1. Molar relationships in chemical reactions
Stoichiometry
Molar relationships in chemical reactions

2. 3.1 Chemical Equations: A Review
Law of conservation of mass
Relationship between reactant and products produces a balanced chemical equation
Reactants  Products
A + B  C + D
Balancing chemical eqns

3. Balancing Chemical Equations
Must have correct chemical formulas
Change coefficients as needed to balance equation
Do not change subscripts of chemical formulas
Indicate physical states by writing in parentheses (s), (l), (g), (aq)
Begin with an element that appears in only one reactant and product
If possible, do not begin with O or H

4. Balancing Chemical Equations
____ N2 + ____ H2  ____ NH3
____ KClO3  ____ KCl + ____ O2
____ NaCl + ____ F2  ____ NaF + ____ Cl2
____ H2 + ____ O2  ____ H2O
____ Pb(OH)2 + ____ HCl  ____ H2O + ____ PbCl2
____ AlBr3 + ____ K2SO4  ____ KBr + ____ Al2(SO4)3
____ CH4 + ____ O2  ____ CO2 + ____ H2O
____ C3H8 + ____ O2  ____ CO2 + ____ H2O
____ C8H18 + ____ O2  ____ CO2 + ____ H2O
10) ____ FeCl3 + ____ NaOH  ____ Fe(OH)3 + ____NaCl

5. Balancing Chemical Equations
____ P + ____O2  ____P2O5
____ Na + ____ H2O  ____ NaOH + ____H2
____ Ag2O  ____ Ag + ____O2
____ S8 + ____O2  ____ SO3
____ CO2 + ____ H2O  ____ C6H12O6 + ____O2
____ K + ____ MgBr  ____ KBr + ____ Mg
____ HCl + ____ CaCO3  ____ CaCl2 + ____H2O + ____ CO2
____ HNO3 + ____ NaHCO3  ____ NaNO3 + ____ H2O + ____ CO2
____ H2O + ____ O2  ____ H2O2
20) ____ NaBr + ____ CaF2  ____ NaF + ____ CaBr2

6. 3.2 Patterns of Chemical Reactivity
Because groups of elements have similar chemical behavior, the periodic table can be used to predict reactions involving elements
Example:
Na (s) + H2O (l)  NaOH (aq) + H2 (g)
Alkalai metals + water  metal hydroxides + H2

7. Types of Chemical Reactions
Synthesis (Combination) reactions
Two or more substances combine to form a new substance
Often involve elements combining to form a compound
A + B  C
4Na (s) + O2 (g)  2Na2O (s)
sodium oxide

8. Types of Chemical Reactions
Decomposition reactions
A single substance breaks down into two or more smaller substances
A  B + C
H2CO3 (aq)  CO2 (g) + H2O (l)
carbonic acid

9. Types of Chemical Reactions
Combustion reactions
Hydrocarbons + O2  CO2 + H2O
Cpds containing C, H, O  CO2 + H2O
Involve…
O2 as reactant
Release of heat & light
Often H2O is a product

10. 3.3 Formula Weights Atomic mass unit (amu)
A unit invented to describe extremely small masses
Defined as 1/12 the mass of an atom of 12C
1 amu = x g

11. Average Atomic Masses
Atomic masses given on the PT are weighted averages of all the isotopes of that element
Equals the sum of the products of the (mass) x (frequency) of the isotopes
mavg = Σ (m1f1) + (m2f2) + (m3f3) + …..

12. Average atomic mass mavg = Σ (m1f1) + (m2f2) + (m3f3) + ….. Example
Chlorine occurs as two isotopes:
Isotope amu freq (m x f)
35Cl % 26.41
37Cl % +9.05
Atomic mass of chlorine = 35.46

13. Formula & Molecular Weights
Formula mass refers to ionic cpds
Molecular mass refers to covalent cpds
Is the sum of atomic masses of the molecule
Calculate mol mass of Ca (NO3)2

14. Percent Composition of Formulas
Percentage by mass contributed by each element in the formula

15. Percent Composition
Determine the % composition of all elements in sucrose, C12H22O11
%C = (12 x 12) / 342 = 42.0%
%H = (1 x 22) / 342 = 6.4%
%O = (16 x 11) / 342 = 51.6%

16. 3.4 The Mole Is a counting unit Avagadro’s number = 6.022 x 1023
Interconvert between moles & particles
Molar mass = the mass of one mole of a substance
Molar mass of elements = atomic mass in grams

17. Converting between particles, moles, and grams
Particles and grams cannot be interconverted directly.
The mole is the bridge between particles and grams.

18. 3.5 Empirical Formula Analysis
To determine empirical formula from percent composition data
Empirical formula gives simplest ratio of atoms in a compound
Therefore, always convert to moles when calculating empirical formulas
Sample exercise 3.13

19. Combustion Analysis
Determines empirical formula based upon analysis of products of combustion
Example: g of a compound of CHO produced 41.7 g CO2 and g H2O
CxHyOz + O2  CO H2O
18.4 g g g
All carbon in CO2 came from the CHO
All hydrogen in H2O came from CHO

20. Combustion Analysis %C in CO2 x g CO2 = g C in CHO
Determine amt of C in CHO by determining amt of C in the CO2 produced.
%C in CO2 x g CO2 = g C in CHO
.2729 x = g C = mol C

21. Combustion Analysis
2. Determine amt of H in CHO by determining amt of H in the H2O produced.
%H in H2O x g H2O = g H in CHO
.1119 x = 2.20 g H = 2.18 mol H

22. Combustion Analysis
Determine amt of O in CHO by determining amt of H in the H2O produced
By law of conservation of mass,
grams of CHO = gC + gH + gO
So grams O = grams CHO – gC – gH
18.40 – – 2.20 = 4.97 gO = .311 mol O
Mole ratio C H 2.18 O 0.311
Empirical formula C3H7O

23. 3.6 Quantitative Information from Balanced Equations
Balanced chemical equations are like specific recipes
Balanced equations tell us ….
the relative amounts of reactants (ingredients)
the relative amounts of products
other relevant conditions needed (e.g. heat)
The unit of measure in a balanced equation is the mole
Indicated by coefficients

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Apple Snapple Oatmeal (4 servings)
1 medium apple or pear
3 cups apple juice or cider
1 and 1/3 cups regular rolled oats
1/4 cup raisins or chopped pitted dates
1/4 teaspoon ground cinnamon
milk (if you like)
brown sugar (if you like)

25. Stoichiometric Calculations
The coefficients in the balanced equation give the ratio of moles of reactants and products.

26. 3.6 Quantitative Information from Balanced Equations
Mole ratio used to calculate actual molar amounts in a balanced eqn
Sample Problem 3.16
Determine mass of water produced from combustion of 1.00 g glucose
C6H12O6 + __ O2  __ CO2 +__ H2O
C6H12O O2  6 CO H2O

27. Solving Stoichiometry Problems

28. Sample problem
Determine mass of water produced from combustion of 1.00 g glucose
C6H12O O2  6 CO2 + 6 H2O
Determine moles glucose
Determine moles water
Determine grams water

29. 3.7 Limiting Reactants Stoichiometrically least abundant reactant
Completely consumed in the reaction
Determines theoretical yield of the reaction
When all reactants are present in stoichiometric amounts, all are limiting reagents

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

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

32. Limiting Reagents
How many moles of ammonia can be produced when 3.0 mol nitrogen and 6.0 mol of hydrogen are permitted to react?
Write & balance the equation
Determine the LR to determine yield
Reactant  least yield is the LR
What is the excess reagent? How much is left over?

33. Theoretical & Actual Yields
Theoretical Yield
Quantity of product produced when 100% of limiting reagent reacts
Determined from balanced equation
Actual Yield
Quantity of product actually produced
No reaction is 100% efficient, therefore….
Theoretical Yield does not equal Actual Yield

34. Sample Problem 3.20 2 C6H12 + 5 O2  2 H2C6H8O4 + 2 H2O
Given 25.0 g cyclohexane and excess oxygen, determine theoretical yield of adipic acid

35. Percent Yield Expresses the efficiency of a reaction
Is ratio of actual yield to theoretical yield
Percent Yield = Actual Y/ Theroretical Y
%Y = AY / TY

36. Sample Problem 3.20 (cont) 2 C6H12 + 5 O2  2 H2C6H8O4 + 2 H2O
Theoretical yield = 43.5 g adipic acid (from first part)
If actual yield of adipic acid was 33.5 g, determine percent yield of reaction
%Y = Actual yield / Theoretical yield
%Y = 33.5g/43.5g = .77 = 77%