1. Stoichiometry
Chapter/Unit 3

2. Significant Figures Review (1.5)
Rules for counting Significant Figures:
1. Nonzero integers always count!
2. Zeroes (3 classes):
a. Leading zeroes do not count (place holders)
b. Captive zeroes do count
c. Trailing zeroes are if there is a decimal point
3. Exact numbers (counting or from definitions)

3. Sig Fig Calculation Review (1.5)
Rules for Significant Figures in calculations:
1. For multiplication or division: the resultant is the same as the LEAST precise number in the calculation.
2. For addition or subtraction: the resultant is the same number of decimal places as the least precise measurement in the calculation.
3. Rounding: DO NOT round until all calculations are completed. Use only the first number to the right of the last significant figure.

4. Stoichiometry
Chemical Stoichiometry: The quantities of materials consumed and those produced in chemical reactions.
How much of A
How much of B
How much of AB
To understand Stoichiometry it is necessary to understand how average atomic mass is calculated.

5. Average Atomic Mass

6. Average Atomic Mass Average Atomic Mass of Carbon
12C = % at 12 amu
13C = 1.11 % at amu
14C = Negligibly small at this level
98.89%*(12 amu) %*( amu)
= amu

7. Average Atomic Mass
What does knowing that carbons average atomic mass is amu do for us?
Even though natural carbon does not contain a single atom with a mass of 12.01, for Stoichiometry purposes, we can consider carbon to be composed of only one type of atom with a mass of amu.
So to obtain 1000 atoms of Carbon, with an average atomic mass of 12.01, we can weigh out 12,010 atomic mass units. (this would be a mixture of 12C and 13C)

8. The Mole
Mole: the number equal to the number of carbon atoms in exactly 12 grams of pure 12C (abbreviated as mol).
Turns out that number is x 1023 which is referred to Avogrado’s number.
One Mole of something equals x 1023 units of that something.
Ex: 1 mol of students contains x 1023 students
Ex: 1 mol of Carbon has x 1023 atoms and has an amu of

9. The Mole
The mole is defined such that a sample of a natural element with a mass equal to the element’s atomic mass expressed in grams contains 1 mole of atoms.
1 mol of Helium = x 1023 atoms = grams
1 mol of Oxygen = x 1023 atoms = grams
1 mol of Zirconium = x 1023 atoms = grams
(6.022 x 1023 atoms)(12 amu) = 12 grams
(atom)
6.022 x 1023 amu = 1 g or x 1023 amu 1g

10. The Mole Determining the mass of a sample of Atoms
Remember that 1 atom of Carbon = amu
If we have 6 atoms of Carbon what is the mass of that sample?
6 atoms x amu x _____1g_____
atom x1023 amu
= x g

11. The Mole Determining the moles of a sample of atoms
Aluminum has mass of amu
If we have 10.0 grams of Al how many moles is in the sample?
10.0 grams Al x 1 mol Al __
26.98 g Al
6.022x1023 atoms x mol Al
1 mol Al
= mol Al Atoms
= 2.23 x 1023 Atoms

12. The Mole Calculating the numbers of Atoms in a sample
Silicon has an amu of 28.09
A silicon ship used in an integrated circuit has a mass of 5.68 mg. How many silicon atoms are present in the chip?
5.68 mg Si x 1 g Si __
1000 mg Si
= 5.68 x 10-3 g Si
5.68 x 10-3 g Si x 1 mol Si
28.09 g Si
= 2.02 x mol Si
2.02 x 10-4 mol Si x x 1023 atoms
1 mol Si
= 1.22 x atoms of Si

13. The Mole Calculating the numbers of Moles and Mass
Cobalt has an amu of 58.93
Calculate both the number of moles in a sample of cobalt containing 5.00 x 1020 atoms and the mass of the sample?
5.00 x 1020 atoms Co x _____1 mol Co_____
6.022x atoms Co
= 8.30 x 10-4 mol Co
8.30 x 10-4 mol Co x ____ grams Co
1 mol Co
= 4.89 x grams Co

14. The Mole Calculating Molar Mass
Molar Mass: the mass in grams of one mole of the compound
Calculating the molar mass of a molecule requires that we sum the molar masses of the individual elements. Example: The molar mass of methane (CH4)?
Mass of 1 mol of C =
Mass of 1 mol of H = g
4 x g
Mass of 1 mol of CH4 = g

15. The Mole Calculating Molar Mass
Example: Juglone, C10H6O3, is a natural herbicide. Calculate the molar mass and how many moles are present in 1.56 x10-2 g of the pure compound.
Mass of 1 mol of C (x10) =
Mass of 1 mol of H (x6)=
Mass of 1 mol of O (x3)=
10 x g
6 x g
3 x g
Mass of 1 mol of C10H6O3 = g
1.56 x 10-2 g C10H6O3 x _1 mol of C10H6O3__ g
= 8.96x mols of C10H6O3

16. The Mole Calculating Molar Mass
Example: Bees release 1 x10-6 grams of Isopentyl acetate (C7H14O2) when they sting. How many molecules are present per sting?
Mass of 1 mol of C (x7) =
Mass of 1 mol of H (x14)=
Mass of 1 mol of O (x2)=
7 x g
14 x g
2x g
Mass of 1 mol of C7H14O2 = g
1 x 10-6 g C7H14O2 x _1 mol of C7H14O2__ g
= 8 x mols of C7H14O2
8 x 10-9 mol C7H14O2 x _6.022 x 1023 molecules__
1 mol of C7H14O2
= 5 x molecules C7H14O2

17. The Mole Calculating Percent Composition
Percent composition: comparison of the mass of individual elements to its parent molecules mass.
Example: What percent mass is each element in ethanol, C2H5OH?
2x g = g C
6 x g = g H
1x g = g O
Mass of 1 mol of C (x2) =
Mass of 1 mol of H (x6)=
Mass of 1 mol of O (x1)=
2x g
6 x g
1x g
Mass of 1 mol of C2H5OH = g
____24.022g C___
g C2H5OH
= % Carbon
X 100%
____ g H___
g C2H5OH
X 100%
= % Hydrogen
= % Oxygen

18. The Mole Calculating Percent Composition
Example: What percent mass is each element in Penacillin C14H20N2SO4?
Mass of 1 mol of C (x14) =
Mass of 1 mol of H (x20)=
Mass of 1 mol of O (x4)=
Mass of 1 mol of N (x2)=
Mass of 1 mol of S (x1)=
14x g
20x g
4x g
2x g
1x g
= g C
= g H
= g O
= g N
= g S
Mass of 1 mol of C14H20N2SO4 = g
____ g C___
g C14H20N2SO4
= % Carbon
= % Oxygen
X 100%
= % Sulfur
____ g H __
g C14H20N2SO4
X 100%
= 6.45 % Hydrogen
= 8.96 % Nitrogen

19. The Mole Calculating empirical and Molecular formula
Example: A compound contains 71.65% Cl, 24.27% C and 4.07% H. The molar mass is known to be g/mol. What is the empirical and molecular formula?
1.) Convert the mass percents to masses in grams (use 100 grams as your conversion)
71.65 g Cl x 1 mol Cl__ g
= mol Cl
24.27 g C x 1 mol C__ g
= mol C
4.07 g H x 1 mol H__ g
= mol H

20. The Mole Calculating empirical and Molecular formula
Example: A compound contains 71.65% Cl, 24.27% C and 4.07% H. The molar mass is known to be g/mol. What is the empirical and molecular formula?
2.) Divide each mole value by the smallest mole number present Cl
= Chlorines
__C
= 1 Carbons
Empirical Formula = ClCH2 _H
= Hydrogens

21. The Mole Calculating empirical and Molecular formula
Example: A compound contains 71.65% Cl, 24.27% C and 4.07% H. The molar mass is known to be g/mol. What is the empirical and molecular formula?
3.) Determine the molar mass of the empirical formula and compare to the given molar mass.
Empirical Formula = ClCH2
Empirical Formula molar mass = g/mol
g/mol
g/mol
= 2
Molecular Formula = (ClCH2)2
Molecular Formula = Cl2C2H4

22. Physical vs. Chemical Changes in physical properties Melting Boiling
Condensation
No change occurs in the identity of the substance
Example: Ice , rain, and steam are all water

23. Physical vs. Chemical In a chemical Reaction:
Atoms in the reactants are rearranged to form one or more different substances.
Old bonds are broken; new bonds form
Examples:
Fe and O2 form rust (Fe2O3)
Ag and S form tarnish (Ag2S)

24. Chemical Reactions

25. Balancing Chemical Reactions
States within a chemical reaction:
Solid: symbolized using a (s)
Liquid: symbolized using a (l)
Gas: symbolized using a (g)
Dissolved in water: symbolized using a (aq) for aqueous
HCl(aq) + NaHCO3(s) CO2(g) + H2O(l) +NaCl(aq)

26. Balancing Chemical Reactions
Rules for balancing chemical equations:
1. Determine what reaction is occurring. What are the reactants and what are the products? What are the physical states involved?
2. Write the unbalanced equation that summarizes the reaction.
3. Balance the equation by inspection, starting with the most complicated molecules. Determine the coefficients necessary. DO NOT change the formulas of the reactants or products.

27. Balancing Chemical Reactions
Balancing by inspection
C2H5OH(l) + O2(g) CO2(g) + H2O(g)
Make a list of what atoms are present on each side:
Reactants
Products
C x 2 = 2C
H x 6 = 6H
O x 3 = 3O
C x 1 = 1C
H x 2 = 2H
O x 3 = 3O
Start with the most complicated molecule
Since C2H5OH contains 2 carbons then we need to have 2 carbons on the reactant side. We place a coefficient of 2 in front of carbon dioxide
C2H5OH(l) + O2(g) CO2(g) + H2O(g)

28. Balancing Chemical Reactions
Balancing by inspection
C2H5OH(l) + O2(g) CO2(g) + H2O(g)
Change the list of what atoms are present on each side after the change:
Reactants
Products
C x 2 = 2C
H x 6 = 6H
O x 3 = 3O
C x 2 = 2C
H x 2 = 2H
O x 5 = 5O
Since C2H5OH contains 6 H’s we can place a coefficient of 3 in front of water. Which will give us a change in the number of H’s and O’s
C2H5OH(l) + O2(g) CO2(g) + 3H2O(g)

29. Balancing Chemical Reactions
Balancing by inspection
C2H5OH(l) + O2(g) CO2(g) + 3H2O(g)
Reactants
Products
C x 2 = 2C
H x 6 = 6H
O x 3 = 3O
C x 2 = 2C
H x 6 = 6H
O x 7 = 7O
All that’s left is to balance the oxygens.
C2H5OH(l) + 3O2(g) CO2(g) + 3H2O(g)
Reactants
Products
C x 2 = 2C
H x 6 = 6H
O x 7 = 7O
C x 2 = 2C
H x 6 = 6H
O x 7 = 7O

30. Chemical Stoichiometry
Rules for calculating masses of reactants and products in chemical equations:
1. Write the equation and balance
2. Convert the known mass of the reactant or product to moles of that substance
3. Use the balanced equation to set up the appropriate mole ratios
4. Use the mole ratio to calculate the number of moles of the desired reactant/product
5. Convert from moles back to grams if the desired by question

31. Chemical Stoichiometry
Example: Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide forming solid lithium carbonate and liquid water. What mass of carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide?
Step 1: Write the equation and balance 2
LiOH(s) + CO2(g) Li2CO3(s) H2O(l)
Step 2: Convert the mass of LiOH to moles
1.00 kg LiOH 1
1000 g LiOH
1 mol LiOH
g LiOH x
= 41.8 moles of LiOH

32. Chemical Stoichiometry
Example: Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide forming solid lithium carbonate and liquid water. What mass of carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide?
Step 3: Write the appropriate mole ratio
1 mol CO2_
2 mol LiOH
Step 4: Calculate the moles of CO2 needed to react with the moles of LiOH
41.8 mol LiOH 1
1 mol CO2_
2 mol LiOH x
= 20.9 moles of CO2

33. Chemical Stoichiometry
Example: Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide forming solid lithium carbonate and liquid water. What mass of carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide?
Step 5: Convert the moles of the desired product to mass using the molar mass of the product
20.9 mol CO2 1
g CO2_
1 mol CO2 x
= 9.20 x 102 g of CO2

34. Limiting Reactants
Limiting Reactant: the reactant that is consumed first and therefore limits the amount of products that can be formed.
The rules for determining Limiting reactants are the same as for mass to mass calculations, but you are given the masses of both reactants.

35. Limiting Reactants
Example: Nitrogen gas can be prepared by passing gaseous ammonia over solid Copper (II) Oxide at high temperatures producing solid copper and water vapor. If a sample containing 18.1 grams of NH3 is reacted with 90.4 grams of Copper (II) Oxide, which is the limiting reactant? How many grams of Nitrogen gas are produced?
Step 1: Write the equation and balance 3
CuO(s) NH3(g) N2(g) Cu(s) + H2O(l) 2 3 3

36. Limiting Reactants CuO(s) + NH3(g) N2(g) + Cu(s) + H2O(l) 2 3
Step 2: Convert the mass of CuO and NH3 to moles
18.1 g NH3 1
1 mol NH3
g NH3 x
= moles of NH3
90.4 g CuO 1
1 mol CuO
g CuO x
= moles of CuO

37. Limiting Reactants CuO(s) + NH3(g) N2(g) + Cu(s) + H2O(l) 2 3
Step 3: Write the a mole ratio for the two reactants
3 mol CuO_
2 mol NH3
Required
= 3/2 = 1.5
1.14 mol CuO_
1.06 mol NH3
Actual
= 1.14/1.06 = 1.08
Since the actual ratio is smaller than the required CuO will be limiting

38. Limiting Reactants CuO(s) + NH3(g) N2(g) + Cu(s) + H2O(l) 2 3
Step 4: Calculate the number of moles that are produced for your target product using a your limiting reactant mole ratio. In this case the ratio between N2 and CuO.
1 mol N2_
3 mol CuO
1.14 moles of CuO 1 x
= moles of N2
Step 5: Convert the moles of the desired product to mass using the molar mass of the product
0.380 moles of N2 1 x
g N2_
1 mol N2
= 10.6 grams of N2