Unit 6: The Mole 6.02 X 10 23 Mr. Blake Review: Atomic Masses Elements occur in nature as mixtures...

Unit 6: The Mole Unit 6: The Mole

6.02 X 6.02 X 10102323

Mr. Blake

ReviewReview: Atomic Masses: Atomic Masses Elements occur in nature as

mixtures of isotopes.

Carbon = 98.89% 12C (6 p+, 6 n) 1.11% 13C (6p+, 7n)

<0.01% 14C (6p+, 8n)

Carbon’s atomic mass = 12.01 amu

Atomic Weights are relative

C: # and mass H: # and mass Mass Ratio

1 C = 12.0 amu 1 H = 1.0 amu 12 to 1

2 C = 24.0 amu 2 H = 2.0 amu 12 to 1

3 C = 36.0 amu 3 H = 3.0 amu 12 to 1

4 C = 48.0 amu 4 H = 4.0 amu 12 to 1

5 C = 60.0 amu 5 H = 5.0 amu 12 to 1

10 C = 120.0 amu 10 H = 10.0 amu 12 to 1

What is the mole?

Not this kind of mole!

A. What is the Mole?A. What is the Mole?

• A counting number (like a dozen).

• 1 mol =

602,000,000,000,000,000,000,000 of

anything.

A large amount!!!!

Similar Words• PairPair: 1 pair of shoelaces = 2

shoelaces.• DozenDozen: 1 dozen oranges = 12

oranges.• GrossGross: 1 gross of spider rings = 144

rings.• ReamReam: 1 ream of paper = 500 sheets

of paper.• MoleMole: 1 mole of Na atoms = 6.02 X

1023 Na atoms.

The Mole• So 1 mole of any element has 6.02 X

1023 particles.

• This is a really big number – It’s so big because atoms are very small!

• Defined as the number of atoms in 12.0 grams of C-12. This is the standard!

• 12.0 g of C-12 has 6.022 X 1023 atoms.

The MoleThe Mole

• This number is named in honor of Amedeo ______________ (1776 – Amedeo ______________ (1776 – 1856)1856), who studied quantities of gases and discovered that no matter what the gas was, there were the same number of molecules present

• 6.02 x 1023 is also called _______________________.

Avogadro

Avogadro’s number (NA)

Amadeo Avogadro

I didn’t discover it. Its just named after

me!

Just How Big is a Mole?Just How Big is a Mole?

• Enough soft drink cans to cover the surface of the earth to a depth of over 200 miles.

• If you had Avogadro's number of unpopped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles.

• If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.

Everybody Knows Avogadro’s Everybody Knows Avogadro’s Number!Number!

But Where Did it Come From?But Where Did it Come From?• It was NOT just

picked! It was measured.

• One of the better methods of measuring this number was the Millikan Oil Drop Experiment.

• Since then we have found even better ways of measuring using x-ray technology.

The MoleThe Mole• 1 dozen cookies = 12 cookies• 1 mole of cookies = 6.02 X 1023 cookies

• 1 dozen cars = 12 cars• 1 mole of cars = 6.02 X 1023 cars

• 1 dozen Al atoms = 12 Al atoms• 1 mole of Al atoms = 6.02 X 1023 atoms

- Note that the NUMBER is always the same, but the MASS is very different!

• Mole is abbreviated mol

= 6.02 x 1023 C atoms

= 6.02 x 1023 H2O molecules

= 6.02 x 1023 NaCl “molecules”(technically, ionics are compounds not molecules so they are called formula units)

6.02 x 1023 Na+ ions and

6.02 x 1023 Cl– ions

A Mole of ParticlesA Mole of Particles Contains 6.02 x 1023

particles 1 mole C

1 mole H2O

1 mole NaCl

Suppose we invented a new collection unit called a widget. One widget contains 8 objects.

1. How many paper clips in 1 widget?a) 1 b) 4 c) 8

2. How many oranges in 2.0 widgets?

a) 4 b) 8 c) 16

3. How many widgets contain 40 gummy bears?

a) 5 b) 10 c) 20

Learning CheckLearning Check

6.02 x 1023 particles 1 mole

or

1 mole

6.02 x 1023 particles

- Note that a particle could be an atom OR a

molecule!

Avogadro’s Number as Avogadro’s Number as Conversion FactorConversion Factor

1. Number of atoms in 0.500 mole of Ala) 500 Al atoms

b) 6.02 x 1023 Al atomsc) 3.01 x 1023 Al atoms

2.Number of moles of S in 1.8 x 1024 S atomsa) 1.0 mole S atomsb) 3.0 mole S atomsc) 1.1 x 1048 mole S atoms

Learning CheckLearning Check

B. Molar Mass• Mass of 1 mole of an element or compound.

• Equal to the numerical value of the average atomic mass (get from periodic table)

• Atomic mass tells the...– atomic mass units per atom (amu)– grams per mole (g/mol)

• Round to 2 decimal places (hundredths)

1 mole of C atoms = 12.01 g

1 mole of Mg atoms =

24.31 g

1 mole of Cu atoms = 63.55

g

Molar Masses of Elements Molar Masses of Elements (Found in Periodic Table)(Found in Periodic Table)

Carried to the hundredths place!!

B. Molar Mass Examples

• carbon

• aluminum

• zinc

12.01 g/mol

26.98 g/mol

65.39 g/mol

Calculating Molecular Mass

Calculate the molecular mass of magnesium Calculate the molecular mass of magnesium carbonate, MgCOcarbonate, MgCO33..

24.31 g + 12.01 g + 3(16.00 g) 24.31 g + 12.01 g + 3(16.00 g) == 84.32 84.32

g/molg/mol

1 mole Ca x 40.08 g/mol = 40.08 g/mol

+2 moles Cl x 35.45 g/mol = 70.90 g/molTherefore,the mass of the entire compound is…

1 mole of CaCl2 = 110.98 g/mol

Molar Mass of Molecules Molar Mass of Molecules and Compoundsand Compounds

A substance’s A substance’s molecular massmolecular mass (molecular (molecular weight) is the mass in grams of one mole of weight) is the mass in grams of one mole of the compound.the compound.

B. Molar Mass Examples

•water

•sodium chloride

–H2O

–2(1.01) + 16.00 = 18.02 g/mol

–NaCl–22.99 + 35.45 = 58.44 g/mol

B. Molar Mass Examples• sodium

bicarbonate

• sucrose

– NaHCO3

– 22.99 + 1.01 + 12.01 + 3(16.00) = 84.01 g/mol

– C12H22O11

– 12(12.01) + 22(1.01) + 11(16.00) = 342.34 g/mol

C. Molar Conversions

molar mass

(g/mol)

MASS

IN

GRAMS

MOLES

NUMBER

OF

PARTICLES

6.02 1023

(particles/mol)

molar mass Grams Moles

Calculations with Molar Calculations with Molar MassMass

Converting moles to grams

Ex. #1 How many grams of lithium are in 3.50 moles of lithium?

= g Li3.50 mol Li

1 mol Li

6.94 g Li 24.3

Ex. #2 Aluminum is often used for the structure of light-weight bicycle frames. How many grams of Al are in 3.00 moles of Al?

Goal: 3.00 moles Al ? g Al

Converting Moles and Converting Moles and GramsGrams

1. Molar mass of Al 1 mole Al = 26.98 g Al

2. Conversion factors for Al 26.98 g Al or 1 mol Al

1 mol Al 26.98 g Al

3. Setup 3.00 moles Al x 26.98 g Al

1 mole AlAnswer = 81.0 g Al

Molar Conversion Examples3. How many moles of carbon

are in 26 g of carbon?

26 g C 1 mol C

12.01 g C= 2.2 mol C

4) How many molecules are in 2.50

moles of C12H22O11?

2.50 mol6.02 1023

molecules

1 mol= 1.51 1024

molecules C12H22O11

molar mass Avogadro’s number Grams Moles particles

Everything must go through Moles!!!

CalculationsCalculations

Ex. #1 Find the mass of 2.1 1024 molecules of NaHCO3. 2.1 1024

molecules 1 mol

6.02 1023

molecules

= 290 g NaHCO3

84.01 g

1 mol

Ex. #2 How many atoms of Cu are present in 35.4 g of Cu?

35.4 g Cu 1 mol Cu 6.02 X 1023 atoms Cu 63.5 g Cu 1 mol Cu

= 3.4 X 1023 atoms Cu

V. Percent Composition: The percent by ______ of each _________ in a compound.mas

selement

ncompositio % 100 x compound of Mass

element of Mass

What is the % composition of Ba(CN)2 ? 1(137.33 g Ba) + 2(12.01 g C) + 2(14.01 g N) = 189.37 g Ba(CN)2

100 x Ba(CN) g 189.37

C g 24.02 C %

2

100 x Ba(CN) g 189.37

N g 28.02 N %

2

72.52 % Ba

12.68 % C

14.80 % N

100 x Ba(CN) g 189.37

Ba g 137.33 Ba %

2

What is the % composition of Iron III sulfate, ___________? Fe2(SO4)3

2(55.85 g Fe) + 3(32.07 g S) + 12 (16.00 g O) = 399.91 g Fe2(SO4)3

100 x )(SOFe g 399.91

Fe) g 2(55.85 Fe %

342

100 x )(SOFe g 399.91

S) g 3(32.07 S %

342

100 x )(SOFe g 399.91

O) g 12(16.00 O %

342

27.93 % Fe

24.06 % S

48.01 % O

Percent Composition (Hydrates):

1. Ba(CN)2 ∙ 2 H2O is what % water by mass? 1(Ba) + 2(C) + 2(N) + 2(H2O) = 225.41 g Ba(CN)2 • 2

H2O

100 x OH 2 Ba(CN) g 225.41

O)H g 2(18.02 OH %

22

22

15.99 % H2O

Formulas

molecular formula = (empirical formula)n [n = integer]

molecular formula = C6H6 = (CH)6

empirical formula = CH

•Empirical formula: the lowest whole number ratio of atoms in a compound.•Molecular formula: the true number of atoms of each element in the formula of a compound.

Formulas (continued)

Formulas for Formulas for ionic compoundsionic compounds are are ALWAYSALWAYS empirical ( empirical (lowestlowest whole number ratio).whole number ratio).

Examples:Examples:

NaCl MgCl2 Al2(SO4)3K2CO3

Formulas (continued)

Formulas for Formulas for molecular compoundsmolecular compounds might be empirical (lowest whole might be empirical (lowest whole number ratio).number ratio).

Molecular:Molecular:

H2O

C6H12O6 C12H22O11

Empirical:

H2O

CH2O C12H22O11

Empirical Formula Determination

1. Base calculation on 100 grams of compound. 2. Determine moles of each element in 100

grams of compound.3. Divide each value of moles by the smallest of

the values.4. Multiply each number by an integer to obtain

all whole numbers.

Empirical Formula Empirical Formula DeterminationDetermination

Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

( ) ( )( )

49.32g C4.107

12.01molC

gC=

( )( )( )

6.8516.78

1.01

gH 1 molHmolH

gH=

( )( )( )

43.842.74

16.00

g O 1molOmolO

gO=

1 mol C

Empirical Formula Determination

(part 2)Divide each value of moles by the smallest Divide each value of moles by the smallest of the values.of the values.

CarbonCarbon::

HydrogenHydrogen::

OxygenOxygen::

4.1071.50

2.74molCmolO

=

6.782.47

2.74molHmolO

=

2.741.00

2.74molOmolO

=

Empirical Formula Determination

(part 3)Multiply each number by an integer to Multiply each number by an integer to obtain all whole numbers.obtain all whole numbers.

Carbon: 1.50Carbon: 1.50 Hydrogen: 2.50Hydrogen: 2.50 Oxygen: 1.00Oxygen: 1.00x 2 x 2 x 2

33 55 22

Empirical formula: C3H5O2

Finding the Molecular Formula

The empirical formula for adipic The empirical formula for adipic acid is Cacid is C33HH55OO22. The molecular mass . The molecular mass of adipic acid is 146 g/mol. What is of adipic acid is 146 g/mol. What is the molecular formula of adipic the molecular formula of adipic acid?acid?

1. Find the empirical mass of 1. Find the empirical mass of CC33HH55OO223(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) =

73.08 g73.08 g

Finding the Molecular Formula

The empirical formula for adipic The empirical formula for adipic acid is Cacid is C33HH55OO22. The molecular mass . The molecular mass of adipic acid is 146 g/mol. What of adipic acid is 146 g/mol. What is the molecular formula of adipic is the molecular formula of adipic acid?acid?

3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g

2. Divide the molecular mass by 2. Divide the molecular mass by the mass given by the emipirical the mass given by the emipirical formula.formula.

1462

73

Finding the Molecular FormulaThe empirical formula for adipic acid The empirical formula for adipic acid

is Cis C33HH55OO22. The molecular mass of . The molecular mass of adipic acid is 146 g/mol. What is the adipic acid is 146 g/mol. What is the molecular formula of adipic acid?molecular formula of adipic acid?

3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g

3. Multiply the empirical formula by 3. Multiply the empirical formula by this number to get the molecular this number to get the molecular formula.formula.

(C(C33HH55OO22) x 2 ) x 2 ==

CC66HH1010OO44

1462

73=

HydratesHydrates A. An _______ compound that has _______ trapped in the

crystal lattice structure.1. ________________: A compound with water (i.e. BaI2 •2 H2O).

2. __________________: A compound without water (i.e. BaI2).

B. Determination of a hydrate:1. Determine the mass of ________.2. Determine the mass of the _________________.3. Find the __________ of water and anhydrous salt.4. Find the mol ratio. X = * Whole number  5. Write the formula.

H2

OHydrated salt

Anhydrous salt

moles

ionic

H2

O anhydrous salt

salt anhydrous molOH mol 2

C. Example:1. A barium iodide hydrate is heated in a crucible. Determinethe formula of the hydrate given the following data:Before heating: 10.407 g BaI2 • X H2O (Hydrated Salt)

After heating: 9.520 g BaI2 (Anhydrous Salt)

Step 1: Mass of water?

_____________

0.887 g H2OStep 2: Mass of Anhydrous salt?

Step 3: Calculate the mol of the water & the anhydrous salt.

Step 4: Calculate the number of waters.

Step 5:

9.520 g BaI2

= 0.0492 mol H2O1 mol H2O

18.02 g H2O

0.887 g H2O

1 mol BaI2

391.13 g BaI2

9.520 g BaI2 = 0.0243 mol BaI2

= 2

BaI2• 2 H2Osalt anhydrous mol

OH mol 2=x

2

2

BaI mol 0.0243

OH mol 0.0492=

2. Data: Before heating: 5.20 g ZnSO3 • X H2O (Hydrated salt)

After heating: 2.60 g ZnSO3 (Anhydrous salt)

_____________

Step 1: Mass of water?

2.60 g H2O

Step 2: Mass of Anhydrous salt?

Step 3: Calculate the mol of the water & the anhydrous salt.

Step 4: Calculate the number of waters.

Step 5:

2.60 g ZnSO3

= 0.144 mol H2O1 mol H2O

salt anhydrous mol

OH mol 2x

3

2

ZnSOmol 0.0179

OH mol 0.144 = 8

ZnSO3 • 8 H2O

18.02 g H2O

2.60 g H2O

= 0.0179 mol ZnSO3

1 mol ZnSO3

145.43 g ZnSO3

2.60 g ZnSO3

IQ #2Using your knowledge of mole calculations and unit conversions, determine how many atoms there are in 1 gallon of gasoline. Assume that the molecular formula for gasoline is C6H14 and that the density of gasoline is approximately 0.85 grams/mL.

Using a conversion factor of 3785 mL per gallon, we can determine that the mass of gasoline in one gallon is 3785 mL x 0.8500 g/mL = 3217 grams.

Because the molar mass of C6H14 is 86 g/mole, there are 3217 / 86 moles of gasoline molecules, or 37.4 moles of molecules present.

Multiplying 37.4 x 20 (the number of atoms per mole of gasoline), there are 748 moles of atoms. Finally, multiplying 748 moles of atoms by 6.02 x 1023 atoms/mole, we can find that there are 4.50 x 1026 atoms present in the sample.

CHEMICAL REACTIONSCHEMICAL REACTIONS

Reactants: Zn + IReactants: Zn + I22 Product: Zn IProduct: Zn I22

Chemical Chemical EquationsEquations

• Their Job: Depict the kind of Their Job: Depict the kind of reactantsreactants and and productsproducts and their and their relative amounts in a reaction.relative amounts in a reaction.

44 Al Al (s)(s) + + 33 O O2 (g)2 (g) ---> ---> 22 Al Al22OO3 (s)3 (s)

• The numbers in the front are The numbers in the front are calledcalled

stoichiometricstoichiometric ________________________..• The letters (s), (g), and (l) are the The letters (s), (g), and (l) are the

physical statesphysical states of compounds. of compounds.

coefficients

–Chemical reactions occur when bonds between the outermost parts of atoms are formed or broken.

–Chemical reactions involve changes in matter, the making of new materials with new properties, and energy changes.

–Symbols represent elements, formulas describe compounds, chemical equations describe a chemical reaction.

IntroductionIntroduction

– Chemical equations show the conversion of reactantsreactants (the molecules shown on the left of the arrow) into productsproducts (the molecules shown on the right of the arrow).•A “+” sign separates molecules on the same side

•The arrow is read as “yields”•Example

C + O2 CO2

•This reads “carbon plus oxygen react to yield carbon dioxide”

Parts of a Reaction Parts of a Reaction EquationEquation

• The charcoal used in a grill is basically carbon. The carbon reacts with oxygen to yield carbon dioxide. The chemical equation for this reaction, C + O2 CO2, contains the same information as the English sentence but has quantitative meaning as well.

Because of the principle of Because of the principle of

the the conservation of matterconservation of matter, ,

an equation must bean equation must be

balancedbalanced.. It must have the sameIt must have the same

number of atoms of the number of atoms of the

same kind on same kind on bothboth sides. sides. Lavoisier, 1788Lavoisier, 1788

Chemical EquationsChemical Equations

•Solid ___•Liquid (l)•Gas ___•Aqueous solution (aq)•Catalyst H2SO4

•Escaping gas ()•Change of temperature ()

Symbols Used in Symbols Used in EquationsEquations

(s)

(g)

– When balancing a chemical reaction you may add coefficients in front of the compounds to balance the

reaction, but you may notnot

change the subscripts.–Changing the subscripts changes the compound.

–Subscripts are determined by the valence electrons (charges for ionic or sharing for covalent)

Balancing EquationsBalancing Equations

Subscripts vs. Subscripts vs. CoefficientsCoefficients

• The subscripts The subscripts tell you how tell you how many atoms of many atoms of a particular a particular element are in element are in a compound. a compound. The coefficient The coefficient tells you about tells you about the quantity, the quantity, or number, of or number, of molecules of molecules of the compound.the compound.

Chemical EquationsChemical EquationsChemical EquationsChemical Equations4 Al(s) + 3 O4 Al(s) + 3 O22(g) (g)

---> 2 Al---> 2 Al22OO33(s)(s)This equation means: This equation means:

4 Al atoms + 3 O2 molecules ---produces--->

2 molecules of Al2O3

AND/ORAND/OR

4 moles of Al + 3 moles of 4 moles of Al + 3 moles of OO22 ---produces--->---produces--->

2 moles of Al2 moles of Al22OO33

There are four basic steps to balancing a chemical equation.1.Write the correct formula for the reactants and

the products. DO NOT TRY TO BALANCE IT YET! DO NOT CHANGE THE FORMULAS!

2.Find the number of atoms for each element on the left side. Compare those against the number of the atoms of the same element on the right side.

3.Determine where to place coefficients in front of formulas so that the left side has the same number of atoms as the right side for EACH element in order to balance the equation.

4.Check your answer to see if:– The numbers of atoms on both sides of the

equation are now balanced.– The coefficients are in the lowest possible

whole number ratios. (reduced)

Steps to Balancing Steps to Balancing EquationsEquations

Some Suggestions to Help Some Suggestions to Help YouYou

• Take one element at a time, working left to right except for H and O. Save H for next to last, and O until last.

•If everything balances except for O, and there is no way to balance O with a whole number, double all the coefficients and try again. (Because O is diatomic as an element)

•(Shortcut) Polyatomic ions that appear on both sides of the equation should be balanced as independent units

Balancing Balancing EquationsEquationsBalancing Balancing EquationsEquations

___ H___ H22(g) + ___ O(g) + ___ O22(g) ---> ___ H(g) ---> ___ H22O(l)O(l)2 2

What Happened to the Other Oxygen Atom?????

This equation is not balanced!Now, follow your rules for balancing equations and see if you can do it!

IQ 2Rewrite these word equations as

balanced chemical equations1) Hydrogen + sulfur hydrogen sulfide

2) Iron III chloride + calcium hydroxide iron III hydroxide + calcium chloride

3) Heating copper (II) sulfide in the presence of oxygen gas produces pure copper and sulfur dioxide gas.

Balancing Balancing Equations EquationsBalancing Balancing Equations Equations

___ Al(s) + ___ Br___ Al(s) + ___ Br22(l) ---> ___ Al(l) ---> ___ Al22BrBr66(s)(s)2 3

Balancing Balancing EquationsEquationsBalancing Balancing EquationsEquations

____C____C33HH88(g) + _____ O(g) + _____ O22(g) ---->(g) ---->

_____CO_____CO22(g) + _____ H(g) + _____ H22O(g)O(g)

____B____B44HH1010(g) + _____ O(g) + _____ O22(g) ---->(g) ---->

___ B___ B22OO33(g) + _____ H(g) + _____ H22O(g)O(g)

?

52

3 4

5Need 11 Oxygen atoms but they only come in pairs! How can we fix this?

Take the number you need and put it over 2 (i.e. 11/2).

Since we can’t have a fraction for a coefficient, we must multiply it out by multiplying all of the coefficients by 2.11

2

4 10

=11

12

Learning Check!Learning Check!Learning Check!Learning Check!Sodium phosphate + iron (III) oxide Sodium phosphate + iron (III) oxide

sodium oxide + iron (III) phosphatesodium oxide + iron (III) phosphate

NaNa33POPO44 + + __________ Fe Fe22OO33 ----> ---->

____ Na ____ Na22O + _____ FePOO + _____ FePO44

Algebraic MethodAlgebraic MethodConsider:

1. Assign letters to unknown coefficients:

A Ca3(PO4)2 + B H2SO4 C CaSO4 + D H3PO4

Ca3(PO4)2 + H2SO4 CaSO4 + H3PO4

2. Make a grid indicating the appearance of element (or ion) in each species of the equation. Use a whole number and the coefficient “letter” to indicate each appearance.

Ca 3a + 0 = c + 0

PO4 2a + 0 = 0 + dH 0 + 2b = 0 + 3d

SO4 0 + b = c + 0

3. Reduce each equation:Ca 3a = cPO4 2a = dH 2b = 3dSO4 b = c

4. Assume a = 1; solve for coefficients:B = 3D = 2C= 3A= 1

5. Write equation with coefficients:

Ca3(PO4)2 + 3 H2SO4 3 CaSO4 +2 H3PO4

Example:

Na2CO3 + C + Sb2S3 Sb + Na2S + CO2

A B C D E F

Na: 2a = 2e C: a + b = f

O: 3a = 2f

Sb: 2c = d

S: 3c = e

a = 1 e = 1

x 6

A = 6 E = 6

F = 9

C = 2 D =4

B = 3

Na2CO3 + C + Sb2S3 Sb + Na2S + CO26 3 2 4 6 9

b = 1/2

d = 2/3

c = 1/3

f = 3/2

Topics for Unit #5

• The Mole & Quantifying Substances (ch. 10)

• Avagadro’s Number (ch. 10)

• Chemical Equations & Balancing (ch. 11)

• Empircal vs. Molecular Formulas (ch. 11)

Introductory Questions #11) How do you measure matter if there’s a lot of it

there? Name the three ways suggested in the text (pg. 287)

2) What is the mass of 90 average-sized apples if 1 dozen of apples has a mass of 2.0 kg?

3) How many representative parts do you find in one mole?

4) Magnesium is a light metal used in the manufacture of aircraft, automobile wheels, tools, and garden furniture. How many moles of magnesium is 1.25 x1023 atoms of magnesium? (see practice problem 10.2 on pg 291)

IQ#1-answers1. Count the items, determine the mass, & Volume

2. We know: 1 dozen apples = 2.0 Kg

# of apples = 90

12 apples = 1 dozen

**Dimensional analysis:

90 apples x 1 doz. apples x 2.0 kg

12 apples 1 doz apples

= 15 kg

3. 1 mol = 6.02 x 1023 parts

IQ#1-answers cont’d4. Count the items, determine the mass, &

Volume**We know: # of atoms present = 1.25 x1023 Mg atoms

1 mol Mg = 6.02 x1023 atoms Mg

Conversion: # atoms → # moles

Use Dimensional analysis:

1.25 x1023 atoms Mg x 1 mol Mg 6.02 x1023 atoms Mg

= 0.208 mol Mg

Answers to Homework Practice Problems #1-5

1) 5.0 kg (pg. 289)

2) 672 seeds (pg. 289)

3) 4.65 mol Si (pg. 291)

4) 0.360 mol Br2 (pg. 291)

5) 2.75 x1024 atoms (pg. 292)

6) 7.72 mol NO2 (pg. 292)

Introductory Questions #21. Using your periodic table determine the

molecular mass of ammonium nitrate and Potassium permanganate.

2. Which compound has higher molar mass ammonium nitrate or Potassium permanganate? Show your work.

3. If I had 38.5 g of potassium permanganate and 58.6g of ammonium nitrate which do I have more of? Convert each into the number ofmoles present to compare.

Homework

• Do Quest. #7-13 from “The Mole Worksheet I”

• Practice problems: pg. 296 #7 & 8

pg. 298 #16 & 17

pg. 299 #18 & 19

Read/Review Percent Composition Pgs 305-307

*Do Practice problems #34 & 35 on pg. 307

Introductory Questions #31. Which of the following substances has

more molecules/compounds present? Be sure to show all calculations and work.

242.38g Heptanitrogen pentoxide

85.56g Aluminum arsenate

154.28g hydrogen perioxide

2. How much would 1000 atoms of magnesium weigh in grams?

3. Determine the % composition of barium perchlorate.

Assignments to Check/Review

• % Composition Worksheet answers• Chemistry WS II: % Comp & Empirical Formulas• Practice Problems from text: #36 & 37 (pg.310)• Hydrate Formula----Worksheet Today’s Handout• Lect/Disc: Chemical Reactions & equation writing

**Homework: Hydrate Formula WS

Chemical Equation Worksheet

Signs of a Chemical Signs of a Chemical ReactionReaction

•Evolution of heat and light•Formation of a gas•Formation of a precipitate•Color change

B.Law of Conservation of Mass

• Mass is neither created nor destroyed in a chemical reaction

4 H

2 O

4 H

2 O4 g 32 g

36 g

• Total mass stays the same

• Atoms can only rearrange

C. Chemical Equations

A+B C+DREACTANT

SPRODUCT

S

Chemical SymbolsChemical Symbols

Writing Equations

• Identify the substances involved.

• Use symbols to show:

2H2(g) + O2(g) 2H2O(g)

– How many? - coefficient

– Of what? - chemical formula

– In what state? - physical state

• Remember the diatomic elements.

D. Writing EquationsD. Writing EquationsTwo atoms of aluminum react with three units of aqueous copper(II) chloride to produce three atoms of copper and two units of aqueous aluminum chloride.

• How many?• Of what?• In what state?

Al2 (s) +3CuCl2(aq) 3 Cu(s)+ 2AlCl3

(aq)

E. Describing Equations• Describing Coefficients:

– individual atom = “atom”

– covalent substance = “molecule”

– ionic substance = “unit”

3 molecules of carbon dioxide

2 atoms of magnesium

4 units of magnesium oxide

3CO2

2Mg

4MgO

E. Describing Equations

to produce

• How many?• Of what?• In what state?

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

One atom of solid zinc reacts withtwo molecules of aqueous hydrochloric acid one unitof aqueous zinc chloride and onemolecule of hydrogen gas.

Unit 5 HW Packet1. Book Notes (15 total) 10-1 practice problems #1-4*

2. Page 315 #50-56, 58-59

3. Moles, Molecules, and Grams WS

4. Percent Composition WS

5. p. 315 # 63-68

6. Percent Comp., Emp. Formula WS

7. Writing Complete Equations WS

8. Balancing Chemical Equations WS

9. Balancing Equations: Algebraic Technique WS

10. Unit 5 Sample Test