section 3.4—counting molecules
DESCRIPTION
Section 3.4—Counting Molecules. So the number of molecules affects pressure of an airbag…how do we “ count ” molecules?. What is a Mole? Ted Ed video. http://ed.ted.com/lessons/daniel-dulek-how-big-is-a-mole-not-the-animal-the-other-one. What is a mole?. Mole – metric unit for counting. - PowerPoint PPT PresentationTRANSCRIPT
Section 3.4—Counting Molecules
So the number of molecules affects pressure of an airbag…how do we “count” molecules?
What is a Mole? Ted Ed video
http://ed.ted.com/lessons/daniel-dulek-how-big-is-a-mole-not-the-animal-the-other-one
What is a mole?
Mole – metric unit for counting
The only acceptable abbreviation for “mole” is “mol”…not “m”!!
We use it just like we use the terms dozen and ream!
What is a counting unit?You’re already familiar with one counting unit…a “dozen”
“Dozen” 12
A dozen doughnuts 12 doughnuts
A dozen books
A dozen cars
12 books
12 cars
A dozen = 12
What can’t we count atoms in “dozens”?
Atoms and molecules are extremely small
We use the MOLE to count particles
A mole = 6.02 1023 particles(called Avogadro’s number)
“mole” 6.02 1023
1 mole of doughnuts 6.02 1023 doughnuts
1 mole of atoms
1 mole of molecules
6.02 1023 atoms
6.02 1023 molecules
6.02 1023 = 602,000,000,000,000,000,000,000
This number was named after Amadeo Avogadro. He did not calculate it!
FUNNY!
Representative ParticlesRemember, matter is broken down into either SUBSTANCES or mixturesSubstances are broken down into either ELEMENTS or COMPOUNDS
Type of Matter Example Representative Particle
Element Fe Atom
Ionic Compound NaCl Formula Unit
Covalent Compound CO2 Molecule
Example: Particles & MolesUse the conversion factor (1 mol = 6.02 x 1023)
particles to convert
Example 1:
How many molecules of water
are in 1.25 moles?
= _______ molecules H2O
Example: Molecules & Moles
1.25 mol H2O
mol H2O
Molecules H2O
6.02 1023
17.531023
1 mol = 6.021023 moleculesExample 1:
How many molecules of water
are in 1.25 moles?
Let’s Practice #2
Example:How many moles are equal to 2.8 × 1022 formula units of KBr?
= _______ moles
Let’s Practice #2
2.8 × 1022 formula units
Formula units
mole1
6.02 1023
0.047
1 mol = 6.021023 formula unitsExample:How many moles are equal to 2.8 × 1022 formula
units KBr?
Let’s Practice #3
Example:How many atoms are equal to
3.56 moles of
Fe?
= _______ atoms
Let’s Practice #3
3.56 moles Fe
moles
atoms6.02 x 10 23
12.14 x 1024
1 mol = 6.021023 moleculesExample:How many atoms are equal to
3.56 moles of
Fe?
Molar Mass
Molar Mass – The mass for one mole of an atom or molecule.
Other terms commonly used for the same meaning:Molecular WeightMolecular MassFormula WeightFormula Mass
Molar Mass for Elements
The average atomic mass = grams for 1 mole
Element Mass
1 mole of carbon atoms (C) 12.01 g
1 mole of oxygen atoms (O2)
1 mole of hydrogen atoms (H2)
16.00 g x 2 = 32.00 g O2
1.01 g x 2 = 2.02 g H2
Unit for molar mass: g/mole or g/mol
Average atomic mass is found on the periodic table
Molar Mass for Compounds
The molar mass for a molecule = the sum of the molar masses of all the atoms
Calculating a Molecule’s Mass
Count the number of each type of atom
Find the molar mass of each atom on the periodic table
Multiply the # of atoms by the molar mass for each atomFind the sum of all the masses
1
2
3
4
To find the molar mass of a molecule:
Example: Molar Mass
Example:Find the
molar mass for
CaBr2
Example: Molar Mass
Count the number of each type of atom1
Ca
Br
1
2
Example:Find the
molar mass for
CaBr2
Example: Molar Mass
Find the molar mass of each atom on the periodic table2
Ca
Br
1
2
40.08 g/mole
79.90 g/mole
Example:Find the
molar mass for
CaBr2
Example: Molar Mass
Multiple the # of atoms molar mass for each atom3
Ca
Br
1
2
40.08 g/mole
79.90 g/mole
Example:Find the
molar mass for
CaBr2
= 40.08 g/mole
= 159.80 g/mole
Example: Molar Mass
Find the sum of all the masses4
Ca
Br
1
2
40.08 g/mole
79.91 g/mole
= 40.08 g/mole
= 159.80 g/mole+
199.88 g/mole
1 mole of CaBr2 =199.90 g
Example:Find the
molar mass for
CaBr2
Example 2: If you see a Parentheses in the Formula
Be sure to distribute the subscript outside the parenthesis to each element inside the parenthesis.
Example:Find the
molar mass for Sr(NO3)2
Example 2: Molar Mass & Parenthesis
Be sure to distribute the subscript outside the parenthesis to each element inside the parenthesis.
1
6
87.62 g/mole
16.00 g/mole
= 87.62 g/mole
= 96.00 g/mole+
211.64 g/mole
1 mole of Sr(NO3)2 =211.64 g
2 14.01 g/mole = 28.02 g/mole
Sr
N
O
Example:Find the
molar mass for Sr(NO3)2
Let’s Practice #3
Example:Find the
molar mass for Al(OH)3
Let’s Practice #2
Be sure to distribute the subscript outside the parenthesis to each element inside the parenthesis.
1
3
26.98 g/mole
1.01 g/mole
= 26.98 g/mole
= 3.03 g/mole+
78.01 g/mole
1 mole of Al(OH)3 =78.01 g
3 16.00 g/mole = 48.00 g/mole
Al
O
H
Example:Find the
molar mass for Al(OH)3
Using Molar Mass in Conversions
Example: Moles to Grams
Example:How many grams are
in 1.25 moles of water?
Example: Moles to Grams
1.25 mol H2O = _______ g H2Omol H2O
g H2O18.02
122.5
When converting between grams and moles, the molar mass is needed
1 mole H2O molecules = 18.02 g
HO
21
1.01 g/mole16.00 g/mole
= 2.02 g/mole= 16.00 g/mole+
18.02 g/mole
Example:How many grams are
in 1.25 moles of water?
Example: Grams to Moles
Example:How many moles are
in 25.5 g NaCl?
25.5 g NaCl
Example: Grams to Moles
= ____ moles NaCl
g NaCl
mol NaCl1
58.44
.436
1 moles NaCl molecules = 58.44 g
NaCl
11
22.99 g/mole35.45 g/mole
= 22.99 g/mole= 35.45 g/mole+
58.44 g/mole
Example:How many moles are
in 25.5 g NaCl
Example: Grams to Molecules
Example:How many
formula units are
in 25.5 g NaCl?
25.5 g NaCl
Example: Grams to Moles
= ____ FU’s NaCl
g NaCl
mol NaCl1
58.44
2.63 x 1023
1 moles NaCl formula units = 58.44 g
NaCl
11
22.99 g/mole35.45 g/mole
= 22.99 g/mole= 35.45 g/mole+
58.44 g/mole
Example:How many
formula units are
in 25.5 g NaCl
1 mol NaCl
6.02 x 1023 FU’s
Percent CompositionDefined as the percent by mass of each element in a
compound
Steps to Finding Percent Composition1.Add up the mass of each element within the compound to
get the mass of the compound.2.Divide each element’s mass by the mass of the compound.3.Multiply by 100
% composition= mass of element x 100 mass of compound
Percent Composition by Mass of Air
Example: Calculate the % composition of each element in calcium carbonate.
CaCO3Molar mass = 100.09 g
% C = 12.01/100.09 x 100 = 12.00 %
%Ca = 40.08/100.09 x 100 = 40.04%
%O = 48.00/100.09 x 100 = 47.96%
Example: What is the % of each element in a compound that contains 29.00g Ag and 4.30g S only?
Total mass of compound = 33.30 g
% Ag = 29.00/33.30 x 100 = 87.09 %
%S = 4.30/33.30 x 100 = 12.9%
Hydrates
A HYDRATE is an ionic compound with water trapped in its crystal.
Examples are: CuSO4 5H2O MgSO4 7 H2O CoCl2 H2O
Heating a hydrate removes the water and leaves behind just the salt which is called the anhydrate.
Example: What is the % water in the hydrate, CuCl2 2H2O
Molar mass of hydrate = 170.48 g
% water = 36.04/170.48 x 100 = 21.14%
http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/flashfiles/stoichiometry/empirical.html
Heating of A Hydrate Animation
Calculating the experimental % composition of water in a hydrate.
Empirical FormulaA chemical formula showing the simplest whole number ratio of moles of elements (subscipts)
Hydrogen Peroxide has an actual formula (molecular formula) of H2O2 but an empirical formula of HO
How to Calculate Empirical Formula RHYME: Percent to Mass
Mass to Mole Divide by Small Multiply til Whole
1.Assume 100 grams of the sample of compound. Switch the percent sign to grams2.Convert each element’s mass into moles.3.Divide each element’s mole amount by the smallest mole amount in the entire problem. The answer is the subscript of the element within the compound.4.OPTIONAL: If mole ratio is not within .1 of a whole number, multiply each amount by the smallest whole number that will produce either a whole number itself or a number within .1 of a whole number.
Example: What is the empirical formula for 40.05% S and 59.95% O?
1. Switch the percent sign to grams & convert each element’s mass into moles
40.05 g S / 32.01g = 1.250 mol S
59.95 g O / 16.00 g = 3.747 mol O
2. Divide each element’s mole amount by the smallest mole amount in the entire problem.
1.250 mol S = 1 3.747 mol O = 2.99 = 31.250 mole 1.250 mol
S1O3 SO3
Example: What is the empirical formula for 43.64% P and 56.36% O?1. Switch the percent sign to grams & convert each element’s mass
into moles
43.64 g P / 30.97g = 1.409 mol S
56.36 g O / 16.00 g = 3.522 mol O
2. Divide each element’s mole amount by the smallest mole amount in the entire problem.
1.409 mol S = 1 3.522 mol O = 2.49 ≠ 31.409 mole 1.409 mol
3. If mole ratio is not within .1 of a whole number, multiply each amount by the smallest whole number that will produce either a whole number itself or a number within .1 of a whole number.
1 x 2 = 2 2.49 x 2 = 4.998 = 5
P2O5
Molecular Formula
Is the ACTUAL, true formula of the compound.They are usually multiples of their empirical formulaN2O4 is the molecular formula; the empirical formula is NO2
Notice that the molecular formula is 2 times larger than the empirical formula
Molecular Formula
How to Calculate Molecular Formula
1. You need to find the empirical formula and calculate its molar mass. Call this empirical formula mass EFM.2. Find the mass of the actual formula which will most likely be given to you in grams. Call this molecular formula mass MFM.3. Divide the MFM by the EFM to get a factor. 4. Multiply the factor by the empirical formula to get the MOLECULAR FORMULA
Example: What is the molecular formula of a compound whose empirical formula is CH4N and the molecular mass is 60.12 g/mol?
1. Empirical Formula Mass (EFM) = 12.01 + 4.04 + 14.01 = 30.06 g
2. Molecular Formula Mass (MFM) = 60.12 g
3. 60.12 / 30.06 = 2
4. 2(CH4N) = C2H8N2
Section 3.5—Gas Behavior How does the behavior of gases affect airbags?
What is PRESSURE?
Force of gas particles running into a surface.
Pressure is measured by a Barometer
If pressure is molecular collisions with the container…
As # of moles increase, pressure increases Think about blowing up a balloon!
Pressure and Moles (# of Molecules)
As number of molecules increases, there will be more molecules to collide with the wall
•Collisions between molecules and the wall increase
Pressure increases
# of Gas Particles vs. Pressure
Pressure & Volume
If pressure is molecular collisions with the container…
As volume increases, pressure decreases. Think about how your lungs work! http://www.youtube.com/watch?v=q6-oyxnkZC0
As volume increases, molecules can travel farther before hitting the wall•Collisions between molecules & the wall decrease
Pressure decreases
What is “Temperature”?
Temperature – measure of the average kinetic energy of the molecules
Energy due to motion(Related to how fast the molecules are moving)
As temperature increases, Average Kinetic Energy Increases and Molecular motion increases
Pressure and Temperature
If temperature is related to molecular motion…and pressure is molecular collisions with the container…
As temperature increases, pressure increases
As temperature increases, molecular motion increases
•Collisions between molecules & the wall increase
Pressure increases
Volume and Temperature
If temperature is related to molecular motion…and volume is the amount of space the gas occupies…
As temperature increases, volume increases Think of liquid nitrogen and the balloon. http://www.youtube.com/watch?v=QEpxrGWep4E
As temperature increases, molecular motion increases•molecules will move farther away from each otherVolume increases
Pressure In Versus Out
Example: A bag of chips is bagged at sea level. What happens if the bag is then brought up to the top of a mountain.
A container will expand or contract until the pressure inside equals atmospheric pressure outside
The internal pressure is higher than the external pressure.
The bag will expand in order to reduce the internal pressure.
The internal pressure of the bag at low altitude is high
At high altitude there is lower pressure
Higher pressure
Lowerpressure
Lower pressure
CanExplodes!
When Expansion Isn’t Possible
Example: An aerosol can is left in a car trunk in the summer. What happens?
Rigid containers cannot expand
The internal pressure is higher than the external pressure.
The can is rigid—it cannot expand, it explodes!
The temperature inside the can begins to rise.
As temperature increases, pressure increases.
Higher pressure
Lowerpressure
Air Pressure Crushing Cans
http://www.csun.edu/scied/4-discrpeant-event/the_can_crush/index.htm
http://www.youtube.com/watch?v=Zz95_VvTxZM
Another cool videohttp://www.youtube.com/watch?v=JsoE4F2Pb20
Air Pressure Crushing “Cans”
Kinetic Molecular Theory(KMT): explains gas behavior based upon
the motion of molecules based on an ideal gas
IDEAL gases are IMAGINARY gases that follow the assumptions of the KMT
1
Assumptions of the KMT
All gases are made of atoms or molecules that are in constant, rapid, random motion
Gas particles are not attracted nor repelled from one another ***All gas particle collisions are perfectly elastic (no kinetic energy is lost to other forms)
The volume of gas particles is so small compared to the space between the particles, that the volume of the particle itself is insignificant***
2
3
4
5
The temperature of a gas is proportional to the average kinetic energy of the particles
So what is a “REAL” gas?Real gases, (like nitrogen), will eventually condense into a liquid when the temperature gets too low or the pressure gets too high BECAUSE:
Assumption #3
Assumption #5
Gas particles do have attractions and repulsions towards one another
Gas particles do take up space
Real Gases Deviate from Ideal Gas Behavior when at high pressure
The gas molecules are compressed making the volume they take up more significant than if they were spread out
Real Gases Deviate from Ideal Gas Behavior when at low temperature.
The lower kinetic energy causes the molecules to move slower and ATTRACTIVE FORCES that really exist start to take effect---------------------------
Polar gases (HCl) deviate more than nonpolar gases (He or H2
At Lower Temperature
Gas Movement: Effusion vs Diffusion
Effusion –gas escapes from a tiny hole in the container
Effusion is why balloons deflate over time!
Diffusion –gas moves across a space from high to low concentration
Diffusion is the reason we can smell perfume across the room
Effusion, Diffusion & Particle Mass
How are particle size (mass) and these concepts related?
As mass of the particles increases, rate of effusion and diffusion is lowered.
As particle size (mass) increases, the particles move slowerit takes them more time to find the hole or to go across the room
Rate of Diffusion & Particle Mass
Watch as larger particles take longer to get to your nose
H2
CO2
Section 3.6—Gas Laws
How can we calculate Pressure, Volume and Temperature of our airbag?
Pressure Units
Several units are used when describing pressure
Unit Symbol
atmospheres atm
Pascals, kiloPascals
millimeters of mercury
pounds per square inch
Pa, kPa
mm Hg
psi
1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi
Conversions Between Different Pressure Units
1 atm = 760 mmHg = 101.3 kPa
Examples
1.Convert 654 mm Hg to atm
1.Convert 879 mm Hg to kPa
1.Convert 15.6 atm to kPa
654 mmHg x 1atm = 760 mmHg
.861 atm
879 mmHg x 101.3 Kpa = 760 mmHg
1.16 Kpa
15.6 atm x 101.3 Kpa = 1atm
1580 Kpa
Temperature Unit used in Gas Laws
Kelvin (K)– temperature scale with an absolute zero
Temperatures cannot fall below an absolute zero
KC 273
Examples1.Convert 15.6 °C into K
2. Convert 234 K into °C
15.6 + 273 = K 288.6 289 K
°C + 273 = 234 -39 °C
Standard Temperature & Pressure (STP)
the conditions of:1 atm (or the equivalent in another unit) 0°C (273 K)
Problems often use “STP” to indicate quantities…don’t forget this “hidden” information when making your list!
GAS LAWS: “Before” and “After”
This section has 5 gas laws which have “before” and “after” conditions.
For example:
2
2
1
1
n
P
n
P
1= initial amount2= final amount
P= PressureV= VolumeT=Temperaturen= moles(molecules)
Boyle’s Law
Pressure Increases as Volume Decreases
Boyles’ Law
Volume & Presssure are INVERSELY proportional when temperature and moles are held constant
P = pressureV = volume2211 VPVP
The two pressure units must match and the two volume units must match!
Example: A gas sample is 1.05 atm when at 2.5 L. What volume is it if the pressure is changed to 0.980 atm?
Boyles’ Law 2211 VPVP ***The two pressure units must match & the two volume units must match!
Example: A gas sample is 1.05 atm when 2.5 L. What volume is it if the pressure is changed to 0.980 atm?
P1 = 1.05 atm
V1 = 2.5 L
P2 = 0.980 atm
V2 = ? L
V2 = 2.7 L
2980.05.205.1 VatmLatm
2980.0
5.205.1V
atm
Latm
Boyles Law: Graph
Charles’ Law
Volume Increases as Temperature Increases
Charles’ LawVolume & Temperature are DIRECTLY proportional when pressure and moles are held constant.
V = VolumeT = Temperature
2
2
1
1
T
V
T
V
The two volume units must match & temperature must be in Kelvin!
Example: What is the final volume if a 10.5 L sample of gas is changed from 25C to 50C?
V1 = 10.5 L
T1 = 25C
V2 = ? L
T2 = 50C
Temperature needs to be in Kelvin!
25C + 273 = 298 K
50C + 273 = 323 K
Charles’ Law2
2
1
1
T
V
T
V
***The two volume units must match & temperature must be in Kelvin!
Example: What is the final volume if a 10.5 L sample of gas is changed from 25C to 50C?
V1 = 10.5 L
T1 = 25C
V2 = ? L
T2 = 50C
V2 = 11.4 L
= 298 K
= 323 K
K
V
K
L
323298
5.10 2
2298
5.10323V
K
LK
Charles Law: Graph
Gay-Lussac’s Law
Temperature decreases as Pressure decreases
Gay-Lussac’s LawPressure & temperature are DIRECTLY proportional when moles and volume are held constant
P = PressureT = Temperature
The two pressure units must match and temperature must be in Kelvin!
Example: A sample of hydrogen gas at 47C exerts a pressure of .329 atm. The gas is heated to 77C at constant volume and moles. What will the new pressure be?
P1 = .329 atm
T1 = 47C
P2 = ? atm
T2 = 77C
Temperature needs to be in Kelvin!
47C + 273 = 320 K
77C + 273 = 350 K
Gay-Lussac’ Law
Example: A sample of hydrogen gas at 47C exerts a pressure of .329 atm. The gas is heated to 77C at constant volume and moles. What will the new pressure be?
P1 = .329 atm
T1 = 47C
P2 = ? atm
T2 = 77C
P2 = .360 atm
= 320 K
= 350 K
Gay Lussac Law: Graph
Avogadro’s Law
Moles and Volume are directly proportional when temp. & pressure are held constant
V = Volumen = # of moles of gas
2
2
1
1
n
V
n
V
Example: A sample with 0.15 moles of gas has a volume of 2.5 L. What is the volume if the sample is increased to 0.55 moles?
The two volume units must match!
Avogadro’s Law2
2
1
1
n
V
n
V
Example: A sample with 0.15 moles of gas has a volume of 2.5 L. What is the volume if the sample is increased to 0.55 moles?
The two volume units must match!
n1 = 0.15 moles
V1 = 2.5 L
n2 = 0.55 moles
V2 = ? L
mole
V
mole
L
55.015.0
5.2 2
215.0
5.255.0V
mole
Lmole
V2 = 9.2 L
Combined Gas Law
P = PressureV = Volumen = # of molesT = Temperature22
22
11
11
Tn
VP
Tn
VP
Each “pair” of units must match and temperature must be in Kelvin!
Example: What is the final volume if a 15.5 L sample of gas at 755 mmHg and 298 K is changed to STP?
Combined Gas Law
P = PressureV = VolumeT = Temperature
Moles is not mentioned so remove it from equation!
Example: What is the final volume if a 15.5 L sample of gas at 755 mmHg and 298K is changed to STP?
P1 = 755 mmHg
V1 = 15.5 L
T1 = 298 K
P2 = 760mmHg
V2 = ? L
T2 = 273 K V2 = 14.1 L
STP is standard temperature (273 K) and pressure (1 atm)
22
22
11
11
Tn
VP
Tn
VP
22
12
11
11
Tn
VP
Tn
VP
The combined gas law can be used for all “before” and “after” gas law problems!
For example, if volume is held constant, then
and the combined gas law becomes:
21 VV
22
2
11
1
Tn
P
Tn
P
Why you really only need 1 of these
Watch as variables are held constant and the combined gas law “becomes” the other 3 laws
22
22
11
11
Tn
VP
Tn
VPHold pressure and
temperature constantAvogadro’s Law
22
22
11
11
Tn
VP
Tn
VPHold moles and
temperature constantBoyles’ Law
22
22
11
11
Tn
VP
Tn
VPHold pressure and
moles constantCharles’ Law
Transforming the Combined Law
Dalton’s Law
Dalton’s Law
Each gas in a mixture exerts its own pressure called a partial pressure = P1, P2….
Total Pressure = PT
Example: A gas mixture is made up of oxygen(2.3 atm) and nitrogen(1.7 atm) gases. What is the total pressure?
....321 PPPPT
PT = 2.3 atm + 1.7 atm
4.0 atm
2KClO3 (s) 2KCl (s) + 3O2 (g)
PT = PO + PH O2 2
Modified Dalton’s Law
When a gas is Collected over water, the total pressure of the mixture collectedis a combination of water vapor and the gas you are collecting!
Modified Dalton’s Law
Example: What is the pressure of the water vapor if the total pressure of the flask is 17.5 atm and the pressure of the oxygen gas is 16.1 atm?
watergas PPPT
17.5 = 16.1 atm + PH2O
1.4 atm
The Ideal Gas Law (an“AT NOW”equation)
The volume of a gas varies directly with the number of moles and its Kelvin temperature
P = PressureV = Volume n = molesR = Gas Law ConstantT = Temperature
nRTPV
There are three possibilities for “R”!
Kmole
atmL
*
*0821.0
Choose the one with units that match your pressure units!
Volume must be in Liters when using “R” to allow the unit to cancel!
The Ideal Gas Law nRTPV Example: A sample with 0.55 moles of gas is at 105.7
kPa and 27°C. What volume does it occupy?
The Ideal Gas Law
Example: A sample with 0.55 moles of gas is at 105.7 kPa and 27°C. What volume does it occupy?
n = 0.55 moles
P = 105.7 kPa
T = 27°C + 273 = 300 K
V = ?
R = 8.31 L kPa / mole K
)300(**31.8)55.0()7.105( KkmolekPaLmoleVkPa
V2 = 13 L
The Ideal Gas Law does not compare situations—it describes a gas in one situation.
)7.105(
)300(**31.8)55.0(
kPa
KkmolekPaLmole
V
nRTPV
Chosen to match the kPa in the “P” above
The Ideal Gas LawExample 2:
What mass of hydrogen gas in grams is contained in a 10.0 L tank at 27°C and 3.50 atm of pressure?
n = ?
P = 3.50 atm
T = 27°C + 273 = 300 K
V = 10.0 L
R = .0821 L atm /mole K
)300(**0821.)0.10()50.3( KkmoleatmLnLatm
)*
*0821(.)300(
0.10)50.3(
kmoleatmLxK
Latmn
nRTPV
Chosen to match the atm in the “P” above
n = 1.42 mol 1.42 mol x 2.02 g = 2.87 g 1 mol