Upload
letuong
View
236
Download
3
Embed Size (px)
Citation preview
Slide 1 / 157
This material is made freely available at www.njctl.org and is intended for the non-commercial use of students and teachers. These materials may not be used for any commercial purpose without the written permission of the owners. NJCTL maintains its website for the convenience of teachers who wish to make their work available to other teachers, participate in a virtual professional learning community, and/or provide access to course materials to parents, students and others.
Click to go to website:www.njctl.org
New Jersey Center for Teaching and Learning
Progressive Science Initiative
Slide 2 / 157
www.njctl.org
Mole / StoichiometryCalculations
Slide 3 / 157
Table of Contents
· Avagadro's Number· Molar Mass
· Emperical Formula
· Molar Volume· Percent Composition
Click on the topic to go to that section
Slide 4 / 157
Return toTable ofContents
Avogadro's Number
Slide 5 / 157
MolesRecall an atom's atomic mass is equal to the number of
protons plus the number of neutrons in the atom.
C
6
12.01
Atomic Number or number of protons (Z)
Atomic Mass in amu
The unit for atomic mass is amu. Carbon-12 has 6 protons and 6 neutrons.
One amu is equal to 1/12 the mass of carbon-12 or
approximately the mass of one proton or neutron
mass of 1 proton = 1 amu
Slide 6 / 157
Moles
What if you wanted to measure the mass of one atom in the laboratory? Would it be possible?
A single atom has a very small mass.
One carbon atom has a mass of about 2.0 x 10-23 grams.
Slide 7 / 157
The Mole
It takes a lot of atoms to give us enough material to directly measure in a lab.
Hydrogen has a mass of 1 amu. How many atoms of hydrogen would be needed to make a 1 gram sample of hydrogen?
602,200,000,000,000,000,000,000The amount 6.02x1023 is called Avogadro's number or a mole.
How big is a mole?
Slide 8 / 157
Calculating Avogadro's NumberThe atomic mass of one carbon atom is 12.01 amu or 2x10-23 g.
How many carbon atoms would it take to get 12.01 grams of carbon?
Givens:mass of 1 carbon atom = 2x10-23 gtotal mass of carbon atoms = 12.01 g
2x10-23 g x ? of atoms = 12.01 g
? of atoms 12.01 g=2x10-23 g
? of atoms 6.02 x 1023 atoms=
Slide 9 / 157
Holy Mole-y!
If you were able to count at the rate of 1 million numbers a second, it would take about 20 billion years to count out one mole!
1 mole of pennies could be distributed to all the
currently-living people of the world so that they could spend a million dollars per hour every
hour (day and night) for the rest of their lives!
One mole of marbles would cover the entire Earth (oceans included) for a
depth of two miles!
Slide 10 / 157
The MoleA mole is just a grouping of numbers...like dozen, ream, etc.
A dozen means 12 of something.
A mole means 6.02 x 1023 of something.
Common Grouping Quantities1 dozen = 12
1 gross =144
1 ream = 500
1 mole = 6.02 x 1023
Slide 11 / 157
1 How many eggs are in two dozen eggs?
A 12
B 24
C 0.0833
D 2
E 6.02 x 1023
Slide 12 / 157
2 How many eggs are in half a dozen eggs?
A 12
B 24
C 6
D 0.5
E 6.02 x 1023
Slide 13 / 157
3 How many particles of sand are in 0.5 moles of sand?
A 1.5 x 1023
B 3.01 x 1023
C 6.02 x 1023
D 1.2 x 1024
E 6.02 x 1024
1 mole = 6.02 x 1023
Slide 14 / 157
4 How many pieces of gold dust are in 2 moles of gold dust?
A 1.5 x 1023
B 3.01 x 1023
C 6.02 x 1023
D 1.2 x 1024
E 6.02 x 1024
1 mole = 6.02 x 1023
Slide 15 / 157
5 How many dozen eggs are in a container of 6 eggs?
A 4
B 0.5
C 1
D 2
E 6.02 x 1023
Slide 16 / 157
6 How many dozen eggs are in a container of 18 eggs?
A 1.5
B 3
C 0.67
D 2
E 6.02 x 1023
Slide 17 / 157
7 The Milky Way Galaxy may have up to 400 billion (4 x 1014 stars). How many moles of stars does it have?
A 1.5 x 109
B 6.7 x 10-10
C 10
D 0.5
E 6.02 x 1023
1 mole = 6.02 x 1023
Slide 18 / 157
8 Mathematicians estimate Earth's beaches contain nearly 5.6 x 1021 grains of sand. How many moles of sand are on Earth's beaches?
A 1.5
B 9.3 x 10-3
C 10
D 107.5
E 6.02 x 1023
1 mole = 6.02 x 1023
Slide 19 / 157
Measuring Matter with Moles
The mole is the SI unit for measuring the amount of particles in a chemical substance.
1 mole of Carbon
Slide 20 / 157
One mole (mol) of a substance is 6.02 x 1023 representative particles of that substance.
Measuring Matter with Moles
n = NA
NWhere:
n is the number of moles
N is the total number of particles
NA is Avogadro's Number 6.02 x 1023
Slide 21 / 157
NA = 6.02 x1023 atoms
How many moles of Gold are there in a sample containing 3.01 x 1023 atoms of Gold?
Measuring Matter with Moles
Slide 22 / 157
9 How many atoms of titanium are in a sample containing 0.5 mole of titanium?
A 1.5 x 1023
B 3.01 x 1023
C 6.02 x 1023
D 1.2 x 1024
n = NA
N
NA = 6.02 x 1023
E 6.02 x 1024
Slide 23 / 157
10 How many atoms of sodium are in a sample containing 2.0 moles of sodium?
A 1.5 x 1023
B 3.01 x 1023
C 6.02 x 1023
D 1.2 x 1024
n = NA
N
NA = 6.02 x 1023
E 6.02 x 1022
Slide 24 / 157
11 How many moles of potassium are in a sample containing 3.01 x 1023 atoms of potassium?
A 1.0
B 2.0
C 0.5
D 0.75
n = NA
N
NA = 6.02 x 1023
E 6.02 x 1024
Slide 25 / 157
12 How many moles of potassium are in a sample containing 1.2 x 1024 atoms of potassium?
B 0.50 mol
C 1.0 mol
D 2.0 molE 3.0 mol
A 0.25 mol
n = NA
N
NA = 6.02 x 1023
Slide 26 / 157
13 How many moles of tungsten atoms are there in a sample containing 1.8 x 1024 atoms of tungsten?
A 0.33 mol
B 0.50 mol
C 1.0 mol
D 1.5 mol
E 3.0 mol
n = NA
N
NA = 6.02 x 1023
Slide 27 / 157
14 How many moles of silver are there in a pure sample containing 1.5 x 1023 atoms of silver?
A 0.10 mol
B 0.25 mol
C 0.50 mol
D 1.0 mol
n = NA
N
NA = 6.02 x 1023
E 1.5 mol
Slide 28 / 157
15 How many atoms are there in 5.00 mol of hafnium?
A 6.02 x1023 atoms
B 1.20 x 1023 atoms
C 1.20 x 1022 atoms
D 3.43 x 1023 atoms
E 3.01 x 1024 atoms
n = NA
N
NA = 6.02 x 1023
Slide 29 / 157
A mole of ANY substance contains Avogadro’s number of representative particles, or 6.02 x 1023 representative particles.
1 mole of C atoms = 6.02 X 1023 atoms of C
1 mole of pick-up trucks = 6.02 x 1023 pick-up trucks
The term representative particle refers to the species or types of particles in the substance
For Example: atoms, molecules, formula units, ions
Measuring Matter with Moles
Slide 30 / 157
In 1 mole of water there are 6.02 x 1023 water molecules.
In 1 mole of NaCl there are 6.02 x 1023 formula units.
In 1 mole of carbon there are 6.02 x 1023 carbon atoms.
molecule of H2O
18.0 amu
Avagadro's number of molecules(6.02 x1023) 1 mol H2O
(18.0 g)
laboratory sample size
H
O
H
Measuring Matter with Moles
Slide 31 / 157
16 Formula units refer to particles of __________ compounds and molecules refer to particles of __________ compounds.
A molecular/covalent, ionic
B ionic, molecular/covalent
C atoms, molecular
D atoms, ionic
E ionic, atomic
Slide 32 / 157
17 How many molecules are there in 2.10 mol CO2 ?
A 3.79 x 1024
B 3.49 x 10-24
C 1.05 x 10-23
D 2.53 x 1024
E 1.26 x 1024
n = NA
N
NA = 6.02 x 1023
Slide 33 / 157
18 How many moles of helium atoms are there in a pure sample containing 6.02 x 1024 atoms of helium?
A 2.0 mol
B 4.0 mol
C 6.0 mol
D 10.0 mol
E 2.4 x 1024 mol
n = NA
N
NA = 6.02 x 1023
Slide 34 / 157
19 How many moles of NaCl are there in a pure sample containing 6.02 x 1023 formula units of sodium chloride, NaCl?
A 1.0 mol
B 2.0 mol
C 4.0 mol
D 6.0 mol
E 6.02 x 1023 mol
n = NA
N
NA = 6.02 x 1023
Slide 35 / 157
20 How many formula units of Pb(NO3)2 are there in 0.5 mole of Pb(NO3)2?
A 0.5 formula units
B 2.0 formula units
C 3.01 x 1023 formula units
D 1.2 x 1024 formula units
E 6.02 x 1023 formula units
n = NA
N
NA = 6.02 x 1023
Slide 36 / 157
Calcium deficiency can cause osteoporosis (weakening of the bones). The minimum amount of calcium in 1 mL of blood should be around 1.3 x 1018 atoms.
A patient has her blood tested and the lab finds there are 3 x 10-5 moles of calcium in the blood. Is this patient at risk for osteoporosis?
(6.02 x 1023 atoms/mole) x (3 x 10-5 moles) = 1.8 x 1023 atoms Ca
This exceeds the normal range so they are OK!
Real World Application
slide for answer
Slide 37 / 157
Review: Ionic and Molecular Compounds
Ionic Compounds Molecular Compounds
H2O2 hydrogen
atoms1 oxygen
atom
C6H6
6 carbon atoms
6 hydrogen atoms
NaCl1 Na+ ion 1 Cl- ion
K2CrO42 K+ ions 1 CrO4
2- ion
Sn(OH)2
1 Sn 2+ ion 2 OH- ions
The total number of atoms or ions in a compound depends on electronegativity and bonding.
Slide 38 / 157
Review: Ionic and Molecular Compounds
Chemicals are composed of more than one molecule or formula unit.
To indicate more than one molecule or formulat unit, add a coefficient in front of the compound.
Example: six molecules of carbon dioxide = 6CO2.
6 atoms of carbon and 12 atoms of oxygenMove to reveal answer
How many atoms of carbon and oxygen are in 6CO2?
Slide 39 / 157
Review: Ionic and Molecular Compounds
Ionic Compounds - Fill in Molecular Compounds - Fill in
H2O__hydrogen
atoms __oxygen
atom
6
C6H6
__carbon atoms
__hydrogen atoms
NaCl__ Na+ ions __ Cl- ions
33 3
K2CrO4
__K+ ions __CrO42- ions
2
Sn(OH)2
__Sn 2+ ions __OH- ions
4
Slide 40 / 157
Converting Moles to Number of Particles
In one mole of water molecules there are:
molecule of H2O
18.0 amu
Avagadro's number of molecules(6.02 x1023 ) 1 mol H2O
(18.0 g)
laboratory sample size
H
O
H
H2O2 moles of
hydrogen atoms1 mole of
oxygen atoms
2 x 6.02x1023 atoms of hydrogen
1 x 6.02x1023 atoms of oxygen
Slide 41 / 157
Converting Moles to Number of Particles
In 3 moles of sodium chloride formula units there are:
3NaCl3 moles of
Na+ ions3 moles of
Cl- ions
Slide 42 / 157
21 How many hydrogen atoms are in six molecules of ethylene glycol, the major component in antifreeze?
The formula for ethylene glycol is: HOCH2CH2OH.
A 6 atoms of HB 36 atoms of H
C 6 x 6.02 x 1023 atoms of H
D 36 x 6.02 x 1023 atoms of H
n = NA
N
NA = 6.02 x 1023
E 6.02 x 1023 atoms of H
Slide 43 / 157
22 How many CO32- ions are in one formula unit of CaCO3?
A 1 ion
B 3 ions
C 6 x 6.02 x 1023 ionsD 36 x 6.02 x 1023 ions
n = NA
N
NA = 6.02 x 1023
E 6.02 x 1023 ions
Slide 44 / 157
23 How many K+ ions are there in two formula units of potassium hydroxide, 2KOH?
A 1 K+ ion
B 2 K+ ions
C 1 x 6.02 x1023 ions of K+
D 2 x 6.02 x 1023 ions of K+
n = NA
N
NA = 6.02 x 1023
E 3.12 x 1023 ions of K+
Slide 45 / 157
24 How many sulfide ions (S2- ) are there in 2.0 moles of ammonium sulfide,(NH4 )2 S?
A 2.0 ions
B 1.2 x 1024 ions
C 2.4 x 1024 ions
D 6.02 x 1023 ions
n = NA
N NA = 6.02 x 1023
E none
Slide 46 / 157
25 How many ammonium ions (NH4 + ) are there in 2.0 moles of ammonium sulfide, (NH4 )2 S?
A 2.0 ions
B 1.2 x 1024 ions
C 2.4 x 1024 ions
D 6.02 x 10 23 ions
n = NA
N
NA = 6.02 x 1023
E 8.0 ions
Slide 47 / 157
Hemoglobin is a protein that carries O2 around your body. The formula for it is
approximately C2800H4800N3200O800S8Fe4.
If a patient has 2 x 1016 atoms of Fe, how many moles of Hb would be
present?
moles of Fe: n = N/Na --> (2 x 1016 atoms)/(6.02 x 1023 atoms/n)= 3.3 x 10-7 moles Fe
moles of Hb: There are 1 Hb/4 Fe --> (3.3 x 10-7 moles Fe)/(4 moles of Fe/Hb)
= 8.2 x 10-8 moles Hb
Real World Application
slide for answer
Slide 48 / 157
Return toTable ofContents
Molar Mass
Slide 49 / 157
Mass of Compounds
The total mass of a chemical compound can be calculated by using the masses on the Periodic Table.
Example: Calculate the mass of the compound
Magnesium Chloride (MgCl2)
Mg = 24.305 amu
2Cl = (2)35.453 amu
Mass of MgCl2 95.211 amu
+
Slide 50 / 157
Masses of Elements/Compounds
Atomic mass
Formula mass or
Formula weight (FW)
Molecular mass or
Molecular weight
used for elements (like F, V, etc..) only
used for ionic compounds (like
NaCl, MgO, etc..) only
used for molecular compounds (like
CO2, H2O, etc..) only
amu
Units
amu
amu
Slide 51 / 157
26 What is the formula weight of sodium bromide?
A 79.904 amu
B 102.894 amu
C 205.780 amu
D 300.120 amu
E 605.102 amu
Slide 52 / 157
27 What is the formula weight of Pb(NO3)2?
A 79.90 amu
B 102.89 amu
C 205.78 amu
D 331.34 amu
E 605.10 amu
Slide 53 / 157
28 What is the molecular mass of 3H2O2?
A 17 amu
B 34 amu
C 68 amu
D 102 amu
E 204 amu
Slide 54 / 157
Molar Mass (M)
The mass in grams of one mole of any substance is its molar mass (M). Each of the bars shown below equals one mole of a pure element.
A mole represents the number of atoms it takes to convert from a single atomic mass in amu to the same mass in grams.
1 mole of Aluminum
= 26.982 g
1 mole of Copper
= 63.546 g
Slide 55 / 157
Molar Mass (M)
One mole of carbon, sulfur and silver are shown.
1 mol of Carbon atoms = 12.0 g
1 mol of sulfur atoms 32.0 g of S
1 mol of silver = 107.9 g of Ag
Slide 56 / 157
Average atomic mass of 1 atom of Kr = 83.8 amu.
1 mole (6.02 x 1023 atoms) of Kr = 83.8 grams.
The atomic mass of an element expressed in grams is the mass of one mole or molar mass (M) of the element.
How is the atomic mass of an element related to the molar mass?
1 mole of Kr = 83.8 grams
Molar mass of Kr = 83.8 grams or 83.8 g/mol1 mol
Slide 57 / 157
Molar Mass
Gram atomic mass
Gram formula mass or
Formula weight (FW)
Gram molecular mass or
Molecular weight
used for elements (like F, V, etc..) only
used for ionic compounds (like
NaCl, MgO, etc..) only
used for molecular compounds (like
CO2, H2O, etc..) only
grams
Units
grams
grams
The molar mass of an element or compound can be used to convert directly from masses in amu to masses in grams.
Slide 58 / 157
To convert from moles to mass, or vice versa, use the following formula.
Molar Mass
n = MmWhere:
n is the number of moles
m is the mass of the sample
M is the molar mass of the substance.
Slide 59 / 157
29 How many moles are in a 64-gram sample of pure sulfur?
n = Mm
Slide 60 / 157
30 How many moles are in a 72-gram sample of pure magnesium?
n = Mm
Slide 61 / 157
31 What is the mass, in grams, of 2 moles of carbon?
n = Mm
Slide 62 / 157
32 What is the mass, in grams, of 5 moles of iron?
n = Mm
Slide 63 / 157
33 How many grams is 2.0 mol neon atoms?
n = Mm
Slide 64 / 157
Diatomic Molecules
(Recall the seven diatomic molecules: HONClBrIF)
The molar mass of these molecules will be twice their atomic mass.
Examples:
M of hydrogen gas: H2 = (2 x 1) = 2 g/mol
M of bromine liquid: Br2 = (2 x 79.9) = 159.8 g/mol
M of fluorine molecules: F2 = (2 x 19) = 38 g/mol
Slide 65 / 157
34 How many grams is 1.0 mole of hydrogen molecules?
n = Mm
Slide 66 / 157
35 How many grams is 0.50 mol of oxygen molecules?
n = Mm
Slide 67 / 157
The molar mass of a compound is the sum of the molar masses of all the elements in the compound.
Molar Mass of a Compound
To calculate the molar mass of a compound, find the number of grams of each element in one mole of the compound. Then add the masses of the elements in the compound.
32.065 g/mol + 3(16.000) g/mol = 80.650 g/mol
Molar mass of SO3 = 80.650 g/mol
Slide 68 / 157
Molar Mass of a Compound
Note the different molar masses of these two compounds.
18 g of H2 O = 1 mole H2 O
180 g of glucose (C6 H12 O6 ) = 1 mole glucose
Slide 69 / 157
36 What is the molar mass of K2O, potassium oxide?[*]
Slide 70 / 157
37 What is the molar mass of CaCO3 , calcium carbonate found in eggshells?
A 50 amu
B 50 grams/molC 100 amu
D 100 grams/mol
Slide 71 / 157
38 What is the molar mass of carbon dioxide?
A 28 amu
B 28 grams/mol
C 44 amu
D 44 grams/mol
Slide 72 / 157
39 What is the molar mass of ammonia, NH3 ?
A 17 amu
B 17 grams/mol
C 45 amu
D 45 grams/mol
Slide 73 / 157
Molar mass is the mass of 1 mole of a substance (measured in g/mol).
The atomic mass will be the same number as the molar mass (measured in amu).
The difference is that the atomic mass refers to only one representative particle and molar mass refers to one mole (6.02 x 1023 ) of representative particles.
Summary of Molar Mass
Slide 74 / 157
40 What is the mass of one formula unit of NaCl?
A 17 amu
B 17 grams/mol
C 58.5 amu
D 58.5 grams/mol
Slide 75 / 157
41 What is the mass of one molecule of water?
A 18 amu
B 18 grams/mol
C 8 amu
D 8 grams/mol
Slide 76 / 157
42 The chemical formula of aspirin is C9H8O4 . What is the mass of 0.200 moles of aspirin?
n = Mm
Slide 77 / 157
43 How many moles of O are in 2.4 X 1024 molecules of SO3?
Slide 78 / 157
How many grams of iron are in a 68 gram sample of Fe2O3?
n = m/M 68 g Fe2O3 = 0.43 mol Fe2O3 x (2 mol of Fe) = 0.86 mol Fe
160 g Fe2O3
then...
n = m/M --> m = n*M = (0.86 n Fe)(56 g/n) = 48 grams Fe
OR use dimensional analysis
M of Fe2O3 = 160 g/mol
M of Fe = 56g/mol
69 g Fe2O3 x 1 mol Fe2O3 x 2 mol Fe x 56g Fe = 48 g Fe160g Fe2O3 1 mol Fe2O3 1 mol Fe
Move to see answer
Slide 79 / 157
44 The molar mass of oxygen (O2 ) is:
A equal to the mass of one mole of oxygen atoms.
B 16.0 g/mol
C 32.0 g/mol
D none of the above
n = Mm
E equal to the mass of one oxygen atom.
Slide 80 / 157
45 There are more moles of CO2 in a 44 g sample of carbon dioxide gas than there are moles of helium atoms in a box containing 2 x 1023 atoms of He.
TrueFalse
Slide 81 / 157
Return toTable ofContents
Molar Volume
Slide 82 / 157
The Mole and the Volume of a Gas
The volume of a gas varies with temperature and pressure. But comparisons between gases can be made by designating a standard temperature and pressure (STP).
Slide 83 / 157
STP (Standard Temperature and Pressure)Standard Temperature (T) is considered 0°C (273 K), the temperature at which water freezes.
Standard Pressure (P) is considered 1 atmosphere (101.3 kPa), or the pressure of the atmosphere at sea level.
At STP, 1 mole of gas occupies a volume of 22.4 liters (L)
Vm = 22.4 L
For the time being, we will treat all gases as if they are at STP.
Slide 84 / 157
The Mole and the Volume of a Gas
The volume occupied by one mole of gas is called the molar volume and has the symbol, Vm . It is the same for all gases.
At STP; Vm = 22.4 liters (L).
1 mole of gas = 6.02 x 1023 particles = 22.4 L
Notice that this statement does not depend on the type of gas.
It's true of all gases.
Slide 85 / 157
The Mole and the Volume of a Gas
At STP; Vm = 22.4 liters (L).
This is true because in a gas the molecules are so far apart that they take up almost no space...the volume of a gas is mostly empty space, regardless of the type of gas.
So when it comes to volume (at STP) all gases are created equal.
1 mole of He gas occupies 22.4 L @STP 1 mole of HCl gas occupies 22.4 L @STP
Slide 86 / 157
46 Which of the following must be true about 2 moles of H2 gas vs. 2 moles of CO2 gas at the same temperature and pressure?
A Each sample will have the same density
B The same # of atoms will be present in each sample
C Each sample will have the same massD Each sample will occupy the same volumeE None of these are true
Slide 87 / 157
At STP, 1 mole or, 6.02 x 1023 representative particles, of any gas occupies a volume of 22.4 L.
The quantity 22.4 L is called the molar volume of a gas.
The Mole and the Volume of a Gas
n = Vm
V
Where: n is the number of moles of gas
V is the volume of the gas
Vm is 22.4 L, at STP
Slide 88 / 157
47 How many moles are there in 44.8 liters (at STP) of fluorine gas?
n = Vm
V Vm = 22.4 L
Slide 89 / 157
48 How many moles of atoms are there in 22.4 liters (at STP) of Xenon?
n = Vm
V Vm = 22.4 L
Slide 90 / 157
49 What is the volume (in liters at STP) of 1.00 mole of sulfur dioxide?
n = Vm
V Vm = 22.4 L
Slide 91 / 157
50 What is the volume (in liters at STP) of 2.50 moles of carbon monoxide?
n = Vm
V Vm = 22.4 L
Slide 92 / 157
51 What is the volume (in liters at STP) of 4.00 moles of Nitrogen?
n = Vm
V Vm = 22.4 L
Slide 93 / 157
Real World Application
An inflated airbag requires 60 L of nitrogen gas (N2) at STP in order to protect the occupant of the vehicle. How many moles of nitrogen gas would need to be created in 40 milliseconds to inflate the airbag?
Slide 94 / 157
Combining the three mole formulas
We now have three formulas for finding the number of moles of a substance. Which one should be used depends on whether you are given (or trying to find):
the number of particles
the mass of a substance
the volume of a gas n = Vm
V Vm = 22.4 L at STP
n = Mm
n = NA
N NA = 6.02 x 1023
Slide 95 / 157
Combining the three mole formulas Sometimes two of these formulas must be used together to solve a problem.
Here's an example:
How many atoms of carbon are present in a 24 g sample of pure carbon?
n = Vm
V Vm = 22.4 L at STP
n = Mm
n = NA
N NA = 6.02 x 1023
Slide 96 / 157
Combining the three mole formulas How many atoms of carbon are present in a 24 g sample?
Step 1: find the number of moles of C present:
Step 2: find the number of atoms in 2.0 moles.
n = m = 24g =M 12g/mol
2.0 mol
n = NNA
N = n(NA)rearrange substitute:
2.0mol x 6.02x1023 atoms/mol = 1.2x1024 atoms
Slide 97 / 157
Mole Road Map
n
N
n= N
NA
NA=6.03*1023 particles/mole
(Particles: atoms, ions, molecules, etc.)
V
n= VVm
Vm=22.4 Liters/mole
(Volume)m (mass)
n= mM
M: molar mass found using the periodic table
Slide 98 / 157
52 How many molecules are there in 44.8 liters (at STP) of oxygen gas?
n = Vm
V Vm = 22.4 L
Slide 99 / 157
53 How many atoms are there in 11.2 liters (at STP) of molecular oxygen ( oxygen gas)?
n = Vm
V Vm = 22.4 L at STP
n = Mm
n = NA
N NA = 6.02 x 1023
Slide 100 / 157
54 What is the mass of 44.8 liters (at STP) of molecular oxygen?
n = Vm
V Vm = 22.4 L at STP
n = Mm
n = NA
N NA = 6.02 x 1023
Slide 101 / 157
55 What is the volume (at STP) of 3.0 x 1023 molecules of fluorine?
n = Vm
V Vm = 22.4 L at STP
n = Mm
n = NA
N NA = 6.02 x 1023
Slide 102 / 157
56 What is the volume (at STP) of 240 g of nitrogen gas?
n = Vm
V Vm = 22.4 L at STP
n = Mm
n = NA
N NA = 6.02 x 1023
Slide 103 / 157
57 How many atoms are present in 30 g of boron ?
n = Vm
V Vm = 22.4 L at STP
n = Mm
n = NA
N NA = 6.02 x 1023
Slide 104 / 157
58 What is the mass of a pure sample of lead which contains 3.0 x 1024 atoms?
n = Vm
V Vm = 22.4 L at STP
n = Mm
n = NA
N NA = 6.02 x 1023
Slide 105 / 157
59 How many moles are there in 11.2 liters (at STP) of chlorine?
n = Vm
V Vm = 22.4 L
Slide 106 / 157
60 What volume will 6.0x1024 molecules of oxygen gas occupy at STP?
n = Vm
V Vm = 22.4 L at STP
n = Mm
n = NA
N NA = 6.02 x 1023
Slide 107 / 157
61 How many molecules are in a 32g sample of SO2 ?
n = Vm
V Vm = 22.4 L at STP
n = Mm
n = NA
N NA = 6.02 x 1023
Slide 108 / 157
62 Determine the volume occupied by 216g of N2O5 at STP?
n = Vm
V Vm = 22.4 L at STP
n = Mm
n = NA
N NA = 6.02 x 1023
Slide 109 / 157
63 What volume will 5.0x1024 atoms of krypton occupy at STP?
n = Vm
V Vm = 22.4 L at STP
n = Mm
n = NA
N NA = 6.02 x 1023
Slide 110 / 157
64 What is the mass of 224 L of hexene (C6H12 ) at STP?
n = Vm
V Vm = 22.4 L at STP
n = Mm
n = NA
N NA = 6.02 x 1023
Slide 111 / 157
65 What is the volume occupied by 6.0x1023 atoms of Hydrogen at STP?
n = Vm
V Vm = 22.4 L at STP
n = Mm
n = NA
N NA = 6.02 x 1023
Slide 112 / 157
66 How many atoms of C are in a 30 milligram sample of calcium cyanide. (Hint: you must remember that milli means 1/1000 and you must first write the proper formula for calcium cyanide)
n = Vm
V Vm = 22.4 L at STP
n = Mm
n = NA
N NA = 6.02 x 1023
Slide 113 / 15767 What is the mass of 44.8L of Argon gas at STP?
n = Vm
V Vm = 22.4 L at STP
n = Mm
n = NA
N NA = 6.02 x 1023
Slide 114 / 157
Return toTable ofContents
Percent Composition
Slide 115 / 157
Identify an Unknown Substance
We have been able to calculate the molar mass of a material if we know its formula already. However, what if we encounter an unidentified substance?
Using the tools we already have, we are able to determine the composition of a substance if we have certain pieces of information.
Slide 116 / 157
Percent Composition
Potassium chromate, K2 CrO4 Potassium dichromate, K2 Cr2 O7
One such piece of information is the percent composition, or the percentage of a compound's mass, made up by its various elements.
Slide 117 / 157
% element =(number of atoms)(atomic weight)
(FW of the compound)x 100
Percent Composition
The percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of
the compound, multiplied by 100%.
Slide 118 / 157
= 80.0%%C = (2)(12.0 amu)30.0 amu
x 100
%H = (6)(1.0 amu) 30.0 amu
= 20.0%x 100
The percent composition of carbon and hydrogen in ethane, (C2H6) is…
Percent Composition of Ethane
Since ethane is made of only Carbon and Hydrogen, if the percent composition of Carbon was 80%, the remaining percent must be Hydrogen. Another method to calculate the %H is:
%H = 100% - %C %H = 100% - 80% %H = 20%
Slide 119 / 157
%H = 20% and %C = 80%
The percent composition of ethane, (C2H6) is…
Note that even though there are more ATOMS of hydrogen in ethane, there is a much less percentage of mass in the compound than that of carbon.
This is because 1 carbon atom (12 amu) is much more massive than 1 hydrogen atom (1 amu).
Therefore, carbon accounts for a much greater percentage of the mass of ethane than hydrogen does.
Percent Composition of Ethane
Slide 120 / 157
68 In water (H2 O), which element do you think accounts for more mass?
A Hydrogen
B Oxygen
C Hydrogen and Oxygen account for the same percent of mass
D Carbon
Slide 121 / 157
69 What is the mass percentage of oxygen in water?
Slide 122 / 157
70 What is the percent by mass of carbon in acetone, C3 H6 O?
Slide 123 / 157
71 What is the percent of Ba in Ba(NO3)2
Slide 124 / 157
72 Iron (II) oxide has a smaller % of iron by mass than iron (III) oxide.
TrueFalse
Slide 125 / 157
Real World ApplicationAluminum is used in soda cans, aircraft frames, and automobile engines. Worldwide aluminum demand is increasing. Aluminum is isolated from an ore called bauxite which is roughly 12% aluminum by mass.
How much bauxite ore (in grams) must be mined to provide enough aluminum to make a car engine requiring 100 kg of aluminum?
bauxite ore
Al block engine
100 kg x (1000 g/kg) x (100 g of ore/12 gram of Al) = 830,000 g Almove for answer
Slide 126 / 157
73 A 2.00 sample of a compound containing only potassium and oxygen is heated. The oxygen gas leaves and the resulting mass of the potassium is 1.66 grams. What is the % by mass of oxygen in the compound? A 17%B 70%C 30%D 45%
E Cannot be determined from the information
Slide 127 / 157
74 What is the % by mass of water in the CaSO4*2H2O crystal?
A 10%B 21%C 19%D 50%E 75%
Slide 128 / 157
Return toTable ofContents
Empirical Formula
Slide 129 / 157
Calculating Empirical Formulas
Now that we are able to calculate the percent of a compound by mass of each element, we can begin to identify unknown substances. First, we must identify the ratio of the number of moles of each element in the substance
This formula, based on whole-number ratio, is called an empirical formula.
Slide 130 / 157
Calculating Empirical Formulas
Empirical formula: The formula that indicates the molar ratio of elements present in a molecular compound reduced to the least common denominator.
For instance, the empirical formula for benzene (C6H6) is CH.
Mass %elements
Grams of each element
Moles ofeach element
Empiricalformula
Assume 100g sample
use molar mass
Calculatemole ratio
Slide 131 / 157
Calculating Empirical FormulasEmpirical formula: There is only one empirical formula for a substance, but two different substances can have the same empirical formula.
That is because an empirical formula is always based on the ratio of the elements given in their lowest common denominator.
Example:
Substance
Hydrogen peroxide
Hydroxide
Molecular Formula
Emperical Formula
H2O2
OH
OH
OH
Slide 132 / 157
75 The empirical formula for C6 H4 (NO3 )2 is C3 H2 N2 O3
True
False
Slide 133 / 157
76 Which of the following has an empirical formula that is the same as its molecular formula? (1) NH4 Cl (2) (NH4 )2 CO3 (3) CH2 Cl2 (4) CHCl2 Br
A 1 and 3 onlyB 1, 2 and 4 onlyC 1 and 4 only
D all of them
Slide 134 / 157
Calculating Empirical Formulas
The compound para-aminobenzoic acid (listed as PABA in bottles of sunscreen) is composed of the following
elements (by mass):
carbon (61.31%),
hydrogen (5.14%), nitrogen (10.21%),
and oxygen (23.33%).
Find the empirical formula of PABA.
Mass %elements
Slide 135 / 157
Step 1: convert the mass percentages to mole amounts in a hypothetical 100 g sample. Assume a 100.00 g amount of PABA to represent the mass of each element:
C: 61.31% x 100g = 61.31 g
H: 5.14 % x 100g = 5.14 g
N: 10.21% x 100g = 10.21 g
O: 23.33 % x 100g = 23.33 g
Mass %elements
Grams of each element
Assume 100g sample
Calculating Empirical Formulas
Slide 136 / 157
Step 2: Now that we know the representative mass, we can use the molar mass of each element to calculate the number of moles that would be present.
C: 61.31 g x = 5.105 mol
H: 5.14 g x = 5.09 mol
N: 10.21 g x = 0.7288 mol
O: 23.33 g x = 1.456 mol
1 mol 12.01 g
1 mol14.01 g
1 mol1.01 g
1 mol16.00 g
Calculating Empirical Formulas
Mass %elements
Grams of each element
Moles ofeach element
Assume 100g sample
use molar mass
Slide 137 / 157
Calculating Empirical Formulas
Step 3: Calculate the mole ratio by dividing each mole value by the smallest number of moles. In this case, by that of nitrogen. Divide them all by 0.7288.
C: 5.105 mol
H: 5.09 mol
N: 0.7288 mol
O: 1.456 mol
0.7288 mol
0.7288 mol
0.7288 mol
5.105 mol
5.09 mol
0.7288 mol
1.456 mol
= 7.005 # 7
= 6.984 # 7
= 1.00 = 1
= 2.001 # 20.7288 mol
Mass %elements
Grams of each element
Moles ofeach element
Empiricalformula
Assume 100g sample
use molar mass
Calculatemole ratio
The empirical formula for PABA is C7H7NO2
Slide 138 / 157
C: 62.1 g x = 5.17 mol = 5.17 mol = 3.01 3
H: 13.8 g x = 13.8 mol = 13.8 mol = 7.96 8
N: 24.1 g x = 1.72 mol = 1.72 mol = 1.00 = 1
1 mol14.01 g
1 mol12 .0 g
1 mol1.0 g
1.72 mol
1.72 mol
1.72 mol
#
#
The empirical formula is C3H8N
1,6 - diaminohexane is used in making nylon.
It is 62.1% C, 13.8% H and 24.1% N. What is the empirical formula?
Calculating Empirical Formulas
Slide 139 / 157
77 What is the empirical formula for a compound with the following percent composition?
A C H4
B C2 H5
C C3 H6
D C3 H8
C: 74.9 %H: 25.1 %
Slide 140 / 157
78 What is the empirical formula for a compound with the following percent composition?
A C2 H O
B C H2 O
C C H O2
D C2 H3 O2
C: 40.0 %H: 6.7 % O: 53.0 %
Slide 141 / 157
The molar ratio works very well for most compounds. However, in some cases the math leaves us without whole numbers:
For Example: iron (?) oxide
Fe: 69.92% -> 69.92g = 1.25 mol = 1
O: 30.08% -> 30.08g = 1.88 mol = 1.5
Mass %elements
Grams of each element
Moles ofeach element
Assume 100g sample
use molar mass
Calculatemole ratio
55.85 g/mol
16.00g/mol
1.25 mol
1.25 mol
FeO1.5
Special Cases
Slide 142 / 157
Special CasesIn these cases, it is necessary to multiply the results by an
integer to ensure the ratio is a whole number ratio...
The empirical formula is Fe2O3
x 2 = 2
x 2 = 3
The name of the compound is iron (III) oxide
Slide 143 / 157
79 What is the empirical formula for a compound with the following percent composition?
A V O1.67
B V2 O5
C V3 O5
D V3 O8
V: 56.02 % O: 43.98 %
Slide 144 / 157
80 A 4.68 gram sample of a sulfur oxide is heated releasing oxygen gas and leaving behind solid sulfur. If the mass of the sulfur left behind was 2.34 grams, what must be the empirical formula of the compound?A SOB SO2
C S2OD S2O3
E SO3
Slide 145 / 157
Hydrated Crystals
Many ionic compounds have a fixed quantity of water molecules bound within their structure.
The mole ratio of water to the dry CuSO4 crystal is 5:1
CuSO4 * 5H2O
Slide 146 / 157
Finding the formula of a hydrate
One can find the formula of a hydrate by heating the crystal to remove the water. Then find the mole ratio between the dry crystal and the water just as you would when finding an empirical formula.
g CuSO4 --> n CuSO4
CuSO4 * ? H2O
g H2O --> n H2O
HEAT Find mole ratio!
Slide 147 / 157
Finding the formula of a hydrate
After heating, a hydrate of MgSO4 was found to be roughly 51.3% water by mass. What is the formula of the hydrate?
Step 1: Find the % of the crystal and water and express in grams.
100 - 51.3% water = 48.7 % MgSO4 =
48.7 g MgSO4 & 51.3 g of water
Step 2: Convert to moles
48.7 g MgSO4 = 0.405 mol MgSO4 51.3 g H2O = 2.85 mol H2O
120 g/mol 18 g/mol
Slide 148 / 157
Step 3: Find the mole ratio of the water to the dry crystal
0.405 mol MgSO4 = 1 2.85 mol H2O = 7
0.405 mol 0.405 mol
.....the formula is MgSO4*7H2O
Finding the formula of a hydrate
Slide 149 / 157
81 When a 2.4 g sample of a hydrated crystal of BaCl2 is heated, the dry anhydrous crystal has a mass of 2.08 grams after heating. What is the formula of the hydrate?A BaCl2*H2OB BaCl2*2H2OC BaCl2*3H2OD BaCl2*4H2OE BaCl2*5H2O
Slide 150 / 157
A molecular formula indicates the number of atoms of each element present in the molecule.
The molecular formula of a compound is either the same as its experimentally determined empirical formula, or it is a simple whole-number multiple of its empirical formula
Molecular Formulas
Slide 151 / 157
Molecular FormulasFormula Name Type of Formula Molar MassCH
C2H2 Ethyne
C6H6 Benzene
CH2O Methanal
C2H4O2 Ethanoic Acid
C6H12O6 Glucose
Empirical
Molecular
Molecular
Empirical & Molecular
Molecular
Molecular
13
(13x2 =) 26
(13x6 =) 78
30
(30x2 =) 60
(30x6 =) 180
Slide 152 / 157
To find the actual molecular formula you need one other piece of information...the molecular weight (mass) of the molecule.
If you know the ratio of the elements in a molecule
and you know the total mass of the molecule
then you can determine the molecular formula.
Molecular Formulas
Slide 153 / 157
To find the molecular formula:
Step 1: Determine the molar mass of the empirical formula.
Step 2: Divide the molecular mass by the empirical mass
Step 3: Multiply the Empirical Formula by the resulting integer
Molecular FormulasWhat is the molecular formula of a compound with an empirical formula of CH2 O that has a mass of 180 g/mol?
Molar mass of CH2O = 30 g/mol
Molecular mass = 180 g/mol = 6Empirical mass 30 g/mol
6 x CH2O = C6H12O6
Slide 154 / 157
82 The molecular mass of Benzene is 78. If the empirical formula of benzene is CH, what is its molecular formula?
A C2 H2
B CH
C C6 H6
D C2 H4
Slide 155 / 157
Maleic acid is an organic compound composed of 41.39 % carbon, 3.47% hydrogen, and the rest is oxygen. It has a molecular mass of 116 g/mole.
Molecular Formulas
Determine the empirical formula for Maleic acid.
Then determine the molecular formula for Maleic acid.
Slide 156 / 157
Real World Application
Clenbuterol is a steroid drug that is illegally used in cattle and in sports like cycling to help the cattle or athlete lose fat and gain lean muscle mass. It can be detected by mass spectroscopy.
What is the empirical and molecular formula for clenbuterol if when a 10.0 g sample is combusted in air, it is found it contains 4.60 g of carbon, 0.613 g of H, 0.51 g of oxygen with the rest being chlorine? The molecular weight is 313 g/mol.
Slide 157 / 157