Chapter 3: Calculations and the Chemical EquationThe Mole Concept and Atoms

Atomic mass unit1 amu = 1.661 X 10-24 g

Because the mass of one amu is so small, chemists deal with a much larger number of atoms while working with chemicals

Mole One mole is defined as 6.022 X 1023. This refers to one mole of anything, eggs, paperclips, atoms. One mole of anything is 6.022 X 1023 items. Much like one dozen of something is 12.This number, 6.022 X 1023 is called Avogadros number, named after the scientist who conducted a series of experiments leading to the mole concept.

The mole conceptThe mole and the amu are related. For atoms, the atomic mass of an element corresponds to the average mass of a single atom in amu

And

The mass of a mole of atoms in grams.

For example:The atomic mass of oxygen is 16.00 amu.

And

One mole of oxygen atoms (6.022 X 1023 oxygen atoms) has a mass of 16.00 grams

Another exampleThe atomic mass of iron (Fe) is 55.85 amu.

And

One mole of iron atoms (6.022 X 1023 oxygen atoms) has a mass of 55.85 grams

And yet another exampleThe atomic mass of radium (Ra) is 226 amu.

And

One mole of radium atoms (6.022 X 1023 radium atoms) has a mass of 226 grams

Molar massThe mass of one mol (mole) of atoms in grams

NoteOne mole of atoms of any element contains 6.022 X 1023 atoms, regardless of the type of element.The mass of one mole of an element depends on what that element is, and is equal to the atom mass of that element in grams.

This meansThere are several conversions regarding atoms, moles, and mass

Converting moles to atomsHow many atoms are in 4 moles of H?

4 moles H X 6.022 X 1023 atoms/mole = 24.088 X 1023 atoms of hydrogen or 2.409 X 1024 atoms

In this case you multiply the number of moles X the number of atoms in each mole.

Converting atoms to molesCalculate the number of moles of copper represented by 3.26 X 1024 atoms.

3.26 X 1024 = 32.6 X 1023 (ok, I did this step to make the math easier.)

32.6 X 1023 / 6.022 X 1023 = 5.413 X 1023 moles of copper.

In this case, to convert atoms to moles, I divide the number of atoms by the number of atoms in one mol (by 6.022 X 1023 )

Converting moles of a substance to mass in grams.What is the mass in grams of 5.6 mol of Neon?

The mass of one mole of Ne is the same as the atomic mass in g (20.18 g)

5.6 mol X 20.18 g/mol = 100.9 g of Ne

Converting grams to numbers of atoms.How many atoms would be in a gold ring that weighs 25 g?First, find the number of moles of Gold in 25 g. Gold has an atomic mass of 107.9. So, 25 g / 107.9 g/mol = 0.2317 mol of gold are in the ring.Next, 0.2317 mol X (6.022 X 1023) atoms/mol =1.395 X 1023 atoms

When dealing with molecules. . .Like O2 or H2 , double the number of atoms, because there are 2 atoms per molecule.Remember, one mole of something is 6.022 X 1023 of whatever it is. If it is molecules, its 6.022 X 1023 of them. If it is atoms, its 6.022 x1023 atoms. If there are 2 atoms per molecule you need to double the number of moles. 2 X (6.022 X 1023 ) = 12.044 X 1023 or 1.204 X 1024

Homework Assignment # 10Read p. 119-123. As you read, complete exercises 1-6. When you are done reading, answer problems 23-36 on p. 146-147

Chapter 4: Calculations and the Chemical EquationSection 4.2: The Chemical Formula, Formula Weight, and Molar Mass

Chemical FormulaA combination of symbols of the various elements that make up the compound.

Formula UnitThe smallest amount of atoms that provides the following informationThe identity of atoms in the compoundThe relative numbers of each type of atomExamples

Molecule vs ion pairCovalent compounds form molecules, and when calculating formula weight all of the atoms in the compound are added together.Ion pairs (ionic compounds) form crystalline structures. Its the smallest group of ions that are listed in the formula for these types of chemicals.

Formula weight vs Molecular WeightThe sum of all of the atomic weights in the compound in an ionic compound its the formula weight. In a covalent compound its the molecular weight.

Molar MassThe mass of one mole of the compound or the formula weight in grams.Examples

Conversions using Formula WeightFinding the number of moles corresponding to a certain number of grams.

Conversions using Formula WeightFinding grams corresponding to a certain number of moles.

Homework Assignment #11Read p. 123-126.On p. 147 Exercises 37-58

Chapter 4: Calculations and the Chemical EquationSection 3: The Chemical Equation and the Information it Conveys

Chemical equationThe shorthand notation for a chemical reaction, where one substance changes chemically into another substance.An example: burning sugar

ReactantsThe starting materials that undergo a chemical change

ProductsThe ending materials that are produced by a chemical reaction.

Additional information in a chemical reactionPhysical state of the substance (solid, liquid, or gas)If the reaction occursIdentifies the solvent, if there is one. (A solvent is the solution the materials are dissolved in, such as water.)Experimental conditions such as heat, light, or electrical energy added

Most importantlyThe chemical equation identifies the relative number of moles of reactants and products.

According to the Law of Conservation of MassMatter cannot be gained or lost in the process of a chemical reactionThe total mass of the products must equal the total mass of the reactantsThe chemical equation must be balanced.

Features of a chemical reaction.CaCO3(s) CaO(s) + CO2(g)Reactants are on the left of the arrow.The arrow indicates the reaction occurs in one direction.The products are on the right of the arrow.

Features of a chemical reaction.CaCO3(s) CaO(s) + CO2(g)s indicates the chemical is a solid substanceg indicates the substance is a gasl would indicate the substance were a liquid.

Features of a chemical reaction.CaCO3(s) CaO(s) + CO2(g)The indicates that energy was necessary for the chemical reaction to occur

Features of a chemical reaction.CaCO3(s) CaO(s) + CO2(g)The main feature of a chemical equation is that it is balanced, with the same number of elements in compounds on both sides of the arrow.

The experimental basis of a chemical equationEvidence for a chemical reaction includes:

The release of a gas resulting in bubbles

The formation of a solid (precipitate) in solution

The production of heat resulting in an increase in temperature

A change in color of a substance

The experimental basis of a chemical equationSometimes instruments must be used to measure subtle changes that indicate a chemical reaction.

Heat or light absorbed or emitted

Changes in the way a sample behaves in an electrical or magnetic field

Changes in electrical properties

Writing Chemical ReactionsMost reactions follow a few simple patterns

Combination reactions

Decomposition reactions

Replacement reactions

Combination reactionsInvolve the joining or combining of two or more compoundsThe general form of the reaction is A + B AB

Combination reactionsExamples include:Combination of a metal and non-metal to form a salt

Ca(s) + Cl2(g) CaCl2(s)

Reaction of magnesium oxide and carbon dioxide to produce magnesium carbonate

MgO(s) + CO2(g) MgCO3(s)

Decomposition ReactionsReactions that produce two or more products from a single reactant.The general form for the reaction is AB A + B

Decomposition reactionsExamples includeThe removal of water from a hydrate (a substance that has water molecules linked in its structure)

CuSO45H2O(s) CuSO4(s) + 5H2O(g)

The heating of calcium carbonate to produce calcium oxide and carbon dioxide gas

CaCO3(s) CaO(s) + CO2(g)

Replacement ReactionsSingle ReplacementSingle replacement reactions is where one atom replaces another in the compoundThe general formula is A + BC AC + B

Replacement ReactionsExamples include

The replacement of copper by zinc in copper sulfate forming zinc sulfate

Zn(s) + CuSO4(aq) Zn SO4(aq) + Cu(s)

Replacement ReactionsDouble ReplacementTwo compounds that switch atoms with each otherThe general formula isAB + CD AD + CB

Replacement ReactionsDouble ReplacementExamples includeThe formation of salt and water with the reaction of a base and an acid

HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)

Types of Chemical ReactionsThere are four main types of chemical reactionsPrecipitation reactions

Reactions with Oxygen

Acid-base reactions

Oxidation-reduction reactions

Precipitation reactionsA chemical change that produces an insoluble product that will form a solid. Usually the solid can be seen falling out of the solution, hence, called precipitation. At other times the solid makes the solution turn from clear to cloudy.

Solubility predictionsNa, K, and ammonium compounds are generally soluble.

Nitrates and acetates are generally soluble

Chlorides, bromides, and iodides are generally soluble. However, iodine compounds that contain lead, silver, and mercury are insoluble.

Carbonates and phosphates are generally insoluble. Sodium, potassium, and ammonium carbonates and phosphates are soluble.

Hydroxides and sulfides are generally insoluble. Sodium, potassium, calcium, and ammonium compounds are however soluble.

Reactions with oxygenMany substances react with oxygen. If the substance contains carbon, then carbon dioxide is usually produced. If the substance contains hydrogen, then water is usually produced.An example is iron turning to rust4Fe(s) + 3O2(g) 2Fe2O3(s)This number is called a coefficient, and indicates the number of molecules or moles that reacts with the other compounds.

Acid-base ReactionsThis involves an acid combining with a base to form a salt.An example would be

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(aq)

Oxidation Reduction ReactionsInvolves the transfer of negative charge from one reactant to another.The reaction of zinc with copper would be an example.

Zn(s) + Cu2+ (aq) Zn2+ (aq) + Cu (s)

These specific reactions will be discuss in further detail at a later time.

Homework Assignment #12p. 147 Answer questions 59-68Read p. 126-136. Answer the exercises in the reading that were not answered in class as examples.

Chapter 4: Calculations and the Chemical EquationSection 4: Balancing Chemical Equations

The chemical equationShows the molar quantity of reactants needed to produce a certain molar quantity of products.

sinceThe number of atoms in a molecule cannot be changed (it would make an entirely different compound)

coefficients(whole numbers that show the numbers of entire molecules) are used to balance a chemical equation

For example, in the equationCaCO3(s) CaO(s) + CO2(g)

Since there are the same numbers of each type of molecules on both sides of the arrow, the equation is balanced.

On the reactant sideOn the product side1 mole Ca1 mole Ca1 mole C1 mole C3 moles of O3 moles of O

HCl(aq) + Ca(s) CaCl2(s) + H2(g)

Since there are not the same number of moles on both sides of this equation, the equation is not balanced.

On the reactant sideOn the product side1 mole H2 mole H1 mole Cl2 moles Cl1 mole Ca1 mole Ca

HCl(aq) + Ca(s) CaCl2(s) + H2(g)

2HCl(aq) + Ca(s) CaCl2(s) + H2(g)To balance the equation, place a coefficient of 2 in front of the HCl.

Steps to balancing a chemical equationStep 1: Count the number of moles of atoms of each element on both product and reactant sideH2(g) + O2(g) H2O(l)On the reactant side:2 moles of H2 moles of OOn the product side:2 moles of H1 mole of O

Steps to balancing a chemical equationStep 2: Determine which elements are not balanced.H2(g) + O2(g) H2O(l)The oxygen atoms are not balanced in this equation.

Steps to balancing a chemical equationStep 3: Balance one element at a time. H2(g) + O2(g) H2O(l)First: H2(g) + O2(g) 2H2O(l)Then 2H2(g) + O2(g) 2H2O(l)

Steps to balancing a chemical equationStep 4: After you believe you have successfully balanced the equation, check to make sure you have the same number of atoms on both sides of the equation.2H2(g) + O2(g) 2H2O(l)4 moles of H2 moles of O4 moles of H2 moles of O

Homework Assignment #13p. 147 Answer questions 69-84Read p. 136-145. Answer the exercises in the reading that were not answered in class as examples.

Chapter 4: Calculations and the Chemical EquationSection 4.5: Calculations Using the Chemical Equation

Calculations Using Chemical EquationsUsing the chemical formulas to calculate amounts of materials needed or produced can be done once you have a balanced chemical equation.

In order to carry out chemical calculations the following guidelines must be followed.The chemical formulas of ALL the products and reactants must be known

The basis for the calculations is the balanced chemical equation. Be sure all of the equations are balanced first.The calculations are performed in terms of moles.

Use of Conversion Factors--conversion between moles and grams.Example: convert 10 moles of NaCl to grams

The formula mass of NaCl is the molecular mass of Na + the molecular mass of Cl (22.99 + 35.45 = 58.44 grams per mole.

10 moles NaCl X 55.44 grams = 554.4 grams of NaCl1 mole

Use of Conversion Factors--conversion between moles and grams.Example: How many moles of CaCl2 would 23 grams contain?

The formula mass of CaCl2 equals the atomic mass of Ca and 2 X the atomic mass of Cl (40.08 + 2(35.45) = 110.98)

23 grams X 1 mole = 0.207 moles 110.98 grams

Use of Conversion FactorsConversion of moles of reactants to moles of products.Once you have a balanced chemical equation, develop a conversion factor of reactants to products. After you have done that you can:Calculate reacting quantitiesCalculate grams of product producedRelate the mass of reactants and products

Converting moles of reactants to moles of product2H2(g) + O2(g) 2H2O(l)

In this equation: 2 moles of H2 will react with 1 mole of O2 to produce 2 moles of H2O.

2H2(g) + O2(g) 2H2O(l)To calculate the number of grams of H2O produced by 1 mole of O2

Convert from moles of O2 to moles of H2O

1 mole O2 x 2 moles H2O = 2 moles H2O produced

2. Convert the moles of H2O to grams of H2O2 moles X 18.016 grams = 1 mole36.032 grams

Relating masses of reactants and productsCaCO3(s) CaO(s) + CO2(g)

How many grams of Ca0 will be produced by 100.0 grams of CaCO3?

First, determine how many moles will be produced.

CaCO3(s) CaO(s) + CO2(g)One mole of CaCO3 will produce one mole of CaO

Next, determine how many grams of each are in each mole of substance.CaCO3(s) CaO(s) + CO2(g)1 mole of CaCO3 has a formula mass of 100.09 grams1 mole of CaO has a mass of 56.06 grams

0.9991 moles CaCO3 will produce 0.9991 moles CaO

0.9991 moles CaO X 56.06 grams = 56.00 grams CaO 1 mole100 grams CaCO3 X 1 mole CaCO3 = 0.9991 moles CaCO3 100.09 grams CaCO3

Theoretical and percent yieldIf a chemical reaction occurs, in theory you can calculate how much of the product is created. This would be the maximum amount that is produced. However, in the real world often not all the possible product are produced in a chemical reaction.

Theoretical yieldThe maximum amount of product that could be produced determined by calculations using the chemical equation.

Percent yieldThe ratio of the actual and theoretical yields determined by the formula

%yield = actual yield X 100%Theoretical yield

Example:2HCl(aq) + Ca(s) CaCl2(s) + H2(g)

Assume the theoretical yield of CaCl2 in this equation were 30 g. If the actual yield of CaCl2 were 25 g, calculate the percentage yield.

%yield = actual yield X 100%Theoretical yield

25 g X 100 = 83.3% 30g

Homework Assignmentp. 147-148 Exercises 85-104 (odd)

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