting moles of one substance into moles of another substance. The numbers that go in front of “moles of given” & “moles of unknown” are the coefficients from the balanced equation!
IT IS ALWAYS MOLES OVER MOLES!
Moles of unknown
Moles of Given

13. What was that again?! # Moles Unknown # Moles Given
The #’s in the numerator & denominator MUST come from the balanced chemical equation

14. I Y Stoichiometry !!! Consider: 4NH3 + 5O2  6H2O + 4NO
Is this equation balanced?
Yes!!
You must always start stoich. problems
with a balanced equation!
Write all of the possible mole ratios that
can be formed from this equation.
YES! NOW!!!

15. 4 moles NH3
5 moles O2
5 moles O2
4 moles NH3
5 moles O2
6 moles H2O
6 moles H2O
5 moles O2
4 moles NH3
6 moles H2O
6 moles H2O
4 moles NH3
6 moles H2O
4 moles NO
4 moles NO
6 moles H2O
ETC.

16. I Y Stoichiometry !!!
Using the coefficients of balanced equations and our knowledge of mole conversions we can perform powerful calculations! A.K.A. stoichiometry.
A balanced equation is essential for all calculations involving amounts of reactants and products.
If you know the number of moles of 1 substance, the balanced eqn allows you to calc. the number of moles of all other substances in a rxn equation.

17. Practice
Write the balanced reaction for hydrogen gas reacting with oxygen gas.
2 H2 + O2  2 H2O
How many moles of reactants are needed?
What if we wanted 4 moles of water?
What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get?
What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced?

18. Mole Ratios
These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical
Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)?
2 Na + Cl2  2 NaCl
5 moles Na 1 mol Cl2
2 mol Na
= 2.5 moles Cl2

19. MOLE – MOLE EXAMPLE
The following rxn shows the synthesis of aluminum oxide.
3O2(g) + 4Al(s)  2Al2O3(s)
3O2(g) + 4Al(s)  2Al2O3(s)
How many moles of aluminum are needed to form 3.7 mol Al2O3?
Given: 3.7 moles of Al2O3
Uknown: ____ moles of Al

20. MOLE – MOLE EXAMPLE Solve for the unknown: 3O2(g) + 4Al(s)  2Al2O3(s)
4 mol Al
3.7 mol Al2O3
2 mol Al2O3
= 7.4 mol Al
Mole Ratio

21. Mole - Mole Problems Consider : 4NH3 + 5O2  6H2O + 4NO
How many moles of H2O are produced if mol of O2 are used?
How many moles of NO are produced in the reaction if 17 mol of H2O are also produced?
6 mol H2O
5 mol O2 x
0.211 mol H2O =
0.176 mol O2
# mol H2O=
4 mol NO
6 mol H2O x
11 mol NO =
17 mol H2O
# mol NO=

22. Mole-Mole Conversions
How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal?

23. Mole-Mass Conversions
Most of the time in chemistry, the amounts are given in grams instead of moles
We still go through moles and use the mole ratio, but now we also use molar mass to get to grams
Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride?
2 Na + Cl2  2 NaCl
5.00 moles Na 1 mol Cl g Cl2
2 mol Na 1 mol Cl2
= 178g Cl2

24. Mole-Mass Problems Consider : 4NH3 + 5O2  6H2O + 4NO
How many grams of H2O are produced if 1.90 mol of NH3 are combined with excess oxygen?
How many grams of O2 are required to produce 0.3 mol of H2O?
6 mol H2O
4 mol NH3 x
18.0 g H2O
1 mol H2O x
51.3 g H2O =
1.90 mol NH3
32.0 g O2
1 mol O2 x
5 mol O2
6 mol H2O x
8 g O2 =
0.3 mol H2O

25. Practice
Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum.

26. Mass-Mole
We can also start with mass and convert to moles of product or another reactant
We use molar mass and the mole ratio to get to moles of the compound of interest
Calculate the number of moles of ethane (C2H6) needed to produce 10.0 g of water
2 C2H6 + 7 O2  4 CO2 + 6 H20
10.0 g H2O 1 mol H2O 2 mol C2H6
18.0 g H2O 6 mol H20
= mol C2H6

27. Practice
Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide

28. Mass-Mass Conversions
Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!)
Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in

29. Mass-Mass Conversion
Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen.
N2 + 3 H2  2 NH3
2.00g N mol N mol NH g NH3
28.0g N mol N mol NH3
= 2.43 g NH3

30. Mass – Mass Problems Consider : 4NH3 + 5O2  6H2O + 4NO
How many grams of NO is produced if 12 g of O2 is combined with excess ammonia?
1 mol O2
32.0 g O2 x
4 mol NO
5 mol O2 x
30.0 g NO
1 mol NO x
12 g O2
9.0 g NO =

31. Practice
How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen?

32. Volume – Volume Problems
Ex. What volume of hydrogen gas will be needed to produce L of ammonia gas (NH3) at STP? N H NH3
256.7 L NH3
1 mole NH3
mole H2
22.4 L H2 3
1 mole H2 2
mole NH3
22.4
L NH3
= L H2
Mole Ratio

33. Caution: this stuff is difficult to follow at first.
Limiting Reagents
Caution: this stuff is difficult to follow at first.
Be patient.

34. Limiting Reactants Available Ingredients Limiting Reactant
4 slices of bread
1 jar of peanut butter
1/2 jar of jelly
Limiting Reactant
bread
Excess Reactants
peanut butter and jelly

35. Grilled Cheese Sandwich
Bread Cheese  ‘Cheese Melt’
2 B C  B2C
100 bread slices ? sandwiches

36. Limiting Reactant: Cookies
1 cup butter
1/2 cup white sugar
1 cup packed brown sugar
1 teaspoon vanilla extract
2 eggs
2 1/2 cups all-purpose flour
1 teaspoon baking soda
1 teaspoon salt
2 cups semisweet chocolate chips
Makes 3 dozen
If we had the specified amount of all ingredients listed, could we make 4 dozen cookies?
What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies?
What if we only had one egg, could we make 3 dozen cookies?

37. Limiting Reactant
Most of the time in chemistry we have more of one reactant than we need to completely use up another reactant.
That reactant is said to be in excess (there is too much).
The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant.

38. Limiting Reactants Limiting Reactant Excess Reactant
used up in a reaction
determines the amount of product
Excess Reactant
More than enough to react with the limiting reagent – some left over!
added to ensure that the other reactant is completely used up
cheaper & easier to recycle
Courtesy Christy Johannesson

39. Limiting Reactants 1. Write a balanced equation.
2. For each reactant, calculate the amount of product formed.
3. Smaller answer indicates:
limiting reactant, and
amount of product formed

40. Real-World Stoichiometry: Limiting Reactants
Ideal
Stoichiometry
Limiting
Reactants
LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 366

41. Real-World Stoichiometry: Limiting Reactants
Fe S FeS
Ideal
Stoichiometry
Limiting
Reactants
S =
Fe =
LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 366

42. Limiting Reactants aluminum + chlorine gas  aluminum chloride
Al(s) Cl2(g)  AlCl3
2 Al(s) Cl2(g)  AlCl3
100 g g ? g
A g B g C g D. ???

43. Limiting Reactant
To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one.
The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it!
Remember! You can’t compare to see which is greater and which is lower unless the product is the same!

44. Limiting Reactant: Example
10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting
2 Al + 3 Cl2  2 AlCl3
Start with Al:
Now Cl2:
Limiting Reactant
10.0 g Al 1 mol Al mol AlCl g AlCl3
27.0 g Al mol Al mol AlCl3
= 49.4g AlCl3
35.0g Cl mol Cl mol AlCl g AlCl3
71.0 g Cl mol Cl mol AlCl3
= 43.9g AlCl3

45. LR Example Continued
We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete

46. Limiting Reactant Practice
15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.

47. Finding the Amount of Excess
By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess.
Can we find the amount of excess potassium in the previous problem?

48. Finding Excess Practice
15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I2  2 KI
We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced.
15.0 g I mol I mol K g K
254 g I mol I mol K
= 4.62 g K USED!
15.0 g K – 4.62 g K = g K EXCESS
Given amount of excess reactant
Amount of excess reactant actually used
Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

49. Excess Reactant 2 Na + Cl2  2 NaCl 50 g 50 g x g 81.9 g NaCl
/ 23 g/mol
/ 71 g/mol
x 58.5 g/mol
1 : 2
1.40 mol
“Have”
2.17 mol
0.70 mol
coefficients
“Need”
1.40 mol
EXCESS
LIMITING

50. Excess Reactant (continued)
2 Na Cl  NaCl
50 g g x g
81.9 g NaCl
All the chlorine is used up…
81.9 g NaCl
50.0 g Cl2
31.9 g
Na is consumed in reaction.
How much Na is unreacted?
50.0 g g =
18.1 g Na
total used “excess”

51. Conservation of Mass is Obeyed
2 Na Cl  NaCl
50 g g x g
81.9 g NaCl
2 Na Cl  NaCl
+ Na
50 g g x g
81.9 g NaCl
18.1 g
31.9 g +
18.1 g
100 g product
100 g reactant

52. Limiting Reactant: Recap
You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT.
Convert ALL of the reactants to the SAME product (pick any product you choose.)
The reactant that gave you the lowest answer is the LIMITING REACTANT.
The other reactant is in EXCESS.
To find the amount of excess left over, subtract the amount used from the given amount.
If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!

53. Percentage Yield

54. Percent Yield
measured in lab
calculated on paper

55. % yield
Experimental yield
Theoretical yield
X 100 = % yield

56. Sample problem 1 mol H2 2.02 g H2 x 2 mol H2O 2 mol H2 x 18.02 g H2O
Q - What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2?
Step 1: write the balanced chemical equation
2H2 + O2  2H2O
Step 2: determine actual and theoretical yield.
Actual is given, theoretical is calculated:
1 mol H2
2.02 g H2 x
2 mol H2O
2 mol H2 x
18.02 g H2O
1 mol H2O x
# g H2O=
16 g H2
143 g =
Step 3: Calculate % yield
actual
theoretical
138 g H2O
143 g H2O =
% yield =
x 100%
x 100%
96.7% =

57. Practice problem
Q - What is the % yield of NH3 if 40.5 g NH3 is produced from 20.0 mol H2 and excess N2?
Step 1: write the balanced chemical equation
N2 + 3H2  2NH3
Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated:
2 mol NH3
3 mol H2 x
17.04 g NH3
1 mol NH3 x
# g NH3=
20.0 mol H2
227 g =
Step 3: Calculate % yield
actual
theoretical
40.5 g NH3
227 g NH3 =
% yield =
x 100%
x 100%
17.8% =

58. Percent Yield
When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl.
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
? g
actual: 46.3 g

59. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g
actual: 46.3 g
Theoretical Yield:
45.8 g
K2CO3
1 mol
K2CO3 g
2 mol
KCl
1 mol
K2CO3
74.55
g KCl
1 mol
KCl
= 49.4
g KCl

60. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g 49.4 g
actual: 46.3 g
Theoretical Yield = 49.4 g KCl
46.3 g
49.4 g
% Yield =
 100 =
93.7%

61. Challenging question 2H2 + O2  2H2O
What is the % yield of H2O if 58 g H2O are produced by combining 60 g O2 and 7.0 g H2?
Hint: determine limiting reagent first
1 mol O2
32 g O2 x
2 mol H2O
1 mol O2 x
18.02 g H2O
1 mol H2O x
# g H2O=
60 g O2
68 g =
1 mol H2
2.02 g H2 x
2 mol H2O
2 mol H2 x
18.02 g H2O
1 mol H2O x
# g H2O=
7.0 g H2
62.4 g =
actual
theoretical
58 g H2O
62.4 g H2O =
% yield =
x 100%
x 100%
92.9% =

62. PARTICLES B
MOLE
6.02X10 A
22.4 L 23
MASS
VOLUME
MOLE MAP
MOLAR MASS
RATIO

63. 6 Na(s) + Fe2O3(s)  3 Na2O(s) + 2 Fe (s)
Air Bag Design
Exact quantity of nitrogen gas must be produced in an instant.
Use a catalyst to speed up the reaction
2 NaN3(s)  2 Na(s) N2(g)
6 Na(s) + Fe2O3(s)  3 Na2O(s) Fe (s)

64. 6 Na(s) + Fe2O3(s)  3 Na2O(s) + 2 Fe(s)
2 NaN3(s)  2 Na(s) N2(g)
6 Na(s) + Fe2O3(s)  3 Na2O(s) + 2 Fe(s)
Airbag Design
Assume that 65.1 L of N2 gas are needed to inflate an air bag to the
proper size. How many grams of NaN3 must be included in the gas
generant to generate this amount of N2?
(Hint: The density of N2 gas at this temperature is about g/L).
65.1 L N2
x g/L N2
X g NaN3 = 59.6 g N2 (1 mol N2) (2 mol NaN3) (65 g NaN3)
(28 g N2) (3 mol N2) (1 mol NaN3)
59.6 g N2
X = g NaN3
How much Fe2O3 must be added to the gas generant for this amount of NaN3?
X g Fe2O3 = g NaN3 (1 mol NaN3) (2 mol Na) (1 mol Fe2O3) (159.6 g Fe2O3)
(65 g NaN3) (2 mol NaN3) (6 mol Na) (1 mol Fe2O3)
X = g Fe2O3

65. Water from a Camel
Camels store the fat tristearin (C57H110O6) in the hump. As well as
being a source of energy, the fat is a source of water, because when
it is used the reaction
takes place. What mass of water can be made from 1.0 kg of fat?
2 C57H110O6(s) O2(g)  114 CO2(g) H2O(l)
X g H2O = 1 kg ‘fat” (1000 g ‘fat’) (1 mol “fat”) (110 mol H2O) (18 g H2O)
(1 kg ‘fat’) (890 g ‘fat’) (2 mol ‘fat’) (1 mol H2O)
X = g H2O
or liters water

66. Rocket Fuel B2H6 + O2  B2O3 + H2O B2H6 + 3 O2  B2O3 + 3 H2O
The compound diborane (B2H6) was at one time
considered for use as a rocket fuel. How many grams
of liquid oxygen would a rocket have to carry to burn
10 kg of diborane completely?
(The products are B2O3 and H2O).
B2H6 + O2  B2O3 + H2O
Chemical equation
Balanced chemical equation
B2H O2  B2O H2O
10 kg X g
X g O2 = 10 kg B2H6 (1000 g B2H6) (1 mol B2H6) (3 mol O2) (32 g O2)
(1 kg B2H6) (28 g B2H6) (1 mol B2H6) (1 mol O2)
X = 34,286 g O2

67. Water in Space
In the space shuttle, the CO2 that the crew exhales is removed from the
air by a reaction within canisters of lithium hydroxide. On average, each
astronaut exhales about 20.0 mol of CO2 daily. What volume of water will
be produced when this amount of CO2 reacts with an excess of LiOH?
(Hint: The density of water is about 1.00 g/mL.)
CO2(g) LiOH(s)  Li2CO3(aq) + H2O(l)
20.0 mol
excess
x g
X mL H2O = mol CO2 (1 mol H2O) (18 g H2O) (1 mL H2O)
(1 mol CO2) (1 mol H2O) (1 g H2O)
X = 360 mL H2O

68. Real Life Problem Solving
Determine the amount of LiOH required for a seven-day mission in space
for three astronauts and one ‘happy’ chimpanzee. Assume each passenger
expels 20 mol of CO2 per day.
Note: The lithium hydroxide scrubbers are only 85% efficient.
(4 passengers) x (10 days) x (20 mol/day) = 800 mol CO2
Plan for a delay
CO2(g) LiOH(s)  Li2CO3(aq) + H2O(l)
800 mol X g

69. Note: The lithium hydroxide scrubbers are only 85% efficient.
CO2(g) LiOH(s)  Li2CO3(aq) + H2O(l)
800 mol X g
x 23.9 g/mol
1:2
800 mol
1600 mol
Needed
(actual yield)
X g LiOH = 800 mol CO2
(2 mol LiOH)
(1 mol CO2 )
(23.9 g LiOH)
( 1 mol LiOH)
= 38,240 g LiOH
Note: The lithium hydroxide scrubbers are only 85% efficient.
Actual Yield
Theoretical Yield
38,240 g LiOH
X g LiOH
% Yield =
0.85 =
Amount of LiOH to
be taken into space
X = 44,988 g LiOH

70. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g ? L 0.90 L 2.5M
79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP?
Zn HCl  ZnCl H2
79.1 g
0.90 L
2.5M
? L

71. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g ? L 0.90 L 2.5M 79.1
g Zn
1 mol Zn
65.39
g Zn
1 mol H2 Zn
22.4 L H2
1 mol
= 27.1 L H2
Courtesy Christy Johannesson

72. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L 2.5M ? L 0.90
2.5 mol
HCl
1 L
1 mol H2
2 mol
HCl
22.4
L H2
1 mol H2
= 25 L H2
Courtesy Christy Johannesson

73. Limiting Reactants Zn: 27.1 L H2 HCl: 25 L H2 Limiting reactant: HCl
Excess reactant: Zn
Product Formed: 25 L H2
left over zinc
Courtesy Christy Johannesson

74. % error Experimental yield – Theoretical yield Theoretical yield
X 100 = % error
What you predicted you would get (use stoichiometry)
What YOU got in the experiment

75. Review: Molar Mass of Compounds
The molar mass (MM)- add up all the atomic masses for the molecule (or compound)
Ex. Molar mass of CaCl2
Avg. Atomic mass of Calcium = 40.1g
Avg. Atomic mass of Chlorine = 35.5g
Molar Mass of calcium chloride = g/mol Ca + (2 X 35. 5) g/mol Cl  g/mol CaCl2 20
Ca 
17 Cl 35.45

76. Review: Flowchart Atoms or Molecules Moles Mass (grams)
Divide by 6.02 X 1023
Multiply by 6.02 X 1023
Moles
Multiply by atomic/molar mass from periodic table
Divide by atomic/molar mass from periodic table
Mass (grams)