Chapter #3 Chemical Composition.

Chapter #3 Chemical Composition

Chemical Reactions A chemical reactions is an abbreviated way to show a physical orchemical change A chemical change alters the physical and chemical properties of a substance Factors that indicate a chemical change Change in color Temperature change Change in odor Change in taste (we do not taste chemicals) Reactions always contain an arrow that separates the reactants from the products Reactants Products

Types of Chemical Reactions
Combination reaction (synthesis) Elements for reactants Examples: H2 + O2 H2O N2 + H2 NH3 Al + O2 Al2O3 The Law of Conservation of matter, states matter cannot be created nor destroyed, that means equations must be balanced.

Combination reaction Continued Balance the first equation H2 + O2 H2O Note two oxygen atoms on the reactant side and only one on the product side, therefore place a two in front of water

Balance the first equation H2 + O2 2H2O Note two oxygen atoms on the reactant side and only one on the product side, therefore place a two in front of water The two now doubles everything in water, thus 4 hydrogen and 2 oxygen. Now place a 2 in front of hydrogen.

Balance the first equation 2H2 + O2 2H2O Note two oxygen atoms on the reactant side and only one on the product side, therefore place a two in front of water The two now doubles everything in water, thus 4 hydrogen and 2 oxygen. Now place a 2 in front of hydrogen.

Now balance the second equation N2 + H2 NH3 Note two nigrogen atoms on the reactant side and only one on the product side. Place a 2 in front of ammonia

Now balance the second equation N2 + H2 2NH3 Note two nitrogen atoms on the reactant side and only one on the product side. Place a 2 in front of ammonia. This makes 2 nitrogen atoms and 6 hydrogen atoms. Now place a 3 in front of hydrogen to balance hydrogen atoms.

Now balance the second equation N H2 2NH3 Note two nitrogen atoms on the reactant side and only one on the product side. Place a 2 in front of ammonia. This makes 2 nitrogen atoms and 6 hydrogen atoms. Now place a 3 in front of hydrogen to balance hydrogen atoms.

Decomposition Reaction Compounds form simpler compounds or elements. Examples H2O H O2

Decomposition Reaction Compounds form simpler compounds or elements. Examples 2H2O H O2

Decomposition Reaction Compounds form simpler compounds or elements. Examples 2H2O 2H O2 Notice decomposition reactions are the opposite of combination reactions

Single Replacement reactions have an element and a compound for reactants. Example: Zn + HCl How do we predict the products? Trade places with the metal or nonmetal with the metal or nonmetal in the compound

Single Replacement reactions have an element and a compound for reactants. Example: Zn + HCl ZnCl H Now make the products stable. Slide with Clyde

Single Replacement reactions have an element and a compound for reactants. Example: Zn + HCl ZnCl H2 Now make the products stable. Slide with Clyde

Single Replacement reactions have an element and a compound for reactants. Example: Zn + HCl ZnCl H2 Now make the products stable. Now Balance

Single Replacement reactions have an element and a compound for reactants. Example: Zn + 2HCl ZnCl H2 Now make the products stable. Now Balance

Single Replacement reactions have an element and a compound for reactants. Another Example: Cl2 + MgBr2 How do we predict the products? Trade places with the metal or nonmetal with the metal or nonmetal in the compound. In this case we are trading nonmetals

Single Replacement reactions have an element and a compound for reactants. Another Example: Cl2 + MgBr2 Br MgCl How do we predict the products? Trade places with the metal or nonmetal with the metal or nonmetal in the compound. In this case we are trading nonmetals

Single Replacement reactions have an element and a compound for reactants. Another Example: Cl2 + MgBr2 Br MgCl2 How do we predict the products? Trade places with the metal or nonmetal with the metal or nonmetal in the compound. In this case we are trading nonmetals

Double Replacement reactions contain compounds as reactants. HCl + Ca(OH)2 CaCl HOH Check formulas, and slide with Clyde when necessary

Double Replacement reactions contain compounds as reactants. 2HCl + Ca(OH)2 CaCl HOH Check formulas, and slide with Clyde when necessary Now Balance!

Combustion Reactions occur when an element or compound combine with oxygen to produce oxides of each element. H2 + O2 CH4 + O2 What is the oxide of hydrogen?

Combustion Reactions occur when an element or compound combine with oxygen to produce oxides of each element. H2 + O2 CH4 + O2 What is the oxide of hydrogen? Water

Combustion Reactions occur when an element or compound combine with oxygen to produce oxides of each element. H2 + O2 H2O CH4 + O2 What is the oxide of hydrogen? Water And the oxide of carbon?

Combustion Reactions occur when an element or compound combine with oxygen to produce oxides of each element. H2 + O2 H2O CH4 + O2 CO H2O What is the oxide of hydrogen? Water And the oxide of carbon? Carbon dioxide

Combustion Reactions occur when an element or compound combine with oxygen to produce oxides of each element. 2H2 + O2 2H2O CH4 + O2 CO2 + H2O Now balance

Combustion Reactions occur when an element or compound combine with oxygen to produce oxides of each element. 2H2 + O2 2H2O CH4 + O2 CO H2O Now balance

Combustion Reactions occur when an element or compound combine with oxygen to produce oxides of each element. 2H2 + O2 2H2O CH O2 CO H2O Now balance

Ionic Solution Formation
KCN (S) K+ (aq) + CN- (aq) Ionic equation Note: Not all ionic solutes are soluble in water. How can we tell if an ionic solute is soluble in water?

Ionic Equations K+ (aq) + CN- (aq) KCN (S)
When some ionic solids are dissolved in water equations (physical or chemical) can be used to explain the process as shown below: KCN (S) K+ (aq) + CN- (aq) Ionic equation Note: Not all ionic solutes are soluble in water. How can we tell if an ionic solute is soluble in water?

Ionic Equations K+ (aq) + CN- (aq) KCN (S)
When some ionic solids are dissolved in water equations (physical or chemical) can be used to explain the process as shown below: KCN (S) K+ (aq) + CN- (aq) Ionic equation Note: Not all ionic solutes are soluble in water. How can we tell if an ionic solute is soluble in water? The solubility rules gives ionic solubility.

Solubility Rules There are some more specific rules that allows us to better estimate the solubility of ionic compounds. You will be given these if you need them.

Ionic Equations AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)
Using the solubility rules write the formula equation, the ionic equation and the net ionic equation when aqueous silver nitrate is combined with aqueous sodium chloride. AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) Formula Equation

Using the solubility rules write the formula equation, the ionic equation and the net ionic equation when aqueous silver nitrate is combined with aqueous sodium chloride. AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) Formula Equation Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl- (aq) AgCl (s) + Na+ (aq) + NO3- (aq) Ionic Equation

Using the solubility rules write the formula equation, the ionic equation and the net ionic equation when aqueous silver nitrate is combined with aqueous sodium chloride. AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) Formula Equation Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl- (aq) AgCl (s) + Na+ (aq) + NO3- (aq) Ionic Equation Spectator ions are ions that are identical on the reactants and products side of the equation.

Using the solubility rules write the formula equation, the ionic equation and the net ionic equation when aqueous silver nitrate is combined with aqueous sodium chloride. AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) Formula Equation Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl- (aq) AgCl (s) + Na+ (aq) + NO3- (aq) Ionic Equation Spectator ions are ions that are identical on the reactants and products side of the equation. Place a around the spectator ions.

Ag+ (aq) + Cl- (aq) AgCl (s)
Ionic Equations Using the solubility rules write the formula equation, the ionic equation and the net ionic equation when aqueous silver nitrate is combined with aqueous sodium chloride. AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) Formula Equation Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl- (aq) AgCl (s) + Na+ (aq) + NO3- (aq) Ionic Equation Spectator ions are ions that are identical on the reactants and products side of the equation. Eliminating the spectator ions produces the netionic equation. Ag+ (aq) + Cl- (aq) AgCl (s) Net ionic equation

Yet Another Ionic Equation
Write the formula, ionic and net ionic equations when aqueous sodium chloride combines with aqueous calcium bromide.

Write the formula, ionic and net ionic equations when aqueous sodium chloride combines with aqueous calcium bromide. NaCl (aq) + CaBr2 (aq) CaCl2 (aq) + NaBr (aq) Now balance

Write the formula, ionic and net ionic equations when aqueous sodium chloride combines with aqueous calcium bromide. 2NaCl (aq) + CaBr2 (aq) CaCl2 (aq) + 2NaBr (aq) Now balance

Write the formula, ionic and net ionic equations when aqueous sodium chloride combines with aqueous calcium bromide. 2 NaCl(aq) + CaBr CaCl2(aq) NaBr(aq) 2 2Na+(aq) + 2Cl-(aq) + Ca2+(aq) + 2Br-(aq) Ca2+(aq) + 2Cl-(aq) + 2Na+(aq) +2 Br-(aq) Ionic equation

Write the formula, ionic and net ionic equations when aqueous sodium chloride combines with aqueous calcium bromide. 2 NaCl(aq) + CaBr CaCl2(aq) NaBr(aq) 2 Now balance 2Na+(aq) + 2Cl-(aq) + Ca2+(aq) + 2Br-(aq) Ca2+(aq) + 2Cl-(aq) + 2Na+(aq) +2 Br-(aq) Ionic equation No net ionic equation No Reaction (NR)

REDOX reactions where the oxidation number changes from reactants to products. Oxidation is when the oxidation number increases, by losing of electrons. Reduction is when the oxidation number decreases by gaining electrons. Consider the following equation: H2 + O2 H2O What are the oxidation numbers of hydrogen and oxygen?

REDOX REACTIONS H2 + O2 H2O How about hydrogen and oxygen in water?
2(1+) 2- = 0 H2 + O2 H2O How about hydrogen and oxygen in water?

2(1+) 2- = 0 H2 + O2 H2O How about hydrogen and oxygen in water? Oxidation is caused by the oxygen molecule, so it is referred to as the oxidizing agent (OA) Reduction is caused by the hydrogen molecule, so it is referred to as the reducing agent (RA) reduced oxidized

REDOX REACTIONS Note: All of the previously discussed reactions are REDOX except the double replacement reactions. The number of electrons lost is equal to the number of electrons gained in a reaction. Why? Most elements have variable oxidation numbers, except for hydrogen, oxygen, and the memorized polyatomic ions.

REDOX REACTIONS Oxidation numbers for a compound must add up to equal zero, while the oxidation numbers for a polyatomic ion must up to equal the charge of that ion. Consider the following chlorine compounds HClO4, HClO3, HClO2, HClO, Cl2, HCl What is the oxidation number of chlorine in each of these compounds, assuming H 1+ and oxygen is 2- 1+ 4(2-)=0

REDOX REACTIONS Oxidation numbers for a compound must add up to equal zero, while the oxidation numbers for a polyatomic ion must up to equal the charge of that ion. Consider the following chlorine compounds HClO4, HClO3, HClO2, HClO, Cl2, HCl What is the oxidation number of chlorine in each of these compounds, assuming H is 1+ and oxygen is 2- 1+ 7+ 4(2-)=0

REDOX REACTIONS Oxidation numbers for a compound must add up to equal zero, while the oxidation numbers for a polyatomic ion must up to equal the charge of that ion. Consider the following chlorine compounds HClO4, HClO3, HClO2, HClO, Cl2, HCl What is the oxidation number of chlorine in each of these compounds, assuming H is 1+ and oxygen is 2- 1+ 7+ 4(2-)=0 5+

REDOX REACTIONS Oxidation numbers for a compound must add up to equal zero, while the oxidation numbers for a polyatomic ion must up to equal the charge of that ion. Consider the following chlorine compounds HClO4, HClO3, HClO2, HClO, Cl2, HCl What is the oxidation number of chlorine in each of these compounds, assuming H is 1+ and oxygen is 2- 1+ 7+ 4(2-)=0 5+ 3+

REDOX REACTIONS Oxidation numbers for a compound must add up to equal zero, while the oxidation numbers for a polyatomic ion must up to equal the charge of that ion. Consider the following chlorine compounds HClO4, HClO3, HClO2, HClO, Cl2, HCl What is the oxidation number of chlorine in each of these compounds, assuming H is 1+ and oxygen is 2- 1+ 7+ 4(2-)=0 5+ 3+ 1+

REDOX REACTIONS Oxidation numbers for a compound must add up to equal zero, while the oxidation numbers for a polyatomic ion must up to equal the charge of that ion. Consider the following chlorine compounds HClO4, HClO3, HClO2, HClO, Cl2, HCl What is the oxidation number of chlorine in each of these compounds, assuming H is 1+ and oxygen is 2- 1+ 7+ 4(2-)=0 5+ 3+ 1+ 1-

REDOX REACTIONS 3(2-)=2- How about sulfur in SO3 2-

REDOX REACTIONS How about sulfur in SO3 2- How about carbon in C6H12O6
4+ 3(2-)=2- How about sulfur in SO3 2- How about carbon in C6H12O6 12(1+) +6(2-)=0

REDOX REACTIONS How about sulfur in SO3 2- How about carbon in C6H12O6
4+ 3(2-)=2- How about sulfur in SO3 2- How about carbon in C6H12O6 0 + 12(1+) +6(2-)=0

REDOX reactions where the oxidation number changes from reactants to products. Oxidation is when the oxidation number increases, by losing of electrons. Reduction is when the oxidation number decreases by gaining electrons. Consider the following equation: H2 + O2 H2O What are the oxidation numbers of hydrogen and oxygen?

2(1-) 2- = 0 H2 + O2 H2O How about hydrogen and oxygen in water?

2(1+) 2- = 0 H2 + O2 H2O How about hydrogen and oxygen in water? Oxidation is caused by the oxygen molecule, so it is referred to as the oxidizing agent (OA) Reduction is caused by the hydrogen molecule, so it is referred to as the reducing agent (RA) reduced oxidized

REDOX REACTIONS Note: All of the previously discussed reactions are REDOX except the double replacement reactions. The number of electrons lost is equal to the number of electrons gained in a reaction. Why? Most elements have variable oxidation numbers, except for hydrogen, oxygen, and the memorized polyatomic ions.

Balancing Redox Reactions
I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O Balance remaining hydrogen atoms by adding H+ Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. HNO Cu2O → Cu(NO3) NO H2O

Balancing REDOX Reactions
I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O Balance remaining hydrogen atoms by adding H+ Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 1+ ? 3(2-)=0 HNO Cu2O → Cu(NO3) NO H2O

I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O Balance remaining hydrogen atoms by adding H+ Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 1+ 5+ 3(2-)=0 2(?)+ 2-=0 HNO Cu2O → Cu(NO3) NO H2O

I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 HNO Cu2O → Cu(NO3) NO H2O

Balancing Redox Reactions
I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. ? + 2(1-)=0 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 HNO Cu2O → Cu(NO3) NO H2O

I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 2+ + 2(1-)=0 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 HNO Cu2O → Cu(NO3) NO H2O

I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 2(1-)=0 ? + 2- =0 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 HNO Cu2O → Cu(NO3) NO H2O

I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 2(1-)=0 =0 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 HNO Cu2O → Cu(NO3) NO H2O

I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 2+ 2(1-)=0 =0 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 HNO Cu2O → Cu(NO3) NO H2O oxidized reduced

I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 2+ 2(1-)=0 =0 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 HNO Cu2O → 2 Cu(NO3) NO H2O oxidized reduced

I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 2+ 2(1-)=0 =0 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 HNO Cu2O → 3 (2) Cu(NO3) NO H2O Oxidized3( -2e) Reduced 2(+3)e

I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 2+ 2(1-)=0 =0 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 2HNO Cu2O → 3 (2) Cu(NO3) NO H2O Oxidized3( -2e) Reduced 2(+3)e

I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 2+ 2(1-)=0 =0 2HNO Cu2O → 3 (2) Cu(NO3) NO + H2O Oxidized3( -2e) Reduced 2(+3)e

I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 2+ 2(1-)=0 =0 14HNO Cu2O → 3 (2) Cu(NO3) NO + 7 H2O Oxidized3( -2e) Reduced 2(+3)e

I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 2+ 2(1-)=0 =0 14 HNO Cu2O → Cu(NO3) NO H2O Oxidized3( -2e) Reduced 2(+3)e

OX # BALANCING EXAMPLE MnO Cl- → Mn Cl2

OX # BALANCING EXAMPLE 7+ MnO Cl- → Mn Cl2

OX # BALANCING EXAMPLE 7+ 1- MnO Cl- → Mn Cl2

OX # BALANCING EXAMPLE MnO4 - + Cl- → Mn2+ + Cl2 oxidized reduced 7+
1- MnO Cl- → Mn Cl2 oxidized reduced

OX # BALANCING EXAMPLE MnO4 - + Cl- → Mn2+ + Cl2
7+ 1- MnO Cl- → Mn Cl2 oxidized reduced Step C, balance atoms oxidized or reduced

7+ 1- MnO Cl- → Mn Cl2 2 oxidized reduced Step C, balance atoms oxidized or reduced

7+ 1- MnO Cl- → Mn Cl2 2 - 2 e- oxidized + 5 e- reduced Step d, balance electrons lost or gained. common denominator between 5 and 2 is 10. Therefore multiply Mn on both sides of the equation by two and Cl on both sides by 5.

7+ 1- MnO Cl- → Mn Cl2 2 5(2) 2 5 - 2 e- oxidized + 5 e- reduced Step d, balance electrons lost or gained. The common denominator between 5 and 2 is 10. Therefore multiply Mn on both sides of the equation by 2 and Cl on both sides by 5.

7+ 1- MnO Cl- → Mn Cl2 2 5(2) 2 5 - 2 e- oxidized + 5 e- reduced Step e, balance remaining elements by inspection. There are 8 oxygen atoms on the left. Oxygen is balanced by adding water to the appropriate side. In this case since there are 8 oxygen atoms on the reactant side which requires adding 8 water molecules to the product side of the equation.

OX # BALANCING EXAMPLE MnO4- + Cl- → Mn2+ + Cl2 + 8H2O
1- 7+ MnO Cl- → Mn Cl2 2 10 2 5 + 8H2O - 2 e- oxidized + 5 e- reduced Step e, balance remaining elements by inspection. There are 8 oxygen atoms on the left. Oxygen is balanced by adding water to the appropriate side. In this case since there are 8 oxygen atoms on the reactant side which requires adding 8 water molecules to the product side of the equation. Now the hydrogen atoms need to be balanced by adding 16 H+ to the reactant side.

OX # BALANCING EXAMPLE MnO4-+ Cl- → Mn2++ Cl2 + 8H2O 16 H++
7+ 1- MnO Cl- → Mn2++ Cl2 + 8H2O 16 H++ 2 10 2 5 - 2 e- oxidized + 5 e- reduced Step e, balance remaining elements by inspection. There are 8 oxygen atoms on the left. Oxygen is balanced by adding water to the appropriate side. In this case since there are 8 oxygen atoms on the reactant side which requires adding 8 water molecules to the product side of the equation. Now the hydrogen atoms need to be balanced by adding 16 H+ to the reactant side.

Balancing REDOX Equations by The Half Reaction Method

Half Reaction Steps 1. Write separate equations (Half-reactions) for oxidized and reduced substances. 2. For each half-reaction balance all elements, except hydrogen and oxygen a. Balance oxygen using H2O b. Balance hydrogen using H+ c. Balance charge in each half-reaction by adding electrons (reduction), or removing electrons (oxidation) to the appropriate half reaction. 3. Multiply each half-reaction by an integer so that the number of electrons lost equal the number of electrons gained a. Add half-reactions, and simplify b. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. c. Combine H+ and OH- ions to make water d. Simplify again if necessary.

Half Reaction Example MnO Fe2+ → Mn Fe3+

Half Reaction Example MnO4- + Fe2+ → Mn2+ + Fe3+
Step 1, Write half reactions

Step 1, Write half reactions MnO4- → Mn2+ → Fe3+ Fe2+

Step 2a, Balance Oxygen by adding water. MnO4- → Mn2+ + 4 H2O → Fe3+ Fe2+

Step 2b, Balance hydrogen by adding H+. 8 H+ + MnO4- → Mn2+ + 4 H2O → Fe3+ Fe2+

Step 2c, Balance charge by adding/removing e’s 8 H+ + MnO4- → Mn2+ + 4 H2O → Fe3+ Fe2+ In the top half equation the reactants have 7+ and the products 2+, adding 5 e’s to the reactant side gives 2+ on both sides. In the bottom half equation the reactants have 2+ and the products have 2+, removing 1 e from the reactant side gives 2+ on both sides.

Step 2c, Balance charge by adding/removing e’s 8 H+ + MnO4- + 5e- → Mn2+ + 4 H2O → Fe3+ Fe2+ In the top half equation the reactants have 7+ and the products 2+, adding 5 e’s to the reactant side gives 2+ on both sides. In the bottom half equation the reactants have 2+ and the products have 2+, removing 1 e from the reactant side gives 2+ on both sides.

Step 2c, Balance charge by adding/removing e’s 8 H+ + MnO4- + 5e- → Mn2+ + 4 H2O - e- → Fe3+ Fe2+ In the top half equation the reactants have 7+ and the products 2+, adding 5 e’s to the reactant side gives 2+ on both sides. In the bottom half equation the reactants have 2+ and the products have 2+, removing 1 e from the reactant side gives 2+ on both sides.

Step 3, The common denominator between 5 and 1 is 5. Multiply the bottom half equation by 5 8 H+ + MnO4- + 5e- → Mn2+ + 4 H2O - e- → Fe3+ Fe2+

Half Reaction Example ) MnO4- + Fe2+ → Mn2+ + Fe3+
Step 4, Add the two half equations together 8 H+ + MnO4- + 5e- → Mn2+ + 4 H2O ) 5( - e- → Fe3+ Fe2+

Half Reaction Example ) MnO4- + Fe2+ → Mn2+ + Fe3+
Step 4, Add the two half equations together 8 H+ + MnO4- + 5e- → Mn2+ + 4 H2O ) 5( - e- → Fe3+ Fe2+ → 8 H+ + MnO4- + 5 Fe2+ 5 Fe3+ + Mn2+ + 4 H2O

Other REDOX Examples HNO2 + Cr2O72- → Cr2+ + NO3- (acidic)
CN MnO → CNO MnO2 (basic) Al(s) + OH- (aq) → Al(OH)4- (aq) + H2 (g) (acidic or basic) Cl2 (g) → Cl- (aq) ClO- (aq) (basic) Ag (s) + CN- + O2 → Ag(CN)2 - (aq) (basic)

Real Life Examples of REDOX
REDOX reactions can be used to generate electricity. REDOX reactions can be used to protect metals from oxidation. REDOX reactions can be used to plate metals on to other metals or surfaces.

The Chemical Package About Packages
The baker uses a package called the dozen. All dozen packages contain 12 objects. The stationary store uses a package called a ream, which contains 500 sheets of paper. So what is the chemistry package?

The Chemical Package About Packages
The baker uses a package called the dozen. All dozen packages contain 12 objects. The stationary store uses a package called a ream, which contains 500 sheets of paper. So what is the chemistry package? Well, it is called the mole (Latin for heap). Each of the above packages contain a number of objects that are convenient to work with, for that particular discipline.

The Mole A mole contains 6.022X1023 particles, which is the number of carbon-12 atoms that will give a mass of grams, which is a convenient number of atoms to work with in the chemistry laboratory. The atomic weights listed on the periodic chart are the weights of a mole of atoms. For example a mole of hydrogen atoms weighs g and a mole of carbon atoms weighs g which are weighted averages of the natural abundance of isotopes for that element.

Moles of Objects Suppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows?

Moles of Objects Suppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows? Would cover the entire 50 states 60 miles deep

Moles of Objects Suppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows? Would cover the entire 50 states 60 miles deep How about a mole of computer paper instead of a ream of computer paper, how far would that stretch?

Moles of Objects Suppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows? Would cover the entire 50 states 60 miles deep How about a mole of computer paper instead of a ream of computer paper, how far would that stretch? Way past the planet Pluto!

Formula Weight Calculation
To calculate the molar mass of a compound we sum together the atomic weights of the atoms that make up the formula of the compound. This is called the formula weight (MW, M). Formula weights are the sum of atomic weights of atoms making up the formula. The following outlines how to find the formula weight of water symbol weight number H 1.01 X 2 = 2.02 O 16.0 X 1 = 16.0 18.0 g/mole

Percent Composition Find the formula weight and the percent composition of glucose (C6H12O6) symbol weight number 72.0 C 12.0 x 6 = H 1.01 x 12 = 12.12 O 16.0 x 6 = 96.0 180.1 g/mole 72.0 %C = X = 40.0 %C 180.1 12.12 %H = X = 6.73 %H 180.1 96.0 X = 53.3 %O %O = 180.1

Mole Concepts A mole of glucose (C6H12O6) contains X 1023 molecules of glucose. And 6 X X 1023 atoms of C. Since a mole is X particles then a mole of glucose must contain 6 moles of C atoms. How many moles of hydrogen atoms are contained in a mole of glucose? In 5 moles of H2SO4 how many moles of oxygen atoms is there?

Mole Concepts A mole of glucose (C6H12O6) contains X 1023 molecules of glucose. And 6 X X 1023 atoms of C. Since a mole is X particles then a mole of glucose must contain 6 moles of C atoms. How many moles of hydrogen atoms are contained in a mole of glucose? 12 Moles of hydrogen. How many moles of oxygen and hydrogen are in one mole of H2O contains:

Mole Concepts A mole of glucose (C6H12O6) contains X 1023 molecules of glucose. And 6 X X 1023 atoms of C. Since a mole is X particles then a mole of glucose must contain 6 moles of C atoms. How many moles of hydrogen atoms are contained in a mole of glucose? 12 Moles of hydrogen. How many moles of oxygen and hydrogen are in one mole of H2O contains: One mole of oxygen atoms Two moles of hydrogen atoms

Mole Concepts A mole of glucose (C6H12O6) contains X 1023 molecules of glucose. And 6 X X 1023 atoms of C. Since a mole is X particles then a mole of glucose must contain 6 moles of C atoms. How many moles of hydrogen atoms are contained in a mole of glucose? 12 Moles of hydrogen. How many moles of oxygen and hydrogen are in one mole of H2O contains: One mole of oxygen atoms Two moles of hydrogen atoms In 5 moles of H2SO4 how many moles of oxygen atoms is there?

Mole Concepts A mole of glucose (C6H12O6) contains X 1023 molecules of glucose. And 6 X X 1023 atoms of C. Since a mole is X particles then a mole of glucose must contain 6 moles of C atoms. How many moles of hydrogen atoms are contained in a mole of glucose? 12 Moles of hydrogen. How many moles of oxygen and hydrogen are in one mole of H2O contains: One mole of oxygen atoms Two moles of hydrogen atoms In 5 moles of H2SO4 how many moles of oxygen atoms is there? 20 moles of O atoms.

Mole Conversions In 50.0g of H2SO4 how many moles of sulfuric acid are there? 50.0g of H2SO4

Mole Conversions In 50.0g of H2SO4 how many moles of sulfuric acid are there? 50.0g of H2SO4 mole H2SO4 = 98.0g of H2SO4

Mole Conversions In 50.0g of H2SO4 how many moles of sulfuric acid are there? 50.0g of H2SO4 mole H2SO4 = mole H2SO4 98.0g of H2SO4

Mole Conversions In 50.0g of H2SO4 how many moles of oxygen atoms are there?

Mole Conversions In 50.0g of H2SO4 how many moles of oxygen atoms are there? 50.0g of H2SO4 =

Mole Conversions In 50.0g of H2SO4 how many moles of oxygen atoms are there? 50.0g of H2SO4 mole H2SO4 = 98.0g of H2SO4

Mole Conversions In 50.0g of H2SO4 how many moles of oxygen atoms are there? 50.0g of H2SO4 mole H2SO4 4mole O = mole H2SO4 98.0g of H2SO4

Mole Conversions In 50.0g of H2SO4 how many moles of oxygen atoms are there? 50.0g of H2SO4 mole H2SO4 4mole O = 2.04 mole O mole H2SO4 98.0g of H2SO4

Mole Conversions In 5 moles of H2SO4 how many atoms of oxygen are present?

Mole Conversions In 5 moles of H2SO4 how many atoms of oxygen are present? 5 moles H2SO4 =

Mole Conversions In 5 moles of H2SO4 how many atoms of oxygen are present? 5 moles H2SO4 4 mole O mole H2SO4

Mole Conversions In 5 moles of H2SO4 how many atoms of oxygen are present? 5 moles H2SO4 4 mole O 6.02 x 1023 atoms O = mole H2SO4 mole O 1.20 x 1025 atoms

Empirical Formulas Empirical formula is the smallest whole number ratio between atoms and can be calculated from the percent composition. Molecular formulas happen to be the exact number of atoms making up a molecule, and may or may no be the simplest whole number ratio. Molecular formulas are whole number multiples of the empirical formula.

Empirical Formula Steps
Assume 100 g of compound. Convert percent to a mass number. Convert the mass to moles. Divide each mole number by the smallest mole number. Rounding: If the decimal is ≤ 0.1, then drop the decimals If the decimal is ≥0.9, then round up. All other decimal need to be multiplied by a whole number until roundable.

Empirical Formula Example
A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #1 Assume 100 g of compound 75.0 g C 25.0 g H

A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #2 Convert grams to moles. Mole C 75.0 g C = mole C 12.01 g 25.0 g H Mole H = mole 1.008 g H

A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #3 Divide each mole number by the smallest. Mole C 75.0 g C = mole C 12.01 g 25.0 g H Mole H = mole 1.008 g H 6.225 24.802 = 1.00 = 3.98 6.225 6.225

A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≤ 0.1, drop decimals Mole C 75.0 g C = mole C 12.01 g 25.0 g H Mole H = mole 1.008 g H 6.225 24.802 = 1.00 = 3.98 6.225 6.225

A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≤ 0.1, drop decimals Mole C 75.0 g C = mole C 12.01 g 25.0 g H Mole H = mole 1.008 g H 6.225 24.802 = 1 C = 3.98 6.225 6.225

A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≥ 0.9, round up Mole C 75.0 g C = mole C 12.01 g 25.0 g H Mole H = mole 1.008 g H 6.225 24.802 = 1 C = 3.98 6.225 6.225

A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≥ 0.9, round up Mole C 75.0 g C = mole C 12.01 g 25.0 g H Mole H = mole 1.008 g H 6.225 24.802 = 1 C = 4 H 6.225 6.225

A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≥ 0.9, round up Mole C 75.0 g C = mole C 12.01 g 25.0 g H Mole H = mole 1.008 g H 6.225 24.802 = 1 C = 4 H 6.225 6.225 Empirical Formula = CH4

Molecular Formulas Empirical formula, is the smallest ratio between atoms in a molecular or formula unit. Molecular formula, is the exact number of atoms in a molecule; a whole number multiple of an empirical formula

Possible Molecular Formulas
Assume an empirical formula of C3H5O Empirical formula Integer Molecular Formula C3H5O C3H5O 1 2 3 4 5 C3H5O C3H5O C3H5O C3H5O

Assume an empirical formula of C3H5O Empirical formula Integer Molecular Formula C3H5O C3H5O 1 2 3 4 5 C3H5O C6H10O C3H5O C3H5O C3H5O

Assume an empirical formula of C3H5O Empirical formula Integer Molecular Formula C3H5O C3H5O 1 2 3 4 5 C3H5O C6H10O2 C9H15O3 C3H5O C3H5O C3H5O

Assume an empirical formula of C3H5O Empirical formula Integer Molecular Formula C3H5O C3H5O 1 2 3 4 5 C3H5O C6H10O2 C9H15O3 C3H5O C3H5O C12H20O4 C3H5O C15H25O5

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #1 Assume 100g of compound 83.6 g C 16.3 g H

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #2 Convert grams to moles 83.6 g C mole 12.01 g C 16.3 g H mole 1.008 g H

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #2 Convert grams to moles 83.6 g C mole = mole 12.01 g C 16.3 g H mole = mole 1.008 g H

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #3 Divide each mole number by the smallest. 83.6 g C mole = mole 12.01 g C 16.3 g H mole = mole 1.008 g H

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #3 Divide each mole number by the smallest. 6.961 83.6 g C mole = 1.00 = mole 6.961 12.01 g C 16.3 g H mole = mole = 2.32 1.008 g H

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #4 Round if---Not Roundable 6.961 83.6 g C mole = mole = 1.00 6.961 12.01 g C 16.3 g H mole = mole = 2.32 1.008 g H Step #4, Multiply by an integer until roundable 1.00 X 3 = 3 2.32 X 3 = 7 Empirical formula C3H7

Molecular Formula Integer
Divide empirical weight into molecular weight 3x12 + 7x1 =43 2 43 86 Now multiply the empirical formula by 2

Molecular Formula Integer
Divide empirical weight into molecular weight 3x12 + 7x1 =43 2 43 86 Now multiply the empirical formula by 2 Molecular Formula is C6 H14

Stoichiometry Sotichiometery is the process of converting quantities of reactants or products to other participants of a chemical or physical change using the coefficients of a balanced equation.

STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Example Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required.

STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Example Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2 H2O

STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2H2O 6.33 g H2

STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Example Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2 H2O 6.33 g H2 Mole H2 2.016 g H2O

STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Example Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2 H2O 6.33 g H2 Mole H2 2 Mole H2O 2.016 g H2O 2 Mole H2

STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Example Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2 H2O 6.33 g H2 Mole H2 2 Mole H2O 18.02 g H2O 2.016 g H2O 2 Mole H2 Mole H2O

STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2 H2O 6.33 g H2 Mole H2 2 Mole H2O 18.02 g H2O = 56.6 g H2O 2.016 g H2O 2 Mole H2 Mole H2O

Excess and Limiting Reactants are substances that can be changed into something else. For example, nails and boards are reactants for carpenters, while thread and fabric are reactants for the seamstress. And for a chemist hydrogen and oxygen are reactants for making water.

Building Houses Suppose, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?

Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? Yes, only one house!

Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we had enough boards?

Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we have enough boards?

Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we have enough boards? Yes, nails are in excess!

Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we have enough boards? Yes, nails are in excess! Nine more houses if we have an adequate amount of boards.

Making Water If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?

Making Water If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.

Making Water If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H2 + O H2O 10.0 g O2

Making Water If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H2 + O H2O 10.0 g O2 mole O2 32.0 g O2

Making Water If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H2 + O H2O 10.0 g O2 mole O2 2 mole H2 32.0 g O2 mole O2

Making Water If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H2 + O H2O 10.0 g O2 mole O2 2 mole H2 2.02 g H2 32.0 g O2 mole O2 mole H2

Making Water If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H2 + O H2O 10.0 g O2 mole O2 2 mole H2 2.02 g H2 = 1.26 g H2 32.0 g O2 mole O2 mole H2

Making Water Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.

Making Water Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water. 10.0 g O2 mole O2 32.0 g O2

Making Water Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water. 10.0 g O2 mole O2 2 mole H2O 32.0 g O2 mole O2

Making Water Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water. 10.0 g O2 mole O2 2 mole H2O 18.0 g H2O 32.0 g O2 mole O2 mole H2O

Making Water Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water. 10.0 g O2 mole O2 2 mole H2O 18.0 g H2O = 11.3 g H2O 32.0 g O2 mole O2 mole H2O

Percentage Yield The percent yield is a comparison of the laboratory answer to the correct answer which is determined by the conversion process. Suppose a student combined 10.0 g of oxygen and 10.0 g of hydrogen in the lab and recovered 8.66 g of water. What would be the percent yield?

Percentage Yield The percent yield is a comparison of the laboratory answer to the correct answer which is determined by the conversion process. Suppose a student combined 10.0 g of oxygen and 10.0 g of hydrogen in the lab and recovered 8.66 g of water. What would be the percent yield? Yield (the lab amount) percent yield = X 100 Theoretical Yield (by conversions) 8.66 percent yield = X 100 = 76.6% 11.3

Combustion Analysis Empirical formulas of hydrocarbons can be determined by combustion analysis. The complete combustion of a hydrocarbon produces carbon dioxide and water. Measuring the mass of the carbon dioxide and water produced can give the mass of hydrogen and carbon present in the compound. Subtracting the mass of carbon and hydrogen from the weight of the starting hydrocarbon gives the mass of the oxygen. Ascarite™ a commercial name for sodium or potassium hydroxide absorbs between 0-1 ppm of the carbon dioxide.

Combustion Analysis CO2 + 2 KOH K2CO3 + 2 H2O
Ascarite the weight increase of Ascarite gives the mass of the carbon dioxide according to the following equation. CO KOH K2CO H2O Vitamin C is essential for the prevention of scurvy. Combustion of a g sample of this hydrocarbon, which may or may not contain oxygen, gave g of carbon dioxide and g of water. What is the empirical formula of vitamin C?

Combustion Analysis Vitamin C is essential for the prevention of scurvy. Combustion of a g sample of this hydrocarbon, which may or may not contain oxygen, gave g of carbon dioxide and g of water. What is the empirical formula of vitamin C? First covert the mass of carbon dioxide and water into grams of carbon and of carbon and hydrogen. Subtract these masses from the sample weight, if the difference is zero then vitamin C does not contain any oxgen.

Combustion Analysis g CO2

Combustion Analysis g CO2 Mole CO2 g CO2

Combustion Analysis g CO2 Mole CO2 Mole C g CO2 Mole CO2

Combustion Analysis 0.2298 g CO2 Mole CO2 Mole C 12.011 g C

g H2O mole H2O 2 mole H 1.008 g H = g H 18.02 g H2O mole H2O mole H

g H2O mole H2O 2 mole H 1.008 g H = g H 18.02 g H2O mole H2O mole H = g O

g H2O mole H2O 2 mole H 1.008 g H = g H 18.02 g H2O mole H2O mole H = g O g C Mole C = mole g C g H mole H = mole 1.008 g H g O Mole O = mole 16.00 g O

g H2O mole H2O 2 mole H 1.008 g H = g H 18.02 g H2O mole H2O mole H = g O g C Mole C = mole = X 3 = 3 g C g H mole H = mole = X 3 = 4 1.008 g H g O Mole O = mole = X 3 = 3 16.00 g O

Combustion Analysis C3H4O3 0.2298 g CO2 Mole CO2 Mole C 12.011 g C
g H2O mole H2O 2 mole H 1.008 g H = g H 18.02 g H2O mole H2O mole H = g O g C Mole C = mole = X 3 = 3 g C g H mole H = mole = X 3 = 4 1.008 g H g O Mole O = mole = X 3 = 3 16.00 g O C3H4O3

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Chapter #3 Chemical Composition.

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