1. Formulas, Equations, and Moles
Chapter 3
Formulas, Equations, and Moles

2. Balancing Chemical Equations
Alphabet – elemental symbols
Words – chemical formulas
Sentences – chemical equations (chemical reactions)
reactants  products
limestone  quicklime + gas
Calcium carbonate  calcium oxide + carbon dioxide
CaCO3(s)  CaO(s) + CO2(g)

3. Balancing Chemical Equations
Chemical reactions include
Reactants
Products
Balanced – Law of Conservation of Mass
# of atoms of an element on the reactant side must equal the # of atoms of that element on the product side.
Indicate the state of matter of each chemical in the reaction (Chapter 4)

4. Balancing Chemical Equations
Write the equation without coefficients
List the elements in each equation
Secret: if the same polyatomic ion exists on both sides, keep it together
Determine the # of each kind of atom on both sides
Balance atoms one element at a time by adjusting coefficients
DO NOT ALTER THE FORMULA OF THE COMPOUND!!!!!
Only coefficients can be altered
Secret:
Balance atoms appearing only once on each side first.
Save compounds comprised of only one type of element till last.
Reduce to lowest terms if necessary

5. Examples Balance the following equations:
Al(s) + Fe2O3(s) → Al2O3 (s) + Fe (l)
Solid copper reacts with aqueous silver nitrate to form aqueous copper (II) nitrate and silver solid
H3PO4 (l) → H2O (l) + P4O10 (s)
C4H10(g) + O2 (g) → CO2(g) + H2O (g)

6. Avogadro’s Number and the Mole
Meaning of a chemical reaction
2 C4H10(g) O2(g) → 8 CO2(g) H2O (g)
2 molecules of C4H10(g) reacts with 13 molecules of O2(g)
to form
8 molecule of CO2(g) and 10 molecules of H2O(g)

7. Avogadro’s Number and the Mole
Molecule’s mass = the sum of the atomic masses of the atoms making up the molecule.
m(C2H4O2) = 2·mC + 4·mH + 2·mO
= 2·(12.01) + 4·(1.01) + 2·(16.00)
= amu

8. Avogadro’s Number of the Mole
One mole (mol) of any substance contains 6.02 x 1023 (Avogadro’s Number) units of that substance.
One mole (mol) of a substance is the gram mass value equal to the amu mass of the substance.
Calculated the same as amu’s for a molecule

9. Avogadro’s Number and the Mole
Calculate the molar mass of the following:
Fe2O3 (Rust)
C6H8O7 (Citric acid)
C16H18N2O4 (Penicillin G)

10. Avogadro’s Number and the Mole
Methionine, an amino acid used by organisms to make proteins, is represented below. Write the formula for methionine and calculate its molar mass. (red = O; gray = C; blue = N; yellow = S; ivory = H)

11. Stoichiometry 4 Conversion units Chemical formula
Balanced chemical equation
Coefficients can read as;
# of molecules
# of moles of that molecules
Allows conversion between compounds in an equation
Avogadro’s # x 1023 of X = 1 mole of X
Molar mass – how many grams of a substance = 1 mole of that substance

12. Stoichiometric Calculations

13. Avogadro’s Number and the Mole
How many grams of oxygen are present in x 1020 molecules of KClO3? How many atoms of oxygen are present?

14. Avogadro’s Number and the Mole
Calculate the number of oxygen atoms in g of sodium sulfate, Na2SO4. 
A.  1.244 × 1023 O atoms
B.  4.976 × 1023 O atoms
C.  2.409 × 1024 O atoms
D.  2.915 × 1024 O atoms
E.  1.166 × 1025 O atoms

15. Problem
Potassium dichromate, K2Cr2O7, is used in tanning leather, decorating porcelain and water proofing fabrics. Calculate the number of chromium atoms in g of K2Cr2O7. 
A.  9.490 × 1025 Cr atoms
B.  2.248 × 1024 Cr atoms
C.  1.124 × 1024 Cr atoms
D.  3.227 × 1023 Cr atoms
E.  1.613 × 1023 Cr atoms

16. Stoichiometry: Chemical Arithmetic

17. Stoichiometry: Equation Arithmetic
Balance the following, and determine how many moles of CO will react with moles of Fe2O3.
Fe2O3(s) + CO(g) → Fe(s) + CO2(g)

18. Stoichiometry: Chemical Arithmetic
Aqueous sodium hydroxide and chlorine gas are combined to form aqueous sodium hypochlorite (household bleach), aqueous sodium chloride and liquid water.
How many grams of NaOH are needed to react with 25.0 g of Cl2?

19. Problem
Sulfur dioxide reacts with chlorine to produce thionyl chloride (used as a drying agent for inorganic halides) and dichlorine monoxide (used as a bleach for wood, pulp and textiles). SO2(g) + 2Cl2(g) → SOCl2(g) + Cl2O(g) If mol of Cl2 reacts with excess SO2, how many moles of Cl2O are formed? 
A.  0.800 mol
B.  0.400 mol
C.  0.200 mol
D.  0.100 mol
E.   mol

20. Problem
Nitrogen gas and hydrogen gas are combined to form ammonia (NH3), an important source of fixed nitrogen that can be metabolized by plants, using the Haber process. How many grams of nitrogen are needed to produce 325 grams of ammonia? 
A.  1070 g
B.  535 g
C.  267 g
D.  178 g
E.  108 g

21. Lab Homework
MISC 486 Problem Set 2 – Due

22. Yields of Chemical Reactions
Yields of Chemical Reactions: If the actual amount of product formed in a reaction is less than the theoretical amount, we can calculate a percentage yield.

23. Yield of Chemical Reactions
Dichloromethane (CH2Cl2) is prepared by reaction of methane (CH4) with chlorine (Cl2) giving hydrogen chloride as a by-product. How many grams of dichloromethane result from the reaction of 1.85 kg of methane if the yield is 43.1%?

24. Problem
What is the percent yield for the reaction PCl3(g) + Cl2(g) → PCl5(g) if g of PCl5 ( MM = g/mol) are formed when 61.3 g of Cl2 (  MM = g/mol) react with excess PCl3? 
A.  195%
B.  85.0%
C.  66.3%
D.  51.4%
E.  43.7%

25. Reactions with Limiting Amounts of Reactants
Limiting Reagents: The extent to which a reaction takes place depends on the reactant that is present in limiting amounts—the limiting reagent.
Process
Convert each reactant into a single product
The one that forms the least is the limiting reactant
Complete all other calculations using the limiting reactant

26. Reactions with Limiting Amounts of Reactants
Limiting Reagent Calculation: Lithium oxide is a drying agent used on the space shuttle. If 80.0 kg of water is to be removed and 65 kg of lithium oxide is available, which reactant is limiting?
Li2O(s) + H2O(l)  LiOH(s)
MM(Li2O) = g/mol
MM(H2O) = g/mol

27. Reactions with Limiting Amounts of Reactants
Limiting Reagent Calculation: Cisplatin is an anti-cancer agent prepared as follows:
K2PtCl4 + 2 NH3  Pt(NH3)2Cl2 + 2 KCl
If 10.0 g of K2PtCl4 and 10.0 g of NH3 are allowed to react: (a) which is the limiting reagent? (b) How many grams of the excess reagent are consumed? (c) How many grams of cisplatin are formed?
MM(K2PtCl4) = g/mol MM(NH3) = g/mol
MM [Pt(NH3)2Cl2] = g/mol

28. Problem
What is the percent yield of H2O if 97.2 g CH4S reacts with 183 g of O2 to produce 58.5 g H2O according to the following chemical reaction? (Reaction may or may not be balanced.)
CH4S + O2 → CO H2O + SO3
Write the balanced chemical reaction?
Calculate the molar mass of CH4S?
Calculate the molar mass of H2O?
Calculate the molar mass of O2?
Calculate the theoretical yield of H2O?
What is the limiting reagent?
Calculate the percent yield of H2O?

29. Lab Homework
MISC 486 Problem Set 4 - Due

30. Concentrations of Reactants in Solution: Molarity
Molarity: The most common way of expressing the amount of a substance dissolved in a solution
Conversion factor between moles and volume
It is important to note that the final volume of solution must be used, not volume of solvent.
solution of
Liters
solute
Moles
(M)
Molarity =

31. Solution Stoichiometry
Known: molarity of solution, volume of solution, and balanced chemical equation

32. Concentrations of Reactants in Solution: Molarity
How many moles of solute are present in 125 mL of 0.20 M NaHCO3?
How many grams of solute would you use to prepare mL of 1.25 M NaOH?

33. Problem
A M sodium chloride solution is referred to as a physiological saline solution because it has the same concentration of salts as normal human blood. Calculate the mass of solute needed to prepare mL of a physiological saline solution. 
A.  41.3 g
B.  31.9 g
C.  16.1 g
D.  8.77 g
E.  2.41 g

34. Concentrations of Reactants in Solution: Molarity
Solution Preparation
Determine the mass of solid needed to obtain the desired # of moles
Mass the solid
Add it to a volumetric flask
Add water to the volumetric flask until it is about half-full
Cap and shake to dissolve the solid
Add water to the line
Cap and shake

35. Diluting Concentrated Solutions
Dilution: process of reducing a solution’s concentration by adding more solvent.
Key concept: # of moles remains constant

36. Diluting Concentrated Solutions
Concentrated solution + Solvent  Dilute solution
Moles of solute (mol) = molarity (M) x volume (V)
Mconcentrated x Vconcentrated = Mdilute x Vdilute
only use if the initial solution and the final solution are the same substance

37. Diluting Concentrated Solutions
What volume of 18.0 M H2SO4 is required to prepare mL of M aqueous H2SO4?
What is the final concentration if 750 mL of 3.50 M glucose is diluted to a volume of mL?

38. Problem
Calcium chloride is used to melt ice and snow on roads and sidewalks and to remove water from organic liquids. Calculate the molarity of a solution prepared by diluting 165 mL of M calcium chloride to mL. 
A.  3.86 M
B.  0.743 M
C.  0.222 M
D.  0.123 M
E.  0.114 M

39. Titration
Titration: A technique for determining the concentration of a solution.
Process –
a carefully measured volume of an unknown solution is allowed to react with of a standard solution (concentration is known).
The volume of the known is measured.
Stoichiometry calculations are performed

40. H2SO4(aq) + KOH(aq)  K2SO4(aq) + H2O(l)
Titration
What is the molarity of a sulfuric acid solution if a 25.0 mL sample is titrated to equivalence with 50.0 mL of M potassium hydroxide solution?
H2SO4(aq) + KOH(aq)  K2SO4(aq) + H2O(l)

41. Problem
How many milliliters of 1.58 M HCl are needed to react completely with 23.2 g of NaHCO3 ( MM  = g/mol)? HCl(aq) + NaHCO3(s) → NaCl(s) + H2O(l) + CO2(g) 
A.  638 mL
B.  572 mL
C.  536 mL
D.  276 mL
E.  175 mL

42. Lab Homework
MISC 486 Problem Set 5 - Due

43. Percent Composition and Empirical Formulas
Percent Composition: Identifies the elements present in a compound as a mass percent of the total compound mass.
The mass percent is obtained by dividing the mass of each element by the total mass of a compound and converting to percentage.

44. Percent Composition and Empirical Formulas
Sugar is 42.1% C, 6.4% H, and 51.5% O
Meaning – out of 100 g of sugar 42.1 g of it is due to C

45. Percent Composition and Empirical Formulas
From the percent composition empirical formulas are developed
empirical formula gives the smallest whole number ratio of the atoms of each element in a compound.
Same as ionic formula for ionic compounds

46. Percent Composition and Empirical Formulas
Compound Formula Empirical Formula
Hydrogen H2O OH
peroxide
Benzene C6H CH
Ethylene C2H CH2
Propane C3H C3H8

47. Percent Composition and Empirical Formulas
A compound’s empirical formula can be determined from its percent composition.
A compound’s molecular formula is determined from the molar mass and empirical formula.

48. Percent Composition and Empirical Formulas
A compound was analyzed to be 82.67% carbon and 17.33% hydrogen by mass. An osmotic pressure experiment determined that its molar mass is g/mol.
What is the empirical formula and molecular formula for the compound?

49. Problem
Gadolinium oxide, a colorless powder which absorbs carbon dioxide from the air, contains mass % Gd. Determine its empirical formula. 
A.  Gd2O3
B.  Gd3O2
C.  Gd3O4
D.  Gd4O3
E.  GdO

50. Problem
Hydroxylamine nitrate contains mass % N, 4.20 mass % H, and mass O. If its molar mass is between 94 and 98 g/mol, what is its molecular formula? 
A.  NH2O5
B.  N2H4O4
C.  N3H3O3
D.  N4H8O2
E.  N2H2O4

51. Determining Empirical Formulas: Elemental Analysis
Combustion analysis is one of the most common methods for determining empirical formulas.
A weighed compound is burned in oxygen and its products analyzed by a gas chromatogram.
It is particularly useful for analysis of hydrocarbons.

52. Determining Empirical Formulas: Elemental Analysis
Terephthalic acid, used in the production of polyester fibers and films, is composed of carbon, hydrogen, and oxygen. When g of terephthalic acid was subjected to combustion analysis it produced g CO2 and g H2O. What is its empirical formula? 
A.  C2H3O4
B.  C3H4O2
C.  C4H3O2
D.  C5H12O4
E.  C2H2O

53. Optional Homework
Text –3.30, 3.32, 3.34, 3.36, 3.40, 3.42, 3.54, 3.58, 3.62, 3.64, 3.70, 3.72, 3.74, 3.76, 3.80, 3.82, 3.86, 3.88, 3.90, 3.92, 3.94, 3.96, 3.100, 3.106, 3.108, 3.112, 3.116
Chapter 3 Homework – from website

54. Required Homework
MISC 486 Problem Set 3 – Due
Assignment #3