1. Stoichiometry
Dr. Ron Rusay

2. Chemical Stoichiometry
Stoichiometry is the study of mass in chemical reactions. It deals with both reactants and products.
It quantitatively and empirically relates the behavior of atoms and molecules in a balanced chemical equation to observable chemical change and measurable mass effects.
It accounts for mass and the conservation of mass, just as the conservation of atoms in a balanced chemical equation.

3. Chemical Reactions Atoms, Mass & Balance: eg. Zn(s) + S(s)

4. Stoichiometry _ C2H5OH + _ O2  _ CO2 + _ H2O Reactants Products
Must begin with a correctly balanced equation:
_ C2H5OH + _ O2  _ CO2 + _ H2O
Reactants Products
C=2; H =5+1=6; O=2+1 C=1; H=2; O=2+1
1 C2H5OH O2  2 CO H2O

5. https://phet. colorado

6. Chemical Equation C2H5OH + 3 O2  2 CO2 + 3 H2O
The balanced equation can be completely stated as:
1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water.

7. Chemical Equation 2 C8H18(l)+ 25 O2(g) 16 CO2(g) +18 H2O(l)
All Balanced Equations relate on a molar mass basis. For example the combustion of octane:
2 C8H18(l)+ 25 O2(g) CO2(g) +18 H2O(l)
moles of octane react with 25 moles of oxygen to produce 16 moles of carbon dioxide and 18 moles of water.

8. The Chemical Equation: Mole & Masses
C2H5OH O2  2 CO H2O
46g (1 mole) of ethanol reacts with 3 moles of oxygen (96g) to produce 2 moles of carbon dioxide and 3 moles of water.
How many grams of carbon dioxide and water are respectively produced from 46g (1 mole) of ethanol ?
2 mol x 44 g/mol = 88g
3 mol x 18 g/mol = 54 g

9. The Chemical Equation: Moles & Masses
C2H5OH O2  2 CO H2O
How many grams of oxygen are needed to react with 15.3g of ethanol in a 12oz. beer ?
15.3gethanol x molethanol /46.0gethanol = 0.333molethanol
0.333molethanol x 3moloxygen / 1molethanol = 0.999moloxygen
32.0goxygen /moloxygen x 0.999moloxygen = 32.0goxygen
NOTE: It takes approximately 1 hour for the biologically equivalent amount of oxygen available from cytochrome p450 to consume the alcohol in a human in 1 beer to a level below the legal limit of 0.08%.

10. Chemical Stoichiometry
Epsom salt (magnesium sulfate heptahydrate) is one of five possible hydrates: mono-, di-, tri-, hexa-, or hepta- hydrate.
How can stoichiometry be used to determine, which hydrate is present in a pure unknown sample, by heating the sample in a kitchen oven at 400 o C for 45 minutes?
MgSO4 . x H2O(s) MgSO4 (s) + x H2O (g)
Refer to % Hydrate Lab Experiment for answer.

11. Mass Calculations 2 C8H18(l)+ 25 O2(g) 16 CO2(g) +18 H2O(l)
All Balanced Equations relate on a molar and mass basis. For the combustion of octane:
2 C8H18(l)+ 25 O2(g) CO2(g) +18 H2O(l)
228 g of octane (2 moles)* will react with 800 g of oxygen (25 moles) to produce (16 moles) 704 g of carbon dioxide and (18 moles) 324 g of water.
*(2 moles octane x 114 g/mol = 228 g )

12. Mass Calculations: Reactants Products

13. Mass Calculations: Reactants Products
1. Balance the chemical equation.
2. Convert mass of reactant or product
to moles.
3. Identify mole ratios in balanced equation:
They serve as the “Gatekeeper”.
4. Calculate moles of desired product or
reactant.
5. Convert moles to grams.

14. Mass Calculations: Reactants Products

15. QUESTION
The fuel in small portable lighters is butane (C4H10). After using a lighter for a few minutes, 1.0 gram of fuel was used. How many moles of carbon dioxide would it produce?
58 moles
0.017 moles
1.7  10–24 moles
0.068 moles

16. Mass Calculations: Reactants Products
How many grams of salicylic acid are theoretically needed to produce 1.80 kg of aspirin?
Balanced Equation: 1

17. Mass Calculations: Product Reactant
How many grams of salicylic acid are theoretically needed to produce 1.80 kg of aspirin?
Balanced Equation: 1 1 1

18. Mass Calculations: Product Reactant
grams (Aspirin)
grams (Salicylic Acid)
Avogadro's Number
Atoms
C7H6O3 MW =
Molecules
Stoichiometry
grams (SA)
(Molecular
1800 grams (A)
1 mol (A)
1 mol SA
Weight SA)
? (SA)
1 mol A
grams (A)
1 mol (SA)
(Molecular
Weight A)
"Gatekeeper"
C9H8O4
MW =

19. Mass Calculations: How many grams of salicylic acid are needed to produce 1.80 kg of aspirin?
Balanced Equation: 1 1 1 1
?g C7 = 1.80 x 103 gC9 x [molC9 /180.15gC9 ] x [1 molC7/ 1 molC9] x gC7/molC7
?g C7 = 1380 grams

20. Mass Calculations: Product Reactant
grams (Aspirin)
grams (Salicylic Acid)
?g C7 = 1.80 x 103 gC9 x [molC9 /180.15gC9 ] x [1 molC7/ 1 molC9] x gC7/molC7
grams (SA)
(Molecular
1800 grams (A)
1 mol (A)
1 mol SA
Weight SA)
? (SA)
1 mol A
grams (A)
1380 grams
1 mol (SA)
(Molecular
"Gatekeeper"
Weight A)

21. Mass Calculations: Reactants Products

22. QUESTION
The fuel in small portable lighters is butane (C4H10). After using a lighter for a few minutes, 1.0 gram (0.017 moles) of fuel was used. How many grams of carbon dioxide would it theoretically produce?
How many grams of carbon dioxide would this produce?
A.) 750 mg B.) 6.0 g
C) 1.5 g D.) 3.0 g

23. Percent Yield
In synthesis as in any experimentation, it is very difficult and at most times impossible to be perfect. Therefore the actual yield (g) is measured and compared to the theoretical calculated yield (g). This is the percent yield:
% Yield = actual (g) / theoretical (g) x 100
Some DVC students may report percent yields greater than 100% in their first synthesis experiment. Hmmm? Why is this not possible? How would Walter White do?

24. Theoretical (Yield) Mass Calculations
Reactant Product
grams (Reactant)
grams (Product)
Moles
Molar Mass
grams (P)
grams (R)
1 mol (R)
? mol (P)
? grams (P)
? mol (R)
grams (R)
1 mol (P)
(Divide)
Mass (R)
(Multiply)
Mass (P)
by Molar
"Gatekeepers”
from
Balanced reaction
by Molar

25. QUESTION
A synthetic reaction produced 2.45g of Ibogaine, C20H26N2O, a natural product with strong promise in treating heroin addiction (at least in Europe), the calculated theoretical yield was 3.05g, what is the % yield?
A) 19.7% B) 39.4% C) 80.3% D) 160.6%

26. Combustion Analysis (Chem 120 Prep)
CxHy + (x + y/4) O2  x CO2 + y/2 H2O
Molecules with oxygen in their formula are more difficult to solve for Oz knowing only the respective masses of CxHyOz sample, CO2 and H2O.
CxHyOz + (x + y/4 - z) O2  x CO2 + y/2 H2O
SEE: COMPARISON to wt % CALCULATIONS

27. Combustion Analysis Calculation Ascorbic Acid ( Vitamin C )
Combustion of a 6.49 mg sample in excess oxygen, yielded 9.74 mg CO2 and mg H2O
Calculate it’s Empirical formula!
C: x10-3g CO2 x(12.01 g C/44.01 g CO2)
= 2.65 x 10-3 g C
H: 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O)
= x 10-4 g H
Mass Oxygen = 6.49 mg mg mg
= mg O

28. Vitamin C: Calculation (continued)
C = 2.65 x 10-3 g C / ( g C / mol C ) =
= 2.21 x 10-4 mol C
H = x 10-3 g H / ( g H / mol H ) =
= 2.92 x 10-4 mol H
O = 3.54 x 10-3 g O / ( g O / mol O ) =
= 2.21 x mol O
Divide each by 2.21 x 10-4
C = Multiply each by = 3.00 = 3.0
H = = 3.96 = 4.0
O = = 3.00 = 3.0
C3H4O3

29. QUESTION Erythrose contains carbon, hydrogen and oxygen
(MM = g/mol). It is an important sugar that is used in many chemical syntheses.
Combustion analysis of a mg sample yielded g CO2 and g H2O. Mass Spectrometry produced a molecular 120 mass units (m/z). What is the molecular formula of erythrose?
CH2O
C6H12O6
C3H6O3
C4H8O4

30. Alternative Solution Method
ANSWER
D) C4H8O4
Alternative Solution Method
Mass ratio of C in CO2 = =
= = g C / 1 g CO2
Mass ratio of H in H2O = =
= = g H / 1 g H2O
Calculating masses of C and H:
Mass of Element = mass of compound x mass ratio of element
mol C x M of C
mass of 1 mol CO2
1 mol C x g C/ 1 mol C
44.01 g CO2
mol H x M of H
mass of 1 mol H2O
2 mol H x g H / 1 mol H
18.02 g H2O

31. Alternative Method 0.2729 g C 1 g CO2
Mass (g) of C = g CO2 x = g C
Mass (g) of H = g H2O x = g H
Calculating the mass of O:
Mass (g) of O = Sample mass -( mass of C + mass of H )
= g g C g H = g O
Calculating moles of each element:
C = g C / g C/ mol C = mol C
H = g H / g H / mol H = mol H
O = g O / g O / mol O = mol O
C H O = CH2O formula weight = 30 g / formula
120 g /mol / 30 g / formula = 4 formula units / cmpd = C4H8O4
g H
1 g H2O