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CHP5-8 Flashcards

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1
Q

From left to right name the charges of the columns

A 
Column 1 A:+1
Column 2 A:+2
Column 5 A:-1
Column 6 A:-2
Column 7 A:-3
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2
Q

What Cations in the Transition metals have specific charges and what are those charges?

A 
Al +3
Ga+3
Zn +2
Cd+2
Ag+
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3
Q

What are the Polyatomic Ions that are needed to know? What are the formula and charge?

A

^Ammonium: (NH_4)+ ^Cyanide: (CN)-
^Hydroxide: (OH)- ^[Acetate: (C_2H_3O_2)-
^Nitrate: (NO_3)- or(CH_3CO_2)]
^Nitrite: (NO_2)- ^[Bicarbonate(Hydrogen
^Carbonate: (CO_3)2- –carbonate): (HCO_3)-]
^Phosphate(PO_4)3- ^Hydrogen Phosphate :
^Sulfate: (SO_4)2- (HPO_4)

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4
Q

Polyatomic Ions_______Continued

A

^Dihydrogen Phosphate: (H_2PO_4
)-
^Sulfate: (SO_4)2- ^Bisulfate: (HSO_4)-
^Sulfite: (SO_3)2- ^Bisulfite: (HSO_3)-

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5
Q

What is the Law of Constant Composition?

A

First reported by Joseph Proust in 1799

“All samples of a given compound have the same proportions of their constituent elements.”

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6
Q

Define Compounds vs. Mixtures

A

The ratios of the elements do NOT vary in compound and are exact.
The ratios of the components of a mixture can vary

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7
Q

Writing Chemical Formulas

A

By convention, the more/most metallic element

is written first

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8
Q

Law of Multiple Proportions Author and Summary

A

Both compounds contain only
carbon and oxygen. In each compound,
the carbon and oxygen are in a fixed proportion….

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9
Q

How many atoms of each element are in a formula unit of Al2(SO4)3?

A

2Al, 3S, 12O

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10
Q

Name the Different Types of Formulas and How the different

A

Empirical Formulas give the ratios of the atoms for each of the elements in their lowest terms
The empirical formula for glucose (blood sugar) is CH2O
Molecular Formulas give the number of each type of atom in a molecule
The molecular formula for glucose is C6H12O6
Structural formulas show us how the atoms are hooked together and give much more information.

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11
Q

Define Atomic and Molecular Elements

A

Atomic Elements are the norm
Single atoms are the basic units
Noble gases and metals such as copper, gold and mercury are examples
Molecular Elements include the diatomic elements, where the natural form is two atoms bonded together
Other examples include phosphorus which can exist as P4 or P8 (among others); sulfur exists as S8

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12
Q

Elements That Occur as Diatomic Molecules

A 
Hydrogen: H_2
Nitrogen: N_2
Oxygen: O_2
Fluorine: F_2
Chlorine: Cl_2
Bromine: Br_2
Iodine : I_2
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13
Q

Define Ionic and Molecular Compounds

A

Compounds contain two or more different elements
The atoms of each element are in very specific ratios

Molecular compounds are composed of two or more nonmetals. Molecules are the basic unit.
Ionic compounds are made of a combination of anions and cations.
Usually, the cations are metals, and the anions are nonmetals.

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14
Q

Define Ionic Compounds

A

Ionic compounds do NOT form molecules
In the solid phase the compounds exist as crystals.
We refer to their structure as being in “formula units”
The formula unit gives the ratio of cations to anions.
The formula must be neutral in charge: The total charges of anions and cations are equal.

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15
Q

Ion Formation Rules

A

Metals lose electrons to develop a positive charge. Those in
Column 1A: loses one electron, 1+ charge
Column 2A: lose two electrons, 2+ charge
Aluminum: loses three electrons, 3+ charge
Note others on next figure
Transition metals: Multiple charges possible
You need to be told, or calculate, the charge on the metal ion

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16
Q

Ion Formation cont’d

A

Nonmetals gain electrons to develop a negative charge. Those in
Column 7A: Gain one electron; 1- charge
Column 6A: Oxygen, sulfur and selenium often gain two electrons; 2- charge
Column 5A: Nitrogen and phosphorus can gain 3 electrons to form a 3- charge

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17
Q

CHP 5 Points to Remember

A

Metals form cations.
Group 1A 1+. Group 2A 2+.
Transition metals may have more than one charge. Often it cannot be predicted UNTIL you know the formula of the compound you are examining.
Nonmetals form anions.
Group 6A 2-. Group 7A 1-.
Noble gases do not readily form ions.
Ions are formed by the gain or loss of electrons.

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18
Q

Naming Metal Ions

A

Alkali and alkaline earth metals:
(Name of metal) ion
sodium ion
potassium ion
aluminum ion
For transition metals, you need to tell what the charge on the metal is
Cu1+ copper(I) ion
Cr3+ chromium(III) ion
Hg2+ mercury(II) ion
We will only be using the newer Stock system to name the ions.
Exceptions with naming: Zn2+ (zinc), Cd2+ (cadmium), Ag+ (silver). Also, Al3+, Ga3+ and In3+. These have only the charges listed. Use Main Group rules. Note their places on the periodic table.

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19
Q

Naming Non-metal Ions

A 
Replace the ending of the element name with the suffix –ide.  They have negative charges.
		oxygen … oxide ion
		nitrogen … nitride ion
		fluorine … fluoride ion
		bromine … bromide ion
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20
Q

Polyatomic Ions

A

Ions that contain more than one atom.
Treat the whole group of atoms as one unit.
The given charge is for the whole unit of atoms.

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21
Q

Polyatomic IonsKnow these

A 
NH4+  ammonium		NO3-  nitrate
OH-  hydroxide		NO2-  nitrite
CN-  cyanide			CO32-  carbonate
C2H3O2-  or                                 HCO3-  bicarbonate
CH3COO- acetate

SO42- sulfate PO43- phosphate
SO32- sulfite HPO42- hydrogen phosphate
HSO4- bisulfate H2PO4- dihydrogen phosphate
HSO3- bisulfite

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22
Q

Atoms Are Smaller than Nails

A

The mole is defined as exactly 6.02214076×1023 constitutive particles, which may be atoms, molecules, ions, or electrons.

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23
Q

The Mole

A

We will say that a mole contains 6.022 x 1023 particles
This is known as Avogadro’s number.
Think of moles like dozens, a defined amount, but a bit bigger.

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24
Q

How many atoms are in 0.457 mol of Helium?

A

2.75 x 10 23 atoms He

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25
Q

How many moles are in 1.00 x 1018 atoms of He?

A

1.66 x 10-6 mol He

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26
Q

The Mole cont’d

A

A mole contains 6.022 x 1023 particles.
That’s a big number BUT you should think of it in the same way you think of dozens, etc.

Use this definition of a mole as a conversion factor in the same way as you would 12 items = 1 dozen items.

1 mol = 6.022 x 1023 particles

The ONLY abbrev for mole is mol

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27
Q

The Mole Also Relates to Mass

A

One mole of an element has a mass of the average atomic mass of that element given in the periodic table. The units are grams.
For a single atom the number is the same but the units are daltons (Da) or unified atomic mass units (u).

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28
Q

Moles and Mass Continued

A

1 mol of carbon contains 12.01 g of C (because of the presence of 13C)
1 mol of sulfur contains 32.07 g of S
1 mol of aluminum contains 26.98 g of Al

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29
Q

How many moles of sulfur are in a 21.02 g sample?

A

0.6554 mol S

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30
Q

How many g of S are in 2.75 mol S?

A

88.2 g S

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31
Q

Moles as Conversion Factors Summary

A

Number of particles <>Moles <>Grams
Conversion factors
1. Avogadro’s number
2.Atomic or molar mass

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32
Q

0.654 g of Al?

A

1.46 x 1022 atoms Al

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33
Q

CHP 6 Summary

A

You cannot convert directly from the number of particles (atoms or molecules) that you have directly to the mass in grams.
In order to do that, you MUST always go thru moles.
Use Avogadro’s number, and the atomic weight or molar mass as conversion factors

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34
Q

Molar Mass

A 
You will see several terms which mean the same thing as molar mass:
Molecular Weight (covalent compounds)
Molecular Mass (covalent compounds)
Formula Weight (ionic compounds)
Atomic Mass (for elements)

All of these terms refer to the mass of one mole of a compound (or element).

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35
Q

Determining Molar Mass

A

To determine the molar mass of compound you combine the atomic weights of the elements.
Yes, you need to take into account the number of each type of atom.

E.g. NaCl Na is 22.99 Cl is 35.45 The molar mass is 58.44 grams.

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36
Q

Other Sample Molar Mass CalculationsThe Units are Grams

A

CO2 is 12.01 + 2(16.00) = 44.01

CH4 is 12.01 + 4(1.01) = 16.05

Ba(NO3)2 is 137.33 + 2(14.01) + 6(16.00) =
261.35

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37
Q

Molar Mass as a Conversion Factor

A

Many times you may need to convert from mass of a compound to moles of a compound.
In this case, use the molar mass of the compound as a conversion factor.
58.44 g NaCl/1 mol NaCl or 1 mol NaCl/58.44 g NaCl

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38
Q

Other Conversion Factors

A

Yes, there are 6.022 x 1023 formula units of NaCl in a mole of NaCl.
There are 6.022 x 1023 sodium ions
There are 6.022 x 1023 chloride ions
There are 6.022 x 1023 formula units of Ba(NO3)2 in a mole of Ba(NO3)2.
There are 6.022 x 1023 barium ions
There are 1.204 x 1024 nitrate ions

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39
Q

Mole Ratios

A

A number of calculations in chemistry will require you to use mole ratios.

For example, with NaCl there is one mole of sodium ions in one mole of sodium chloride.
There is also 1 mol of chloride ions in one mole of sodium chloride.
1 mol Na +/1 mol NaCl
1 mol NaCl/1 mol Cl-

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40
Q

Sample Problems

A

If you have 2.44 moles of SO2,
How many grams of SO2 do you have?
How many molecules of SO2 do you have?
How many atoms of oxygen do you have?

If you have 3.50 g of H2O,
How many molecules of water do you have?
How many hydrogen atoms do you have?
How many moles of hydrogen atoms do you have?

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41
Q

Mass On The Atomic Level In Daltons

A

The mass of one atom (on average) is equal to the average atomic mass of that element in daltons. You can read the mass directly off the periodic table.
For a chemical compound, the mass in daltons is numerically equal to the average atomic mass of one molecule or formula unit of the compound in daltons.

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42
Q

Mass Percent Calculations

A 
Percent means “parts per 100”
Calculations always are in the form
Mass of element/Mass of compound
=Percent of element/100 g of compound
100/1 * Mass of element / mass of compound =mass percent of element
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43
Q

Example #2 Mass Percent

A

What is the mass percent of chlorine in the compound CCl2F2?
Here we assume that we are working with one mole of the compound. (It makes things easier.)
100/1* 70.90 g Cl/120.91 g CCl_2 F_2T =58.64%
he compound is 58.64% Cl by mass

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44
Q

Problem CHP6

A

What is the mass percent of oxygen in CO2?

Silver chloride contains 75.27% silver by mass. If you want to plate out 4.8 g of pure silver, how much silver chloride must you start with?

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45
Q

Chemical Reactions

A

Every time we add different substances together, one of two possible things will happen.
They can form a mixture.
They can react to form a new substance.

How do we tell if a chemical reaction has occurred?

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46
Q

Chemical Reactions, cont’d

A

We have stated before that in chemical reactions the atoms combine, “change partners” or a substance decomposes into more simple substances.
HOWEVER, that doesn’t tell us how to recognize when a chemical reaction occurs.

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47
Q

Evidence that a Chemical Reaction Has Occurred

A

Color change
Formation of a solid (precipitate)
If a solution is not clear, it contains a solid.
There is a difference between clear and colorless.
Formation of a gas (Usually you will see bubbles, or you might smell something new.)
These phenomena always indicate a chemical reaction

48
Q

Evidence cont’d

A 
Also a good possibility  Include as evidence
Heat given off
Heat taken in 
(or something
 gets cold)
Light given 
off
49
Q

Chemical Equations

A

Using symbols is a lot more efficient, and less confusing, than using words.

50
Q

Abbreviations indicating the states of reactants and products in chemical equations

A

(g) gas
(l) Liquid
(s) Solid
(aq) aqueous (Water solution) : the (aq) designation stands for aqueous, which indicates that a substance is dissolved in water when a substance dissolves in water, the mixture is called a solution.

51
Q

Balanced Equations

A

Law of Conservation of Matter tells us that the mass of the material on the reactant side must equal the mass of the material on the product side.

We now know that the number of atoms of each element must be the same on each side of the equation.

52
Q

Balancing Equations

A

DO NOT CHANGE CHEMICAL FORMULAS ONCE THEY ARE CORRECTLY WRITTEN!!
Use coefficients. Place them in front of the formulas.

53
Q

Balancing Equations

A

Your book has several suggestions for balancing equations. These may, or may not, be helpful for you.
There is nothing that can substitute for practice.
There is a worksheet on Blackboard.

54
Q

Balancing Equations cont’d.

A
  1. Write the unbalanced equation using correct formulas for reactants and products.
  2. Add coefficients to balance the numbers of atoms of each element.
    Notes: A. If you have polyatomic ions in the equation, generally you can keep them together.
    B. If you have an odd number of atoms of one of the elements on one side, and an even number on the other side, use an even coefficient to make the numbers even on both sides
    C. Do the most complicated-looking parts of the equation first.
  3. Check your work. Make sure the numbers and types of atoms are the same on both sides of the equation.
  4. Make sure the coefficients are in lowest terms.
55
Q

Example CHP 7

A

SiO2 + C <> SiC + CO
Note that we have two O on the left, and one on the right
SiO2 + C <> SiC + 2CO
Now we have 1 C on the left, and 3 on the right
SiO2 + 3C <> SiC + 2CO

Balanced!

56
Q

Aqueous Solutions

A

If you have an aqueous solution, something is dissolved in water.
Solutions are homogeneous mixtures.
Aqua means water in Spanish.
An aqueous solution may be colored
But, there is no cloudiness, or visible solid
Many, but not all, ionic compounds are soluble in water.

57
Q

Dissolved and Soluble vs. ….

A

If something is dissolved, you will only see one phase.
There are degrees of solubility (coming up later), but if something is soluble at a certain concentration, it will be dissolved.
Insoluble means that something will not dissolve. You will see two phases.

58
Q

Dissolved Ionic Compounds Allow Electricity to Flow

A

If an ionic compound dissociates completely, it is called a strong electrolyte.
The term “strong” has nothing to do with the concentration of the compound.

59
Q

How Can We Tell if a Compound Will Be Soluble?

A

There are rules which can be memorized

You are responsible only for the first 2 lines on tests. You will be able to assume that any other compounds are insoluble. (Tests only, not MChemistry)

60
Q

Solubility Rules

A

LI+, Na+, K+, NH_4+, NO_3-, C_2H_3O_2-

no exceptions will dissolve

61
Q

Ways to Organize Reactions

A

There are an number of ways to organize chemical reactions by type.
We will look at these in a different order than the author of your textbook does.

Section Order: 7.1 thru 7.5, 7.10, 7.6 thru 7.9

62
Q

A Way to Organize Chemical Reactions

A

Most chemical reactions can be grouped into one of four categories:

Combination A + B → AB

Decomposition AB → A + B

Single Replacement AB + C → AC + B
AB + C → CB + A
(Metals replace metals; nonmetals replace nonmetals.)

Double Replacement AB + CD → AD + CB

Then, the above reactions are either redox or not redox.

63
Q

Combination

A

In a combination reaction
two or more elements form one product.
or simple compounds combine to form one product.

+

	2Mg(s)  +  O2(g) 		2MgO(s)

	2Na(s)  +  Cl2(g)		  2NaCl(s)

	SO3(g)  +  H2O(l)		   H2SO4(aq)
64
Q

Decomposition

A

In a decomposition reaction

one substance splits into two or more simpler substances.

	2HgO(s)		    2Hg(l)   +  O2(g)

	2KClO3(s)		    2KCl(s)  + 3 O2(g)
65
Q

Sample Problems

A

Classify the following reactions as
1) combination or 2) decomposition.

\_\_\_ A.  H2(g)  + Br2(g)  	          2HBr(l)

\_\_\_ B.  Al2(CO3)3(s)		    Al2O3(s)  + 3CO2(g)

\_\_\_ C.  4Al(s)   + 3C(s)     	     Al4C3(s)
66
Q

Single Replacement Reactions

A

Single replacement reactions

		A  +  BC   AC  +  B
		A  +  BC   BA  +  C

Metal replace metals; non-metals replace non-metals.
You may see this type of reaction called a single displacement reaction.

67
Q

Single Replacement

A

In a single replacement reaction,

One element takes the place of a different element in a reacting compound.

Zn(s)  + 2HCl(aq)	                  ZnCl2(aq)   +  H2(g)  
Br2(g)  + 2NaCl(aq)		   2NaBr(aq)  + Cl2(g)
68
Q

Double Replacement

A

In a double replacement,
Elements in the reactants exchange partners.

AgNO3(aq)  + NaCl(aq)		AgCl(s)   + NaNO3(aq)  
ZnS(s)     +   2HCl(aq)		ZnCl2(aq) +  H2S(g)
69
Q

Types of Double Replacement Reactions

A

You may see this type of reaction referred to as a double displacement reaction
In this type of reaction ions of the original ionic compounds exchange partners.
Your author breaks these reactions down into several categories.
Often, in this type of reaction, the product is a solid, AKA a precipitate.

70
Q

Precipitation

A

General Equation:
AX + BY  AY + BX
Specific Example:

2 AgNO3(aq) + Na2CO3(aq)  
				Ag2CO3(s) + 2 NaNO3(aq)

Note: We used the skills of putting together neutral compounds and balancing equations here.

71
Q

Your Turn

A

Pb(C2H3O2)2(aq) + Na2SO4(aq) 

Write proper formulas for the products
Balance the equation
Predict the precipitate

You should be able to predict the products and write equations for double replacement reactions.

72
Q

Net Ionic Equations

A

Net Ionic Equations Do NOT Show Full Formulas.
Only the ions reacting to form the product (precipitate, liquid or gas) are shown. (Product is also shown with correct phase.)
Spectator Ions (those not involved in the reaction) are not included.
All equations must be balanced.

73
Q

Net Ionic Equations

A

Molecular Equation:
AgNO3(aq) + NaCl(aq)  NaNO3(aq) + AgCl(s)

Complete Ionic Equation:
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)  Na+(aq) + NO3-(aq) + AgCl(s)

Only aqueous ionic compounds are broken up into their constituent ions. Solids, liquids and gases are left as complete compounds.
The complete ionic equation allows us to write the net ionic equation.

74
Q

Net Ionic Equations … Example

A

Molecular equation:
AgNO3(aq) + NaCl(aq)  NaNO3(aq) + AgCl(s)

Complete Ionic Equation:
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)  Na+(aq) + NO3- (aq) + AgCl(s)

For the net ionic equation, drop the spectator ions:
Net Ionic Equation:
Ag+(aq) + Cl-(aq)  AgCl(s)

75
Q

Other Types of Double Displacement Reactions

A

Your text mentions two other types of double replacement reactions
Acid-base neutralization
Gas evolution

We determine the products in the same way as we do for precipitation reactions

76
Q

Acids and Bases

A

Acids liberate H+ ions when dissolved in water
Bases either liberate or form OH- ions when dissolved in water
Basic solutions are also known as alkaline solutions
A solution cannot be both acidic and basic at the same time
The net ionic equation for an acid-base reaction is ALWAYS: H+ + OH-  H2O
Acid-base reactions are also known as neutralization reactions

77
Q

Acid-Base Reactions

A

HCl(aq) + NaOH(aq)  HOH(l) + NaCl(aq)

H+(aq) + OH-(aq)  H2O

78
Q

Acid-Base Reactions Cont’d

A

Note that water is always a product in an acid-base reaction
The other product is an aqueous ionic compound.
A generic term for an ionic compound is salt.
Hence, an acid reacts with a base to form water plus a salt.

79
Q

Gas Evolving Reactions

A

The gas evolving reactions in your text are also double displacement reactions

We are only covering the reactions that form sulfides and
the reactions of acids with carbonates and bicarbonates,
both of which make CO2.

80
Q

Example 1: Formation of H2S CHP7

A

H2S is the stuff that smells like rotten eggs

It is related to the chemicals put in natural gas to let you know if there is a leak

81
Q

Example 2: Formation of Carbon Dioxide Chp 7

A

This is actually a 2-step reaction

The carbonic acid (H2CO3) initially formed is unstable
and breaks down almost immediately.

82
Q

Formation of CO2 from bicarbonate

A

Net Ionic Equation
HCO3- + H+(aq) <>
CO2(g) + H2O(l)
H+ is often written as H3O+.

83
Q

Redox Reactions

A

Redox stands for oxidation-reduction (Can’t have one without the other!)
An important category of reactions Electrons are transferred in these reaction
Reactions are either redox or not redox.

Reduction : The gain of one or more electrons by an atom
Example: S0 + 2e-  S2-

Oxidation: The loss of one or more electrons by an atom
Example: Na0  Na+ + e-

84
Q

Remembering Which is Which

A

Reduction is the GAIN of electrons, so oxidation must be the loss

or

OIL RIG: Oxidation is Loss
Reduction is Gain

85
Q

Oxidizing and Reducing Agents

A

The oxidizing agent causes the other substance to be oxidized. It is, in fact, reduced.
The reducing agent causes the other substance to be reduced. It is, in fact, oxidized.

86
Q

Other Examples of Redox Reactions

A

Corrosion:
4 Fe(s) + 3 O2(g) + H2O  2 Fe2O3.H2O(s) (rust)
Combustion:
CH4(g) + O2  CO2 + 2H2O (l)
Respiration:
C6H12O6 + 6 O2  6 CO2 + 6 H2O + energy
Bleaching:
H2O2 and NaOCl are both oxidizing agents
Metallurgy:
Fe2O3(s) + 3CO(g)  2 Fe(s) + 3 CO2(g)

87
Q

Combustion Reactions

A

Type of redox reaction
Important in society … Energy usage. Burning stuff.
Combustion is the rapid reaction of a substance with oxygen.
Always exothermic
When burning hydrocarbons, or compounds with C,H and O in them, the only products are carbon dioxide and water.

88
Q

Balancing Combustion Reactions

A 
Products always carbon dioxide and water
if reactant has C&H or C,H &O
Always balance in the order
Carbon
Hydrogen
Lastly … oxygen
89
Q

Practice CHP7

A

____ C3H8 + ____ O2  ____ CO2 + ____ H2O

____ C2H5OH + ____ O2  ____ CO2 + ____ H2O

____ C8H18 + ____ O2  ____ CO2 + ____ H2O

Remember, you can only use coefficients. Don’t change any formulas!

90
Q

In a Balanced Equation

A

The coefficients tell us about the ratio of the different compounds in the reaction.
One molecule of nitrogen reacts with three molecules
of hydrogen to give 2 molecules of ammonia.
One mole of nitrogen reacts with three moles
of hydrogen to give 2 moles of ammonia

91
Q

Mole Ratios

A

In Chapter 6, when looking at the relative amounts of the different elements in a compound, we were introduced to the idea of mole ratios.

When working with chemical reactions we also use mole ratios. However, in the case of chemical reactions we use the coefficients in the equations to tell us the mole ratio.

To do this our reactions must be correctly balanced.

92
Q

N_2+3H_2 <>2NH_3

A

For the reaction above, solve the following using
mole ratios and the factor-label method.

A. If we have 0.50 moles of hydrogen gas,
how many moles of nitrogen would we
use in the reaction?

B. If we have 0.50 moles of hydrogen gas,
how many moles of ammonia can we make?

93
Q

Mass Relationships

A

Stoichiometry: The study of mass relationships in chemical reactions.
In order to determine the masses involved when doing chemical reactions, it is necessary to
Be able to covert from mass to moles
Be able to use mole ratios
Be able to convert back from moles to mass

One CANNOT do mass-mass conversions directly. (I.e. One can’t go directly from mass of a reactant to the mass of a specific product, etc.)

94
Q

Your Book’s Guide

A

We must start with a balanced chemical equation.
To predict how much Product B we can make
from Reactant A, we must go thru moles.

95
Q

The Box Method

A

Write the equation for the chemical reaction.
Make sure it is balanced.
Put in the information given to you in the problem directly below the appropriate compound(s).
Indicate what quantity you are looking for.
Draw a box and label it so you go through moles.
Solve the problem by the factor-label method using the appropriate conversion factors.

96
Q

Example 1

CHP8

A

How many grams of N2 are necessary to produce 7.50 g of NH3?

  1. Start with a balanced equation.
    Put in the information given to you.

Indicate what quantity you are
looking for.
Draw a box so that moles of the reagent
is directly underneath the information you put in.

Use arrows to show you the way to go.
Determine your conversion factors.
Set up your calculation.

97
Q

Problem Set-up

A

2 C8H18 + 25 O2  16 CO2 + 18H2O

It’s balanced.

Determine molecular weights of octane and carbon dioxide.
Determine their mole ratio. (No need to simplify it.)

98
Q

Problem Set-up CONT

A

2 C8H18 + 25 O2  16 CO2 + 18 H2O
750 g octane ?? g carbon dioxide

MW octane = 114 MW carbon dioxide = 44.0
Mole ratio: 2 octane to 16 carbon dioxide

99
Q

More Complicated Problems In This Class CHP8

A

When solving a problem, often the most difficult thing is figuring out what is being asked.

Most times, it is possible to break the problem down into a series of steps.

In these cases, the individual steps will be made up of things you have seen before.

100
Q

Limiting Reagent

A

Sometimes when given problems, and
all the time in the laboratory
Chemical reactions are often set up so that one reagent runs out before the other
This may be because one reagent is more expensive than the other, or more toxic
The reagent that one runs out of first is the Limiting Reagent.
One must do more than one box method calculation to determine which one this is

101
Q

Limiting Reagent Problems

A

Limiting Reagent Problems
Extension of stoichiometry problems
All stoichiometry problems are handled the pretty much the same way
Use the “Box Method”

Another stoichiometry problem on next slide

102
Q

Stoichiometry Problem

A

How many grams of mercury(II) iodide
can be produced
if you start with 60.0 g of
potassium iodide?

Your balanced reaction equation is:
2 KI + Hg(NO3)2  2 KNO3 + HgI2

103
Q

Set up the Problem as Below

A

2 KI + Hg(NO3)2  2 KNO3 + HgI2

104
Q

Questions CHP8

A

What are we assuming about the amount of mercury(II) nitrate at the start of the experiment?
We have more than enough.
What does this mean about the amount of potassium iodide present?
We will run out of it before we run out of mercury(II) nitrate.
Potassium iodide is what we refer to as the LIMITING REAGENT: We will run out of KI and still have Hg(NO3)2 left.

105
Q

Change the Problem Slightly

A

What if we start with 60.0 g if KI and
50.0 g of Hg(NO3)2 in the reaction mixture?

How will that effect how much HgI2 we can end up with?

How would we determine that?
Do a second stoichiometry problem and compare the results.

106
Q

Set up the Second Part of the

Problem as Below

A

2 KI + Hg(NO3)2  2 KNO3 + HgI2

107
Q

Answer to Limiting Reagent Problem

A

With 50.0 g Hg(NO3)2 AND 60.0 g KI in the reaction mixture
Which of the two is the limiting reagent?
Why?

Hg(NO3)2
Because we would run out of it before we ran out of KI (70.0 g vs. 82.1 g HgI2 product)

108
Q

Limiting Reagent Problems

A

We are given amounts for both starting materials.
One of them will run out before the other.
That reagent limits the amount of product we can get (limiting reagent).
The other reactant is there in excess. There is still some left after the reaction stops.
Treat the overall problem as two stoichiometry problems.
If you need to you can calculate how much of the reagent is excess is left over.

109
Q

Percent Yield

A 
One calculates the theoretical yield
What one gets in the laboratory is the actual yield
This is usually less than the theoretical yield
Possibilities for this are
Impure reagents
Reaction did not go to completion
Losses when transferring materials
Losses during purifications
110
Q

The formula for percent yield is

A

actual yield/theoretical yield *100
Actual yield is what you got.

Theoretical yield is what you thought you would get (if everything went perfectly).
It is the lower of the 2 amounts if you are
doing a limiting reagent problem

111
Q

Percent Yield Calculation

A

Say that in our earlier problem, instead of 70.0 g HgI2 we actually got 56.7 g. What is our percent yield?

56.7g actual/70.0 g theoretical *100 =81.0%

112
Q

Practice Problem CHP8

A

Cu2O(s) + C(s)  2 Cu(s) +CO(g)

If you react 11.5 g of carbon with 114.5 g of Cu2O, 87.4 g of Cu are obtained.

What are the limiting reactant, theoretical yield and percent yield?

113
Q

Enthalpy

A

We learned in Chapter 3 that all processes are either exothermic or endothermic.
The measure of heat given off or taken in is called the enthalpy.
We often measure enthalpy changes in reactions.
The symbol for enthalpy change is ΔH.
Enthalpy changes can be negative or positive.

114
Q

Enthalpy changes can be negative or positive.

A

When energy is released,
ΔH is negative.
When energy is absorbed,
ΔH is positive.

115
Q

Enthalpy

A

To get the correct sign for enthalpy, put the value on the side of the reactants.

C3H8(g) + 5 O2(g) - 2044 kJ  3 CO2(g) + 4H2O(g)

ΔH rxn = -2044 kJ

116
Q

The amount of heat given off or taken in varies with the size of the reaction.

A

C3H8(g) + 5 O2(g)  3 CO2(g) + 4H2O(g)
ΔH rxn = -2044 kJ

Is the reaction exothermic or endothermic?
If you only made 2 moles of water, how much heat would be given off?
If you started with 64.0 g of propane, how much heat would you produce? (assume plenty of oxygen)

Chem 101 (1 decks)

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