1. Chapter 3 Chemical Reactions and Reaction Stoichiometry

2. Stoichiometry
Area of study that examines the quantities of substances consumed and produced in chemical reactions
Based on the Law of Conservation of Mass (Antoine Lavoisier, 1789)
“We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.”
—Antoine Lavoisier

3. Chemical Equations
Chemical equation – An abbreviated way to describe a chemical reaction.
Arrows separate the reactants (on the left) from the products (on the right).
The same number of atoms of a given element appear on both sides.

4. Balancing Equations Follow the Law of Conservation of Mass, but how?
Start with an element that is only in one reactant and product (C below).
Balance it by changing coefficients, NOT subscripts. (Like in math, a “1” is not written, but it is assumed.)
Move on to other elements, without changing coefficients that are set, until complete, checking all elements at the end (H, then O here; totals below).
Is the following equation balanced? C4H10 + O2  CO2 + H2O
Other aids in balancing a chemical equation:
Write a skeletal equation. Co2O3 + C  Co + CO2
Balance atoms in more complex substances first:
2 Co2O3 + C  Co + 3 CO2 (since 3O  2O)
Balance free elements last:
2 Co2O3 + 3 C  4 Co + 3 CO2
Clear fractions by dividing through. NA (Balance the equation with coefficients using the lowest whole number possible.)
Count atoms to ensure equation is balanced.

5. Why Do We Add Coefficients Instead of Changing Subscripts to Balance?
Hydrogen and oxygen can make water OR hydrogen peroxide
2 H2(g) + O2(g) → 2 H2O(l)
H2(g) + O2(g) → H2O2(l)
We don’t change the formula because we don’t drink hydrogen peroxide.

6. Other Symbols in Chemical Equations
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) Δ
The states of matter for the reactants and products are often written in parentheses to the right of each formula or symbol.
(g) = gas; (l) = liquid; (s) = solid;
(aq) = dissolve in aqueous (water) solution

7. Other Symbols in Chemical Equations
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) Δ
Other symbols can be used to represent conditions during the chemical reaction. One example is the use of Δ over the reaction arrow, which means heat is needed for the reaction to take place.

8. Example – Balance the following Equations
Au2S3(s) + H2 (g)  Au(s) + H2S (g) SiO2 (s) + C (s)  Si (s) + CO (g) C3H8 (g) + O2 (g)  CO2 (g) + H2O (g) C6H12O6 (s) + O2 (g)  CO2 (g) + H2O (l)

9. Example
Crystals of sodium hydroxide (lye) react with carbon dioxide in the air to form a white powder, sodium carbonate, and a colorless liquid, water. Write a balanced equation for this chemical reaction.

10. Simple Patterns of Chemical Reactivity
Types of reactions, which can be predicted at this point
Combination reactions
Decomposition reactions
Combustion reactions

11. Combination Reactions
In a combination reaction, two or more substances react to form one product.

12. Combination Reaction Predictions: Metal and Nonmetal
You should be able to predict the product of a combination reaction between a metal and a nonmetal, like the one below. (Hint: Remember common charges for Groups!)

13. Decomposition Reactions
In a decomposition reaction one substance breaks down into two or more substances.
In an air bag, solid sodium azide releases nitrogen gas quickly.
2NaN3(s)  2Na(s) + 3N2(g)

14. Decomposition Reaction Predictions: Heating a Metal Carbonate
Metal carbonates decompose when heated to give off carbon dioxide and a metal oxide.
Balancing these equations is based on the charge of the metal. Δ
CaCO3(s) CaO(s) + CO2(g)

15. Combustion Reactions
Combustion reactions are rapid reactions that produce a flame.
Combustion reactions most often involve oxygen in the air as a reactant.

16. Combustion Reaction Predictions
When burning compounds with C and H in them, the products are CO2 and H2O.
C3H8(g) + 5 O2(g) CO2(g) + 4 H2O(g)
What is the stoichiometry of the combustion of octane, C8H18? Of glucose?

17. Example - Reaction Types
Balance the following equations and state the type of reactions:
C3H6 (g) O2 (g)  CO2 (g) H2O (g)
NH4NO3 (s)  N2O (g) H2O (g)
N2 (g) + H2 (g)  NH3 (g)

18. Formula Weight (FW)
A formula weight is the sum of the atomic weights for the atoms in a chemical formula.
For an element like calcium, Ca, the formula weight is the atomic weight (40.08 amu).
So, the formula weight of calcium chloride, CaCl2, would be
Ca: 1(40.08 amu)
+ Cl: 2( amu)
amu
Formula weights are generally reported for ionic compounds.
The formula weight of sulfuric acid, H2SO4, would be
2(AW of H) + 1(AW of S) + 4(AW of O)
2(1.0 amu) amu + 4(16.0 amu)
FW (H2SO4) = amu

19. Molecular Weight (MW)
If the substance is a molecule, the formula weight is also called its molecular weight.
A molecular weight is the sum of the atomic weights of the atoms in a molecule.
For glucose, which has a molecular formula of C6H12O6, the molecular weight is
6(AW of C) + 12(AW of H) + 6(AW of O)
6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu)
MW (C6H12O6) = amu

20. Percent Composition
One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation:
% Element =
(number of atoms)(atomic weight)
(FW of the compound)
× 100

21. Percent Composition So the percentage of carbon in glucose is:
(6)(12.0 amu)
(180.0 amu)
72.0 amu
180.0 amu =
× 100
= 40.0%
What is the percentage of hydrogen in glucose? Of oxygen?

22. Avogadro’s Number
In a lab, we cannot work with individual molecules. They are too small.
One mole (abbreviated: mol) is the amount of particles found in exactly 12 g of C-12.
6.02 × 1023 atoms or molecules is the number of particles in one mole.
6.02 × 1023 is Avogadro’s Number

23. Example Calculate the number of H atoms in 0.251 mol of ethane, C2H6
Calculate the number of molecules in mol HCHO2

24. Molar Mass
A molar mass is the mass of 1 mol of a substance (i.e., g/mol).
The molar mass of an element is the atomic weight for the element from the periodic table. If it is diatomic, it is twice that atomic weight.
The formula weight (in amu) will be the same number as the molar mass (in g/mol).

25. Mole Relationships
One mole of atoms, ions, or molecules contains Avogadro’s number of those particles.
The number of atoms of an element in a mole is the subscript in a formula (number of atoms of that element in the formula) times Avogadro’s number.

26. Converting Amounts
Moles provide a bridge from the molecular scale to the real-world scale.
Using equalities, we can convert from mass to atoms or from atoms to mass.
How many atoms in 3 g of copper (Cu)?
3 g Cu x (1 mol Cu/63.5 g Cu) x
(6.02 x 1023 atoms/1 mol Cu) = 3 x 1022 atoms

27. Examples Find the mass in grams of 1.906 x 10-2 mol BaI2
Find the mass in grams of 4.88 x 10-3 mol Al(NO3)3

28. Determining Empirical Formulas
One can determine the empirical formula from the percent composition by following these three steps.
What is the empirical formula for glucose?

29. Example
Give the empirical formula for the following compound whose sample contained 87.5 % N and 12.5% H by mass.

30. Determining Empirical Formulas— an Example
The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

31. Determining Empirical Formulas— an Example
Assuming g of para-aminobenzoic acid and converting to moles:
C: g × = mol C
H: g × = 5.09 mol H
N: g × = mol N
O: g × = mol O
1 mol
12.01 g
14.01 g
1.01 g
16.00 g

32. Determining Empirical Formulas— an Example
Calculate the mole ratio by dividing by the smallest number of moles:
C: =  7
H: =  7
N: = 1.000
O: =  2
5.105 mol
mol
5.09 mol
1.458 mol
C7H7NO2

33. Determining a Molecular Formula
The number of atoms in a molecular formula is a multiple of the number of atoms in an empirical formula.
If we find the empirical formula and know a molar mass (molecular weight) for the compound, we can find the molecular formula.

34. Determining a Molecular Formula— an Example
The empirical formula of a compound was found to be CH. It has a molar mass of 78 g/mol. What is its molecular formula?
Solution:
C + H = 1(12) + 1(1) = 13
Whole-number multiple = 78/13 = 6
The molecular formula is C6H6.

35. Combustion Analysis
Compounds containing C, H, and O are routinely analyzed through combustion in a chamber.
Mass of C is determined from the mass of CO2 produced.
Mass of H is determined from the mass of H2O produced.
Mass of O is determined by the difference of the mass of the compound used and the total mass of C and H.
Note: The mass of O in the compound can NOT be determined from CO2 and H2O because oxygen is added during the combustion.

36. Combustion Analysis
Mass of C is determined from the mass of CO2 produced.

37. Combustion Analysis
Mass of H is determined from the mass of H2O produced.

38. Indirect or Combustion Analysis
The combustion of a g sample of a compound of C, H, and O in pure oxygen gave g CO2 and g of H2O. Calculate the empirical formula of the compound. C H
H2O
CO2
MM (g/mol)
12.011
1.008
18.015
44.01
1. Calculate mass of C from mass of CO2.
mass CO2  mole CO2  mole C  mass C
= g C

39. Indirect or Combustion Analysis
The combustion of a g sample of a compound of C, H, and O gave g CO2 and g of H2O. Calculate the empirical formula of the compound.
2. Calculate mass of H from mass of H2O.
mass H2O  mol H2O  mol H  mass H
= g H
3. Calculate mass of O from difference.
5.217 g sample – g C – g H
= g O

40. Indirect or Combustion AnalysisH O
At. mass
12.011
1.008
15.999 g
2.021
0.5049
2.691
4. Calculate mol of each element
= mol C
= mol H
= mol O

41. Indirect or Combustion Analysis
Preliminary empirical formula
C0.1683H0.5009O0.1682
5. Calculate mol ratio of each element
Because all values are close to integers, round to
= C1.00H2.97O1.00
Empirical Formula = CH3O

42. Quantitative Information from a Balanced Equation
The coefficients in the balanced equation show
Relative numbers of molecules of reactants and products
Relative numbers of moles of reactants and products, which can be converted to mass

43. Stoichiometric Calculations
Starting with the mass of Substance A, you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant).

44. An Example of a Stoichiometric Calculation
How many grams of water can be produced from 1.00 g of glucose?
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
There is 1.00 g of glucose to start.
The first step is to convert it to moles.

45. An Example of a Stoichiometric Calculation
The new calculation is to convert moles of one substance in the equation to moles of another substance.
The mole ratio comes from the coefficients in the balanced equation.

46. An Example of a Stoichiometric Calculation
In the last step, moles of water is converted to grams of water.
This gives the answer we wanted.
You can do each step separately OR you can do them in one calculation, as seen above.

47. Stoichiometric Example
Ammonia used to make fertilizers for lawns and gardens is made by reacting nitrogen of the air with hydrogen. The balanced chemical equation is:
Find:
Mass in grams of ammonia when 1.34 mol N2 reacts
Mass in grams of N2 required to form 1.00 kg NH3
Mass in grams of H2 required to react with 6.00 g N2
3H2 (g) + N2 (g)  2NH3 (g)

48. Another Stoichiometric Example
Iron(III) oxide is a solid ore of iron that reacts with gaseous carbon monoxide to form solid iron and gaseous carbon dioxide.
Write the equation that describes the reaction indicating each reactant and product phase.
b) The reaction uses 2.35 g of iron(III) oxide . Determine how many moles of iron(III) oxide are present.
c) Determine how many grams carbon dioxide are produced.

49. Limiting Reactants
The limiting reactant is the reactant present in the smallest stoichiometric amount.
In other words, it is the reactant you will run out of first (in this case, the H2).

50. Limiting Reactants
In the example below, the O2 would be the excess reagent.

51. Limiting Reactants
The limiting reactant is used in all stoichiometry calculations to determine amounts of products that are produced and amounts of any other reactant(s) that are used in a reaction.

52. Limiting Reactants
Definition – least abundant reactant based on the equation for a reaction that limits the maximum product formed
Strategy:
Find amount of product formed if first reactant is completely consumed
Find amount of product formed if second reactant is completely consumed
Whichever reactant produces the smallest amount of products is the limiting reactant

53. Baking cookies Recipe: 1 cup flour 2 cups sugar 1 cup butter
12 oz chocolate chips
Flour + eggs + sugar + butter + chocolate chips  chocolate chip cookies
Determine limiting reactant:
1 cup flour
1 cups sugar
1 cup butter
12 oz chocolate chips
Determine limiting reactant:
1 cup flour
2 cups sugar
1 cup butter
7 oz chocolate chips

54. Example Determine the limiting reactant for:
2Sb (s) + 3I2 (s)  2SbI3 (s)
When 1.20 mol Sb and 2.40 mol I2 are present.
Answer the question, “how much product is produced from each of these amounts of reactants?”
Which reactant produces the least amount of product? This is the limiting reactant.

55. Theoretical Yield
The theoretical yield is the maximum amount of product that can be made.
In other words, it is the amount of product possible as calculated through the stoichiometry problem.
This is different from the actual yield, which is the amount one actually produces and measures.

56. Percent Yield
One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield):
Percent yield = × 100
actual yield
theoretical yield

57. Percent Yield Analysis:
A chemist set up a synthesis of solid phosphorus trichloride by mixing 12.0 g of solid phosphorus with g chlorine gas and obtained 42.4 g of solid phosphorus trichloride. Calculate the percentage yield of this compound.
Analysis:
Write balanced equation
P(s) + Cl2(g)  PCl3(s) Is this balanced?
Determine Limiting Reagent
Determine Theoretical Yield
Calculate Percentage Yield

58. Learning Check: Percentage Yield
Assembling the Tools:
1 mol P = g P
1 mol Cl2 = g Cl2
3 mol Cl2  2 mol P
Solution
Determine Limiting Reactant
But you only have 35.0 g Cl2, so Cl2 is limiting reactant
Note: The determination of the limiting reagent in this problem is done differently than previously but is still acceptable. Often it is better to determine the amount of a particular product from each reagent and choose as the limiting reagent that reagent which produces the least product.
In this problem they compared the relative amounts of two reactants required to produce a product.
= 41.2 g Cl2

59. Learning Check: Percentage Yield
Solution
Determine Theoretical Yield
Determine Percentage Yield
Actual yield = 42.4 g
= 45.2 g PCl3
= 93.8 %

60. Your Turn!
When 6.40 g of CH3OH was mixed with 10.2 g of O2 and ignited, 6.12 g of CO2 was obtained. What was the percentage yield of CO2?
A. 6.12%
8.79%
100%
142%
69.6%