1. Fig. 3.11

2. 3 general classes of chemical reactions:
Precipitation reactions
Ex: geology, heavy metal analysis
Solid formation from ionic compounds
2) Acid/base reactions
Ex: many biochemical reactions
Proton transfer in polar covalent compounds
3) Oxidation/Reduction (“redox”) reactions
Ex: batteries, metabolic energy production
Electron transfer in ionic & molecular compounds

3. Classifying Chemical Reactions
Fundamental rxn. similarities
Precipitation (Ppt)
Acid/base (A/B)
Redox (R/O)
Textbook
Similarities in written equations
Combination reactions (CR)
often redox
Decomposition reactions (DR)
Single replacement (SR)
Double displacement (DD)
Acid/base, precipitation, redox
Lab

4. Reactions Involving Ions: Molecular vs. Ionic Equations
Chemical Reaction can be expressed by:
Molecular Equation (balanced chemical equation)
Complete Ionic Equation (showing all ions in reaction)
Net Ionic Equation (showing only those ions directly involved in reaction)
Consider
Copper (III) sulfate reacts with sodium hydroxide to form copper (III) hydroxide and sodium sulfate (all in water).
Express reaction in molecular, complete ionic,
and net ionic equations

6. The Solubility of Ionic Compounds in Water
The solubility of ionic compounds in water depends upon the relative
strengths of the electrostatic forces between ions in the ionic compound
and the attractive forces between the ions and water molecules in the
solvent. There is a tremendous range in the solubility of ionic
compounds in water! The solubility of so called “insoluble” compounds
may be several orders of magnitude less than ones that are called
“soluble” in water, for example:
Solubility of NaCl in water at 20oC = 365 g/L
Solubility of MgCl2 in water at 20oC = g/L
Solubility of AlCl3 in water at 20oC = 699 g/L
Solubility of PbCl2 in water at 20oC = 9.9 g/L
Solubility of AgCl in water at 20oC = g/L
Solubility of CuCl in water at 20oC = g/L

7. Precipitation Reactions: Will a Precipitate Form?
If we add a solution containing potassium chloride to a solution
containing ammonium nitrate, will we get a precipitate?
KCl(aq) + NH4NO3 (aq) = K+(aq) + Cl-(aq) + NH4+(aq) + NO3-(aq)
By exchanging cations and anions we see that we could have potassium
chloride and ammonium nitrate, or potassium nitrate and ammonium
chloride. In looking at the solubility table it shows all possible
products as soluble, so there is no net reaction!
KCl(aq) + NH4NO3 (aq) = No Reaction!
If we mix a solution of sodium sulfate with a solution of barium nitrate,
will we get a precipitate? From the solubility table it shows that barium
sulfate is insoluble, therefore we will get a precipitate!
Na2SO4 (aq) + Ba(NO3)2 (aq) BaSO4 (s) + 2 NaNO3 (aq)

8. Solubility Soluble = ability to dissolve in a liquid
Insoluble = inability to dissolve in a liquid
Not all Ionic Compounds are water soluble
Not all molecular compounds are insoluble!

9. 8 Simple Rules For Common Ionic Compounds: Predicting Precipitation

10. AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
Molecular Equation
A chemical equation in which the reactants and products are written as if they were molecular substances, even though they may actually exist in solution as ions.
State symbols are include: (s), (l), (g), (aq).
For example:
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
Although AgNO3, NaCl, and NaNO3 exist as ions in aqueous solutions, they are written as compounds in the molecular equation.

11. AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
Complete Ionic Equation
A chemical equation in which strong electrolytes are written as separate ions in the solution. Other reactants and products are written in molecular form. State symbols are included: (s), (l), (g), (aq).
For example:
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
In ionic form:
Ag+(aq) + NO3-(aq) + Na+(aq)Cl-(aq) 
AgCl(s) + Na+(aq) + NO3-(aq)

12. Spectator Ion
An ion in an ionic equation that does not take part in the reaction. It appears as both a reactant and a product.

13. Ag+(aq) + Cl-(aq)  AgCl(s)
Net Ionic Equation
A chemical equation in which spectator ions are omitted. It shows the reaction that actually occurs at the ionic level.
For example:
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) 
AgCl(s) + Na+(aq) + NO3-(aq)
In net ionic form:
Ag+(aq) + Cl-(aq)  AgCl(s)

15. Figure 4. 6: Reaction of magnesium chloride and silver nitrate
Figure 4.6: Reaction of magnesium chloride and silver nitrate. Photo courtesy of American Color.
Write molecular
and ionic equations
for this reaction.
Ionic equation: Ag+(aq) + Cl-(aq) AgCl(s)

16. Decide whether the following reaction occurs
Decide whether the following reaction occurs. If it does, write the molecular, ionic, and net ionic equations.
KBr + MgSO4 
Determine the product formulas:
K+ and SO42- make K2SO4
Mg2+ and Br- make MgBr2
Determine whether the products are soluble:
K2SO4 is soluble
MgBr2 is soluble
KBr + MgSO4  no reaction

17. Decide whether the following reaction occurs
Decide whether the following reaction occurs. If it does, write the molecular, ionic, and net ionic equations.
NaOH + MgCl2 
Determine the product formulas:
Na+ and Cl- make NaCl
Mg2+ and OH- make Mg(OH)2
Determine whether the products are soluble:
NaCl is soluble
Mg(OH)2 is insoluble

18. 2OH-(aq) + Mg2+(aq)  Mg(OH)2(s)
Molecular Equation
(Balance the reaction and include state symbols)
2NaOH(aq) + MgCl2(aq) 
2NaCl(aq) + Mg(OH)2(s)
Ionic Equation
2Na+(aq) + 2OH-(aq) + Mg2+(aq) + 2Cl-(aq) 
2Na+(aq) + 2Cl-(aq) + Mg(OH)2(s)
Net Ionic Equation
2OH-(aq) + Mg2+(aq)  Mg(OH)2(s)

19. Decide whether the following reaction occurs
Decide whether the following reaction occurs. If it does, write the molecular, ionic, and net ionic equations.
K3PO4 + CaCl2 
Determine the product formulas:
K+ and Cl- make KCl
Ca2+ and PO43- make Ca3(PO4)2
Determine whether the products are soluble:
KCl is soluble
Ca3(PO4)2 is insoluble

20. 2PO43-(aq) + 3Ca2+(aq)  Ca3(PO4)2(s)
Molecular Equation
(Balance the reaction and include state symbols)
2K3PO4(aq) + 3CaCl2(aq) 
6KCl(aq) + Ca3(PO4)2(s)
Ionic Equation
6K+(aq) + 2PO43-(aq) + 3Ca2+(aq) + 6Cl-(aq) 
6K+(aq) + 6Cl-(aq) + Ca3(PO4)2(s)
Net Ionic Equation
2PO43-(aq) + 3Ca2+(aq)  Ca3(PO4)2(s)

21. Mass - Mole Relationships of a Compound
For an Element
For a Compound
Mass (g) of
Element
Mass (g)
of compound
Moles
of Element
Amount (mol)
of compound
Amount (mol)
of compound
Molecules
(or formula units
of compound)
Atoms
of Element

22. Calculating the Number of Moles and Atoms in a Given Mass of Element
Problem: Tungsten (W) is the element used as the filament in light
bulbs, and has the highest melting point of any element
3680oC. How many moles of tungsten, and atoms of the
element are contained in a 35.0 mg sample of the metal?
Plan: Convert mass into moles by dividing the mass by the atomic
weight of the metal, then calculate the number of atoms by
multiplying by Avogadro’s number!
Solution: Converting from mass of W to moles:
Moles of W = mg W x = mol
1.90 x mol
NO. of W atoms = 1.90 x mol W x =
= 1.15 x 1020 atoms of Tungsten
1 mol W
183.9 g W
6.022 x 1023 atoms
1 mole of W

23. Calcite is a mineral composed of calcium carbonate, CaCO3
Calcite is a mineral composed of calcium carbonate, CaCO3. A sample of calcite composed of pure calcium carbonate weighs 23.6 g. How many moles of calcium carbonate is this?
First, find the molar mass of HNO3:
1 Ca 1(40.08) = 40.08
1 C 1(12.01) = 12.01
3 O 3(16.00) = 48.00
100.09
2 decimal places
g/mol 23

24. Next, find the number of moles in 23.6 g:24

25. The daily requirement of chromium in the human diet is 1. 0 × 10-6 g
The daily requirement of chromium in the human diet is 1.0 × 10-6 g. How many atoms of chromium does this represent? 25

26. (2 significant figures)
First, find the molar mass of Cr:
1 Cr 1(51.996) =
Now, convert 1.0 x 10-6 grams to moles:
= x 1016 atoms
1.2 x 1016 atoms
(2 significant figures) 26

27. Chemical Formulas
Empirical Formula - Shows the relative number of atoms
of each element in the compound. It is the simplest
formula, and is derived from masses of the elements.
Molecular Formula - Shows the actual number of atoms
of each element in the molecule of the compound.
Structural Formula - Shows the actual number of atoms,
and the bonds between them ; that is, the arrangement
of atoms in the molecule. 27

28. Percentage Composition
The mass percentage of each element in the compound
The composition is determined by experiment, often by combustion. When a compound is burned, its component elements form oxides—for example, CO2 and H2O. The CO2 and H2O are captured and weighed to determine the amount of C and H in the original compound. 28

29. Lead(II) chromate, PbCrO4, is used as a paint pigment (chrome yellow)
Lead(II) chromate, PbCrO4, is used as a paint pigment (chrome yellow). What is the percentage composition of lead(II) chromate?
First, find the molar mass of PbCrO4:
1 Pb 1(207.2) = 207.2
1 Cr 1(51.996) =
4 O 4(16.00) = 64.00
(1 decimal place)
323.2 g/mol 29

30. Now, convert each to percent composition:
Check: = 30

31. Flow Chart of Mass Percentage Calculation
Moles of X in one
mole of Compound
M (g / mol) of X
Mass (g) of X in one
mole of compound
Divide by mass (g) of one mole
of compound
Mass fraction of X
Multiply by 100
Mass % of X

32. Calculating Mass Percentage and Masses of
Elements in a Sample of a Compound - I
Problem: Sucrose (C12H22O11) is common table sugar.
( a) What is the mass percent of each element in sucrose?
( b) How many grams of carbon are in g of sucrose?
(a) Determining the mass percent of each element:
mass of C = 12 x g C/mol = g C/mol
mass of H = 22 x g H/mol = g H/mol
mass of O = 11 x g O/mol = g O/mol
g/mol
Finding the mass fraction of C in Sucrose & % C :
Total mass of C g C
mass of 1 mole of sucrose g Cpd
Mass Fraction of C = =
= To find mass % of C = x 100% = %

33. Calculating Mass Percents and Masses of
Elements in a Sample of Compound - II
(a) continued
Mass % of H = x 100% = x 100%
= 6.479% H
Mass % of O = x 100% = x 100%
= % O
(b) Determining the mass of carbon:
Mass (g) of C = mass of sucrose X( mass fraction of C in sucrose)
Mass (g) of C = g sucrose X = g C
mol H x M of H x g H
mass of 1 mol sucrose g
mol O x M of O x g O
mass of 1 mol sucrose g
g C
1 g sucrose

34. Empirical and Molecular Formulas
Empirical Formula - The simplest formula for a compound
that agrees with the elemental analysis! The
smallest set of whole numbers of atoms.
Molecular Formula - The formula of the compound as it
exists, it may be a multiple of the Empirical
formula.

35. Steps to Determine Empirical Formulas
Mass (g) of Element
M (g/mol )
Moles of Element
use no. of moles as subscripts
Preliminary Formula
change to integer subscripts
Empirical Formula

36. Empirical Formula (Simplest Formula)
The formula of a substance written with the smallest integer subscripts
For example:
The empirical formula for N2O4 is NO2.
The empirical formula for H2O2 is HO 36

37. Some Examples of Compounds with the Same Elemental Ratio’s
Empirical Formula Molecular Formula
CH2(unsaturated Hydrocarbons) C2H4 , C3H6 , C4H8
OH or HO H2O2
S S8
P P4
Cl Cl2
CH2O (carbohydrates) C6H12O6

38. Determining the Empirical Formula Beginning with percent composition:
Assume exactly 100 g so percentages convert directly to grams.
Convert grams to moles for each element.
Manipulate the resulting mole ratios to obtain whole numbers. 38

39. Manipulating the ratios:
Divide each mole amount by the smallest mole amount.
If the result is not a whole number:
Multiply each mole amount by a factor.
For example:
If the decimal portion is 0.5, multiply by 2.
If the decimal portion is 0.33 or 0.67, multiply by 3.
If the decimal portion is 0.25 or 0.75, multiply by 4. 39

40. Benzene is composed of 92. 3% carbon and 7. 7% hydrogen
Benzene is composed of 92.3% carbon and 7.7% hydrogen. What is the empirical formula of benzene?
Empirical formula: CH 40

41. Molecular Formula
A formula for a molecule in which the subscripts are whole-number multiples of the subscripts in the empirical formula 41

42. To determine the molecular formula:
Compute the empirical formula weight.
Find the ration of the molecular weight to the empirical formula weight.
Multiply each subscript of the empirical formula by n. 42

43. Combustion Train for the Determination of the Chemical Composition of Organic Compounds.2 m 2
CnHm + (n+ ) O2 = n CO2(g) H2O(g)
Fig. 3.4

44. Determine % composition by combustion
Benzene is a liquid compound composed of carbon and hydrogen; it is used in the preparation of polystyrene plastic. A sample of benzene weighing 342 mg is burned in oxygen and forms 1158 mg of carbon dioxide. What is the percentage composition of benzene? 44

45. 1. Use the mass of CO2 to find the mass of carbon from the benzene.
Strategy
1. Use the mass of CO2 to find the mass of carbon from the benzene.
2. Use the mass of benzene and the mass of carbon to find the mass of hydrogen.
3. Use these two masses to find the percent composition. 45

46. First, find the mass of C in 1156 mg of CO2:
= mg C 46

47. Now, we can find the percentage composition:
Next, find the mass of H in the benzene sample:
342 mg benzene
mg C
26.5 mg H
(the decimal is not significant)
Now, we can find the percentage composition: 47

48. Determining a Chemical Formula from Combustion Analysis - I
Problem: Erthrose (M = 120 g/mol) is an important chemical
compound as a starting material in chemical synthesis, and
contains Carbon Hydrogen, and Oxygen. Combustion
analysis of a mg sample yielded g CO2 and
g H2O.
Plan: We find the masses of Hydrogen and Carbon using the mass
fractions of H in H2O, and C in CO2. The mass of Carbon and
Hydrogen are subtracted from the sample mass to get the mass
of Oxygen. We then calculate moles, and construct the empirical
formula, and from the given molar mass we can calculate the
molecular formula.

49. Determining a Chemical Formula from Combustion Analysis - II
Calculating the mass fractions of the elements:
Mass fraction of C in CO2 = =
= = g C / 1 g CO2
Mass fraction of H in H2O = =
= = g H / 1 g H2O
Calculating masses of C and H
Mass of Element = mass of compound x mass fraction of element
mol C x M of C
mass of 1 mol CO2
1 mol C x g C/ 1 mol C
44.01 g CO2
mol H x M of H
mass of 1 mol H2O
2 mol H x g H / 1 mol H
18.02 g H2O

50. Determining a Chemical Formula from Combustion Analysis - III
g C
1 g CO2
Mass (g) of C = g CO2 x = g C
Mass (g) of H = g H2O x = g H
Calculating the mass of O:
Mass (g) of O = Sample mass -( mass of C + mass of H )
= g g C g H = g O
Calculating moles of each element:
C = g C / g C/ mol C = mol C
H = g H / g H / mol H = mol H
O = g O / g O / mol O = mol O
C H O = CH2O formula weight = 30 g / formula
120 g /mol / 30 g / formula = 4 formula units / cpd = C4H8O4
g H
1 g H2O

51. Some Compounds with Empirical Formula
CH2O (Composition by Mass 40.0% C, 6.71% H, 53.3%O)
Molecular M
Formula (g/mol) Name Use or Function
CH2O Formaldehyde Disinfectant; Biological
preservative
C2H4O Acetic acid Acetate polymers; vinegar
( 5% solution)
C3H6O Lactic acid Causes milk to sour; forms
in muscle during exercise
C4H8O Erythrose Forms during sugar
metabolism
C5H10O Ribose Component of many nucleic
acids and vitamin B2
C6H12O Glucose Major nutrient for energy
in cells

52. Ascorbic Acid ( Vitamin C ) - I Contains C , H , and O
Upon combustion in excess oxygen, a mg sample yielded 9.74 mg CO2 and mg H2O
Calculate its Empirical formula!
C: x10-3g CO2 x(12.01 g C/44.01 g CO2)
= 2.65 x 10-3 g C
H: 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O)
= x 10-4 g H
Mass Oxygen = 6.49 mg mg mg
= mg O

53. Vitamin C Combustion - II
C = 2.65 x 10-3 g C / ( g C / mol C ) =
= 2.21 x 10-4 mol C
H = x 10-3 g H / ( g H / mol H ) =
= 2.92 x 10-4 mol H
O = 3.54 x 10-3 g O / ( g O / mol O ) =
= 2.21 x mol O
Divide each by 2.21 x 10-4
C = Multiply each by = 3.00 = 3.0
H = = 3.96 = 4.0
O = = 3.00 = 3.0
C3H4O3

54. H H H
C6H8O6

55. Atoms Molecules Moles Molecular Formula Avogadro’s Number 6.022 x 1023

56. Chemical Equations Reactants Products 2 H2 (g) + O2 (g) 2 H2O (g)
Qualitative Information:
Reactants
Products
States of Matter: (s) solid
(l) liquid
(g) gaseous
(aq) aqueous
2 H2 (g) + O2 (g) 2 H2O (g)

57. Chemical Equation Calculation - I
Atoms (Molecules)
Avogadro’s
Number
6.02 x 1023
Molecules
Reactants
Products

58. Chemical Equation Calculation - II
Mass
Atoms (Molecules)
Molecular
Weight
Avogadro’s
Number
g/mol
6.02 x 1023
Molecules
Reactants
Products
Moles

59. Interpreting a Chemical Equation
Stoichiometry
The calculation of the quantities of reactants and products involved in a chemical reaction
Interpreting a Chemical Equation
The coefficients of the balanced chemical equation may be interpreted in terms of either (1) numbers of molecules (or ions or formula units) or (2) numbers of moles, depending on your needs. 59

60. Information Contained in a Balanced Equation
Viewed in Reactants Products
terms of: C2H6 (g) O2 (g) = 4 CO2 (g) + 6 H2O(g) + Energy
Molecules 2 molecules of C2H6 + 7 molecules of O2 =
4 molecules of CO2 + 6 molecules of H2O
Amount (mol) mol C2H mol O2 = 4 mol CO mol H2O
Mass (amu) amu C2H amu O2 =
amu CO amu H2O
Mass (g) g C2H g O2 = g CO g H2O
Total Mass (g) g = g

61. To find the amount of B (one reactant or product) given the amount of A (another reactant or product):
1. Convert grams of A to moles of A
 Using the molar mass of A
2. Convert moles of A to moles of B
 Using the coefficients of the balanced chemical equation
3. Convert moles of B to grams of B
 Using the molar mass of B 61

62. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
How many grams of CO2 are produced when 20.0 g of propane is burned? 62

63. (3 significant figures)
Molar masses
C3H8: 3(12.01) + 8(1.008) = g
CO2: 1(12.01) + 2(16.00) = g
59.9 g CO2
(3 significant figures) 63

64. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
How many grams of O2 are required to burn 20.0 g of propane? 64

65. (3 significant figures)
Molar masses:
O2 2(16.00) = g
C3H8 3(12.01) + 8(1.008) = g
72.6 g O2
(3 significant figures) 65

66. Once one reactant has been completely consumed, the reaction stops.
Limiting Reactant
The reactant that is entirely consumed when a reaction goes to completion
Once one reactant has been completely consumed, the reaction stops.
Any problem giving the starting amount for more than one reactant is a limiting reactant problem. 66

67. How can we determine the limiting reactant?
All amounts produced and reacted are determined by the limiting reactant.
How can we determine the limiting reactant?
Use each given amount to calculate the amount of product produced.
The limiting reactant will produce the lesser or least amount of product. 67

68. ZrCl4(g) + 2Mg(s)  2MgCl2(s) + Zr(s)
Magnesium metal is used to prepare zirconium metal, which is used to make the container for nuclear fuel (the nuclear fuel rods):
ZrCl4(g) + 2Mg(s)  2MgCl2(s) + Zr(s)
How many moles of zirconium metal can be produced from a reaction mixture containing 0.20 mol ZrCl4 and 0.50 mol Mg? 68

69. ZrCl4 is the limiting reactant.
0.20 mol Zr will be produced. 69

70. Urea, CH4N2O, is used as a nitrogen fertilizer
Urea, CH4N2O, is used as a nitrogen fertilizer. It is manufactured from ammonia and carbon dioxide at high pressure and high temperature:
2NH3 + CO2(g)  CH4N2O + H2O
In a laboratory experiment, 10.0 g NH3 and 10.0 g CO2 were added to a reaction vessel. What is the maximum quantity (in grams) of urea that can be obtained? How many grams of the excess reactant are left at the end of the reactions? 70

71. CO2 is the limiting reactant.
Molar masses
NH3 1(14.01) + 3(1.008) = g
CO2 1(12.01) + 2(16.00) = g
CH4N2O 1(12.01) + 4(1.008) + 2(14.01) + 1(16.00) = g
CO2 is the limiting reactant.
13.6 g CH4N2O will be produced. 71

72. To find the excess NH3, we find how much NH3 reacted:
Now subtract the amount reacted from the starting amount:
10.0 at start
reacted
2.27 g remains
2.3 g NH3 is left unreacted.
(1 decimal place) 72

73. Theoretical Yield
The maximum amount of product that can be obtained by a reaction from given amounts of reactants. This is a calculated amount. 73

74. Actual Yield
The amount of product that is actually obtained. This is a measured amount.
Percentage Yield 74

75. (2 significant figures)
2NH3 + CO2(g)  CH4N2O + H2O
When 10.0 g NH3 and 10.0 g CO2 are added to a reaction vessel, the limiting reactant is CO2. The theoretical yield is 13.6 of urea. When this reaction was carried out, 9.3 g of urea was obtained. What is the percent yield?
Theoretical yield = 13.6 g
Actual yield = 9.3 g
= 68% yield
(2 significant figures) 75

76. Other Resources
Visit the student website at college.hmco.com/pic/ebbing9e 76

77. The chemical name of table sugar is sucrose, C12H22O11
The chemical name of table sugar is sucrose, C12H22O11. How many grams of carbon are in 68.1 g of sucrose.
First, find the molar mass of C12H22O11:
12 C 12(12.01) =
11 O 11(16.00) =
22 H 22(1.008) =
(2 decimal places)
g/mol 77

78. Now, find the mass of carbon in 61.8 g sucrose:78

79. Sodium pyrophosphate is used in detergent preparations
Sodium pyrophosphate is used in detergent preparations. It is composed of 34.5% Na, 23.3% P, and 42.1% O. What is its empirical formula?
Empirical formula
Na4P2O7 79

80. Hexamethylene is one of the materials used to produce a type of nylon
Hexamethylene is one of the materials used to produce a type of nylon. It is composed of 62.1% C, 13.8% H, and 24.1% N. Its molecular weight is 116 amu. What is its molecular formula?
Empirical formula
C3H8N 80

81. Sodium pyrophosphate is used in detergent preparations
Sodium pyrophosphate is used in detergent preparations. It is composed of 34.5% Na, 23.3% P, and 42.1% O. What is its empirical formula?
Empirical formula
Na4P2O7 81

82. Hexamethylene is one of the materials used to produce a type of nylon
Hexamethylene is one of the materials used to produce a type of nylon. It is composed of 62.1% C, 13.8% H, and 24.1% N. Its molecular weight is 116 amu. What is its molecular formula?
Empirical formula
C3H8N 82

83. Molecular formula: C6H16N2
The empirical formula is C3H8N.
Find the empirical formula weight:
3(12.01) + 8(1.008) + 1(14.01) = amu
Molecular formula: C6H16N2 83

84. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
How many grams of CO2 are produced when 20.0 g of propane is burned? 84

85. (3 significant figures)
Molar masses
C3H8: 3(12.01) + 8(1.008) = g
CO2: 1(12.01) + 2(16.00) = g
59.9 g CO2
(3 significant figures) 85

86. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
How many grams of O2 are required to burn 20.0 g of propane? 86

87. (3 significant figures)
Molar masses:
O2 2(16.00) = g
C3H8 3(12.01) + 8(1.008) = g
72.6 g O2
(3 significant figures) 87

88. Converting a Concentrated Solution to a Dilute Solution