1. Stoichiometry
Section 7.2 pg

2. Stoichiometry
A method of problem-solving using known quantities of one entity in a chemical reaction to find out unknown quantities of another entity in the same reaction
Always requires a balanced chemical equation
Series of steps for measurement calculations
Always involves a mole ratio to “switch substances”
Writing chemical reactions, net ionic equations and dissociation equations is essential
Converting grams to moles (and vice versa) is also essential

3. Gravimetric Stoichiometry
Gravimetric Stoichiometry – method used to calculate the masses of reactants or products in a chemical reaction
Gravimetric = mass measurement
Using gravimetric stoichiometry, we can apply our knowledge of balanced chemical reactions and mass to mol conversions to:
Predict the mass of product we will get from a reaction
Estimate the amount of reactant we need for a reaction to produce a certain mass of product

4. Gravimetric Stoichiometry Steps:
Step 1: Write a balanced chemical reaction equation
List the measured mass (given), the unknown quantity (required) and conversion factors (M = molar mass)
Step 2: Convert the mass of the measured substance to moles
Step 3: Calculate the moles of the unknown substance using the
mole ratio required
given
Step 4: Convert from moles of the required substance to its mass.

5. Stoichiometry Calculations
(Measured quantity) solids/liquids m  n (Required quantity)
mole ratio

6. The balanced chemical equation is:
How many grams of oxygen are required to completely burn 10.0 g of propane (C3H8(g))?
The balanced chemical equation is:
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
The number of moles of propane reacting is: (MASS to MOLES)
10.0 g x 1 mol = mol
44.11 g
The number of moles of oxygen produced is: (MOLE RATIO)
0.227 mol x 5 mol = mol
1 mol
The mass of oxygen produced is: (MOLES TO MASS)
mol x g = g

7. Practice #2
20.0 g of butane (C4H10(g)) are completely burned in a lighter. How many grams of CO2(g) are produced?
Start with a balanced chemical equation:
2 C4H10(g) O2(g)  8 CO2(g) H2O (g)
m = 20.0g m = ?
M = g/mol M = g/mol

8. Practice #2 (Team Unit Analysis)
2 C4H10(g) + 13 O2(g)  8 CO2(g) H2O (g)
m = 20.0g m = ?
M = g/mol M = g/mol
2) Mass to moles, mole ratio, moles to mass
20.0 g x mol x mol CO x g = g
58.14 g mol C4H mol

9. Practice #3 (Unit Analysis)
What mass of iron (III) oxide is required to produce g of iron?
Fe2O3(s) CO(g)  2 Fe(s) + 3 CO2(g)
m = ? m = 100.0g
M = g/mol M = g/mol
m Fe2O3(s): g x mol x 1 mol x g = g Fe2O3
55.85 g mol mol

10. Remember to keep the unrounded values in your calculator for further calculation until the final answer is reported.
Pg. 290 #8-14
(Answers pg. 785)

11. Percent Yield for Reactions
We can use stoichiometry to test experimental designs, technological skills, purity of chemicals, etc. We evaluate these by calculating a percent yield.
This is the ratio of the actual (experimental) quantity of product obtained and the theoretical (predicted) quantity of product obtained from a stoichiometry calculation
Percent yield = actual yield x
predicted yield
Some forms of experimental uncertainties:
All measurements (limitations of equipment)
Purity of chemical used ( % purity)
Washing a precipitate (some mass is lost through filter paper)
Estimation of reaction completion (qualitative judgements i.e. color)

12. Percent Yield Example #1
Example: In a chemical analysis, 3.00 g of silver nitrate in solution was reacted with excess sodium chromate to produced g of filtered, dried precipitate.
Predicted value: 2AgNO3(aq) + Na2CrO7(aq) Ag2CrO7(s) +2NaNO3(aq) m = 3.00 g m = M = g/mol M = g/mol
3.00g x mol x mol x g = g
169.88g mol mol
Percent yield = actual yield x % = 2.81g x 100% = %
predicted yield g

13. Testing the Stoichiometric Method
Stoichiometry is used to predict the mass of precipitate actually produced in a reaction.
Filtration is used to separate the mass of precipitate actually produced in a reaction

14. Testing the Stoichiometric Method
What mass of lead is produced by the reaction of 2.13 g of zinc with an excess of lead(II) nitrate solution?
Design: A known mass of zinc is place in a beaker with an excess of lead(II) nitrate solution. The lead is produced in the reaction is separated by filtration and dried. The mass of the lead is determined
Prediction: Zn(s) Pb(NO3)2(aq)  Zn(NO3)2(aq) + Pb(s)
m = 2.13 g m = ?
M = g/mol M = g/mol
2.13 g x 1 mol x x g = g
65.41 g mol

15. Testing the Stoichiometric Method
Prediction: g of lead will be produced (Stoichiometric calculation)
Evidence: In the beaker, crystals of a shiny black solid were produced, and all the zinc disappeared.
Mass of filter paper = 0.92 g Mass of dried filter paper plus lead = 7.60g
Analysis:
mass of lead product = 7.60g – 0.92g = 6.68g
According to the evidence collected, 6.68 g of lead precipitated.
Evaluation: % difference = (experimental – predicted) x 100 %
predicted
6.68 – x 100% = 1%
6.75
Given such a small difference, which can readily be accounted for by normal sources of error, the prediction is clearly verified, and so the stoichiometric method for predicting the mass of lead formed in this experiment is judged to be acceptable.

16. Testing the Stoichiometric Method
Prediction: g of lead will be produced (Stoichiometric calculation)
Analysis:
mass of lead product = 7.60g – 0.92g = 6.68g
According to the evidence collected, 6.68 g of lead precipitated.
Evaluation:
Could you also calculate the % yield ?
Percent yield = actual yield x % = g x 100% = 99.0%
predicted yield g

17. Homework
Investigation 7.2 – Guided Lab Report
Pg. 293 #1-2, 6-10