1. Stoichiometry: Calculations with Chemical Formulas and Equations

2. EVIDENCE OF A CHEMICAL REACTION
A gas is released.
An insoluble substance (called a precipitate) is produced.
A permanent color change is observed.
Changes in energy.
Heat, light, sound, and electrical.
Exothermic reaction: releases heat,light and sound.
Endothermic reaction: absorbs heat and light.
pH change
Odor given off

3. EVIDENCE OF A CHEMICAL REACTION USING YOUR SENSES:
Sight- Change in color, formation of solid, formation of gasses (bubbles), light emission
Hearing- reaction leads to a popping noise, fizzes
Smell- pungent odor, change in smell
Touch- heat absorption and emission

4. Law of Conservation of Mass
“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, 1789

5. Chemical Equations
Concise representations of chemical reactions

6. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

7. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Reactants appear on the left side of the equation.

8. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Products appear on the right side of the equation.

9. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
The states of the reactants and products are written in parentheses to the right of each compound.

10. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Coefficients are inserted to balance the equation.

11. Subscripts and Coefficients Give Different Information
Subscripts tell the number of atoms of each element in a molecule

12. Subscripts and Coefficients Give Different Information
Subscripts tell the number of atoms of each element in a molecule
Coefficients tell the number of molecules

13. Coefficients in Chemical Reactions
Coefficients tell how many molecules or atoms there are.
Coefficients are numbers placed in front of each molecule or atom and are used to balance equations.

14. Conservation of Mass and Atoms
Law of Conservation of Matter:
Mass is neither created or destroyed in chemical reactions.
Lavoisier ( ) proved this.
The number and kinds of atoms present in the products is the same number and kind of atoms present in the reactants.
The atoms just rearrange.

15. Writing and Balancing Equations
Determine the reactants and products.
Write the equation.
Use the correct formulas for each compound—watch for the diatomics; those that end in –ine and –gen. (H2, N2, O2, F2, Cl2, Br2, I2, At2).
Balance the equation (same # and kind of elements on both sides of the arrow).

16. Balancing Examples
Solid copper(II) oxide reacts with hydrogen gas to form solid copper and liquid water.
CuO (s) + H2 (g) ---> Cu (s) + H2O (l)
Aluminum metal reacts with oxygen gas to form solid aluminum oxide.
Al (s) + O2 (g) ---> Al2O3 (s)

17. Balancing Examples
When balancing an equation, ONLY the coefficients can be changed.
NEVER change the subscripts.
For example: 3H2O
3 is the coefficient. 2 and 1 are the subscripts.
Changing the subscripts changes the compound.
H2O2  is not water but hydrogen peroxide.

18. HELPFUL HINTS
Keep polyatomic ions together as a unit if they show up on both sides of the equation.
-- NO3-1 , C2H3O2-1 , NH4+1 , ….
-- Ca(NO3)2 + NaCl ---> CaCl2 + NaNO3
Save H and O until last.
-- They will tend to balance out together with the H2O.

19. Example
Aluminum sulfide reacts with water to produce aluminum hydroxide and hydrogen sulfide.
Al2S3 + H2O --> Al(OH)3 + H2S
What should we look for immediately?
Save O and H for last - there is water.
Al2S3 + H2O ---> 2 Al(OH)3 + H2S
Al2S3 + H2O ---> 2 Al(OH)3 + 3 H2S
The Al and S are balanced - now  H & O.
Al2S3 + 6 H2O > 2 Al(OH)3 + 3 H2S

20. Balance The Following Equations
Al + S Al2S3
16 Al + 3 S Al2S3
CO + O CO2
2 CO + O CO2
C2H4 + O CO2 + H2O
C2H4 + 3 O CO2 + 2 H2O
Pb(NO3)2 + NaCl PbCl2 + NaNO3
Pb(NO3)2 + 2 NaCl PbCl2 + 2 NaNO3
Mg2C3 + H2O Mg(OH)2 + C3H4
Mg2C3 + 4 H2O Mg(OH)2 + C3H4
Ca(OH)2 + HBr CaBr2 + H2O
Ca(OH)2 + 2 HBr CaBr2 + 2 H2O

21. Reaction Types

22. Combination or (Synthesis) Reactions
Two or more substances react to form one product
Examples:
N2 (g) + 3 H2 (g)  2 NH3 (g)
C3H6 (g) + Br2 (l)  C3H6Br2 (l)
2 Mg (s) + O2 (g)  2 MgO (s)

23. 2 Mg (s) + O2 (g)  2 MgO (s)

24. Decomposition Reactions
One substance breaks down into two or more substances
Examples:
CaCO3 (s)  CaO (s) + CO2 (g)
2 KClO3 (s)  2 KCl (s) + O2 (g)
2 NaN3 (s)  2 Na (s) + 3 N2 (g)

25. Combustion Reactions Rapid reactions that produce a flame
Most often involve hydrocarbons reacting with oxygen in the air
Examples:
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)

26. Formula Weights

27. Formula Weight (FW)
Sum of the atomic weights for the atoms in a chemical formula
So, the formula weight of calcium chloride, CaCl2, would be
Ca: 1(40.1 amu)
+ Cl: 2(35.5 amu)
111.1 amu
These are generally reported for ionic compounds

28. Molecular Weight (MW)
Sum of the atomic weights of the atoms in a molecule
For the molecule ethane, C2H6, the molecular weight would be
C: 2(12.0 amu)
+ H: 6(1.0 amu)
30.0 amu

29. 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)
x 100

30. Percent Composition So the percentage of carbon in ethane is…
(2)(12.0 amu)
(30.0 amu)
24.0 amu
30.0 amu =
x 100
= 80.0%

31. Moles

32. Avogadro’s Number
6.02 x 1023
1 mole of 12C has a mass of 12 g

33. The Mole—How big is 6. 022 x 1023; Avogadro ’s number

34. It would take 10 billion chickens laying 10 eggs per day more than 10 billion years to lay a mole of eggs.

35. Avogadro's number of inches is 1,616,434 light years, or across our galaxy and back 8 times.

36. Avogadro's number of seconds is about 19 quadrillion years
Avogadro's number of seconds is about 19 quadrillion years! That is 4,240,666 times the age of the earth, or 954,150 times the age of the universe itself.

37. Avogadro's number of kilograms is just over 20 times the mass of the earth.

38. Avogadro's number of cents could repay the United States National Debt 86 million times.

39. Molar Mass
By definition, these are the mass of 1 mol of a substance (i.e., g/mol)
The molar mass of an element is the mass number for the element that we find on the periodic table
The formula weight (in amu’s) will be the same number as the molar mass (in g/mol)

40. Using Moles
Moles provide a bridge from the molecular scale to the real-world scale

41. Mole Relationships
One mole of atoms, ions, or molecules contains Avogadro’s number of those particles
One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound

42. Finding Empirical Formulas

43. Calculating Empirical Formulas
One can calculate the empirical formula from the percent composition

44. Calculating Empirical Formulas
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.

45. Calculating Empirical Formulas
Assuming g of para-aminobenzoic acid,
C: g x = mol C
H: g x = 5.09 mol H
N: g x = mol N
O: g x = mol O
1 mol
12.01 g
14.01 g
1.01 g
16.00 g

46. Calculating Empirical Formulas
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

47. Calculating Empirical Formulas
These are the subscripts for the empirical formula:
C7H7NO2

48. Combustion Analysis
Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this
C is determined from the mass of CO2 produced
H is determined from the mass of H2O produced
O is determined by difference after the C and H have been determined

49. Elemental Analyses
Compounds containing other elements are analyzed using methods analogous to those used for C, H and O

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

51. Stoichiometric Calculations
From 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)

52. Stoichiometric Calculations
C6H12O6 + 6 O2  6 CO2 + 6 H2O
Starting with 1.00 g of C6H12O6…
we calculate the moles of C6H12O6…
use the coefficients to find the moles of H2O…
and then turn the moles of water to grams

53. Limiting Reactants

54. How Many Cookies Can I Make?
You can make cookies until you run out of one of the ingredients
Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat)

55. How Many Cookies Can I Make?
In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make

56. Limiting Reactants
The limiting reactant is the reactant present in the smallest stoichiometric amount

57. 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)

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

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

60. Percent Yield Actual Yield Theoretical Yield Percent Yield = x 100
A comparison of the amount actually obtained to the amount it was possible to make
Actual Yield
Theoretical Yield
Percent Yield = x 100