unit 5: the mole 6.02 x 10 23 mr. blake / mr.gower/ mr. heaton

Unit 5: The Mole Unit 5: The Mole

6.02 X 6.02 X 10102323

Mr. Blake / Mr.Gower/

Mr. Heaton

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 = 120.1 amu 10 H = 10.1 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 =

602,200,000,000,000,000,000,000 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 = 6.022 X

1023 Na atoms.

The Mole• So 1 mole of any element has 6.022 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

• 6.022 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 = 6.022 X 1023

cookies

• 1 dozen cars = 12 cars• 1 mole of cars = 6.022 X 1023 cars

• 1 dozen Al atoms = 12 Al atoms• 1 mole of Al atoms = 6.022 X 1023

atoms

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

• Mole is abbreviated mol

= 6.022 x 1023 C atoms

= 6.022 x 1023 H2O molecules

= 6.022 x 1023 NaCl formula units(technically, ionics are compounds not molecules so they are called formula units)

6.022 x 1023 Na+ ions and

6.022 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

6.022 x 1023 particles 1 mole

or

1 mole

6.022 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) 6.022 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 = 40.08 g

+2 moles Cl = 70.90 g

Therefore,the mass of the entire compound is…

1 mole of CaCl2 = 110.98 g

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.

–H2O

–2(1.01) + 16.00 = 18.02 g/mol

–NaCl–22.99 + 35.45 = 58.44 g/mol

B. Molar Mass Examples

1) water

2) sodium chloride

– NaHCO3

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

– C6H12O6

– 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

B. Molar Mass Examples3) sodium bicarbonate

4) glucose

C. Molar Conversions

molar mass

(g/mol)

MASS

IN

GRAMS

MOLES

NUMBER

OF

PARTICLES

6.022 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?

3.50 mol Li

1 mol Li

6.94 g Li

= g Li24.3

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 = 80.9 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 mol

6.022 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 formula units of NaHCO3. 2.1

1024

f.u.s 1 mol

6.022 1023

f.u.s= 290 g NaHCO3

84.01 g

1 mol

NaHCO3.

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

35.4 g Cu 1 mol Cu 6.022 X 1023 atoms Cu 63.55 g Cu 1 mol Cu

= 3.35 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

g Ba137.33 Ba%g Ba(CN)189.37 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

S) g 3(32.07 S %

342

100 x )(SOFe g 399.91

O) g 12(16.00 O %

342

24.06 % S

48.01 % O

27.93 % Fe100 x g Fe2(55.85)

Fe%399.91 g Fe2(SO4)3

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.

100% = 100 g

Empirical Formula Empirical Formula DeterminationDetermination

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

49.32g C

12.01 g C4.107 mol C=

6.85 g H

1.01 g H6.78 mol H=

1 mol Cx

x1 mol H

43.84 g O

16.00 g O2.74 mol O=x

1 mol O

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

Using 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.

x 1 gal4 qt1 gal x 1.057 qt

1 L x 10-3 L1 mL x mL

0.85 g C6H14 x 200.59 g1 mol Hg

= 123 mol Hg

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.

Answers for Station ReviewStation 1

1. Aluminum Oxide 2. Iodine 3. Iron (II) oxide

4. Nitrogen trioxide 5. Manganese (III) oxide

6. Calcium phosphide

Station 2

1. 197.84 g/mol 2. 164.10 g/mol

3. 18.02 g/mol 4. 78.05 g/mol

Station 3

591 g Br2

Station 40.966 mol KBr

Station 5All of the same

Station 63.04 x1024 atoms

Station 7

8.26 g Na2SO4

Station 8Na: 46.91%C: 24.51 % N: 28.59%

I. Determination of Empirical Formula:1. Laboratory analysis shows that a compound is 5.80

% H and 94.2 % S by weight (mass). What is its E.F.?

Step 1: Calculate the moles of each element present. (Assume 100 % = 100 g)

H2S2 atoms H : 1 atom S

2 mol H : 1 mol S

5.80 g H x1.01 g

1 mol H = 5.74 mol H

94.2 g S x32.07 g

1 mol S = 2.94 mol S

Step 2: Determine the ratio between moles (smallest whole numbers)

by dividing by the smallest value in step 1.

1.95 94.2

74.5 H

H2S

1 94.2

94.2 S

2

Name?

Hydrosulfuric acid

+/- 0.05 of the whole number

2. Remember that these values are measurements and need not be exact. They should be close. What would be the formulas for elements “X” and “Y”. (Step 2 is done)

  X1.00 Y1.03 X1.5 Y1.0 X1.00 Y1.33 X1.00 Y1.25

3. (a) Laboratory analysis shows that a compound has the following composition: Zinc 52.0 %,Carbon 9.60 %, & Oxygen 38.4 %. Determine its empirical formula.

2X3Y2XY

3

X3Y4

4

X4Y5

52.0 g Zn x65.39 g

1 mol Zn = 0.80 mol Zn

9.60 g C x12.01 g

1 mol C = 0.80 mol C

38.4 g O x16.00 g1 mol O = 2.4 mol O

0.80

0.80

0.80= 1

= 1

= 3

ZnCO3

Name?

Zinc carbonate

+/- 0.05 of the whole number

(b) What is the empirical formula of a compound that is 52.9 % Al and 47.1 % oxygen?

J. Determination of the molecular formula:

1. You must be given the _________ weight.

2. What is the relationship between the empirical formula

weight and the molecular formula weight?

3. Molecular formula = Empirical formula x _________

52.9 g Al x26.98 g

1 mol Al = 1.96 mol Al

47.1 g O x16.00 g

1 mol O = 2.94 mol O1.96

1.96= 1

= 1.5

Al2O3

Name?

Aluminum oxide

x 2

molecular

Molecular formula is a multiple of empirical formula!

multiple

weightempirical g

weightmolecular g Multiple given

calculate

4. Analysis shows that a compound is 30.4 % Nitrogen and 69.6 % Oxygen. If it has a molecular weight of 92 g, what is its Molecular formula?

Step 1: Calculate the Empirical formula.  

Step 2: Calculate the Empirical formula weight.

30.4 g N x14.01 g

1 mol N = 2.17 mol N

69.6 g O x16.00 g

1 mol O = 4.35 mol O2.17

2.17= 1

= 2

NO2

Name?

Nitrogen dioxide

1 (14.01) + 2 (16.00) = 46.01 g

Step 3: Divide the Empirical formula weight into molecular weight (must be given).

 Step 4: Determine the Molecular formula.

weightempirical g

weightmolecular g Multiple

g 46.01

g 92 Multiple = 2

2(NO2) = N2O4 Name?

Dinitrogen tetroxide

Note: Follow normal rounding rules to round to a whole number.