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