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 We use it just like we use the terms dozen and ream! The only acceptable abbreviation for “mole” is “mol”…not “m”!!

What is a counting unit? You’re already familiar with one counting unit…a “dozen” A dozen = 12 “Dozen” 12 A dozen doughnuts 12 doughnuts A dozen books 12 books A dozen cars 12 cars

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) 6.02  1023 = 602,000,000,000,000,000,000,000 “mole” 6.02  1023 1 mole of doughnuts 6.02  1023 doughnuts 1 mole of atoms 6.02  1023 atoms 1 mole of molecules 6.02  1023 molecules This number was named after Amadeo Avogadro. He did not calculate it!

FUNNY!

Representative Particles Remember, matter is broken down into either SUBSTANCES or mixtures Substances 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

How many molecules of water are in 1.25 moles? Example: Particles & Moles Use the conversion factor (1 mol = 6.02 x 1023) particles to convert Example 1: How many molecules of water are in 1.25 moles?

Example: Molecules & Moles How many molecules of water are in 1.25 moles? 1 mol = 6.021023 molecules 1.25 mol H2O Molecules H2O 6.02  1023 = _______ molecules H2O 7.531023 1 mol H2O

How many moles are equal to 2.8 × 1022 formula units of KBr? Let’s Practice #2 Example: How many moles are equal to 2.8 × 1022 formula units of KBr?

How many moles are equal to 2.8 × 1022 formula units KBr? Let’s Practice #2 Example: How many moles are equal to 2.8 × 1022 formula units KBr? 1 mol = 6.021023 formula units 2.8 × 1022 formula units 1 mole = _______ moles 0.047 6.02  1023 Formula units

How many atoms are equal to 3.56 moles of Fe? Let’s Practice #3 Example: How many atoms are equal to 3.56 moles of Fe?

How many atoms are equal to 3.56 moles of Fe? Let’s Practice #3 Example: How many atoms are equal to 3.56 moles of Fe? 1 mol = 6.021023 molecules 3.56 moles Fe 6.02 x 10 23 atoms = _______ atoms 2.14 x 1024 1 moles

Molar Mass Molar Mass – The mass for one mole of an atom or molecule. Other terms commonly used for the same meaning: Molecular Weight Molecular Mass Formula Weight Formula Mass

Molar Mass for Elements The average atomic mass = grams for 1 mole Average atomic mass is found on the periodic table Element Mass 1 mole of carbon atoms (C) 12.01 g 1 mole of oxygen atoms (O2) 16.00 g x 2 = 32.00 g O2 1 mole of hydrogen atoms (H2) 1.01 g x 2 = 2.02 g H2 Unit for molar mass: g/mole or g/mol

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 To find the molar mass of a molecule: 1 Count the number of each type of atom 2 Find the molar mass of each atom on the periodic table 3 Multiply the # of atoms by the molar mass for each atom 4 Find the sum of all the masses

Find the molar mass for CaBr2 Example: Molar Mass Example: Find the molar mass for CaBr2

Find the molar mass for CaBr2 Example: Molar Mass 1 Count the number of each type of atom Example: Find the molar mass for CaBr2 Ca 1 Br 2

Find the molar mass for CaBr2 Example: Molar Mass 2 Find the molar mass of each atom on the periodic table Example: Find the molar mass for CaBr2 Ca 1 40.08 g/mole Br 2 79.90 g/mole

Find the molar mass for CaBr2 Example: Molar Mass 3 Multiple the # of atoms  molar mass for each atom Example: Find the molar mass for CaBr2 Ca 1  40.08 g/mole = 40.08 g/mole Br 2  79.90 g/mole = 159.80 g/mole

Find the molar mass for CaBr2 Example: Molar Mass 4 Find the sum of all the masses Example: Find the molar mass for CaBr2 Ca 1  40.08 g/mole = 40.08 g/mole Br 2  79.91 g/mole = + 159.80 g/mole 199.88 g/mole 1 mole of CaBr2 =199.90 g

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. Example: Find the molar mass for Sr(NO3)2 Sr 1  87.62 g/mole = 87.62 g/mole N 2  14.01 g/mole = 28.02 g/mole O 6  16.00 g/mole = + 96.00 g/mole 211.64 g/mole 1 mole of Sr(NO3)2 =211.64 g

Find the molar mass for Al(OH)3 Let’s Practice #3 Example: Find the molar mass for Al(OH)3

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. Example: Find the molar mass for Al(OH)3 Al 1  26.98 g/mole = 26.98 g/mole O 3  16.00 g/mole = 48.00 g/mole H 3  1.01 g/mole = + 3.03 g/mole 78.01 g/mole 1 mole of Al(OH)3 =78.01 g

Using Molar Mass in Conversions What is molar mass?

Example: Moles to Grams How many grams are in 1.25 moles of water?

Example: Moles to Grams When converting between grams and moles, the molar mass is needed Example: How many grams are in 1.25 moles of water? H O 2 1 1.01 g/mole 16.00 g/mole  = 2.02 g/mole + 18.02 g/mole 1 mole H2O molecules = 18.02 g 1.25 mol H2O 18.02 g H2O = _______ g H2O 22.5 1 mol H2O

Example: Grams to Moles How many moles are in 25.5 g NaCl?

Example: Grams to Moles How many moles are in 25.5 g NaCl Na Cl 1 22.99 g/mole 35.45 g/mole  = + 58.44 g/mole 1 moles NaCl molecules = 58.44 g 25.5 g NaCl 1 mol NaCl 58.44 g NaCl = ____ moles NaCl .436

Example: Grams to Molecules How many formula units are in 25.5 g NaCl?

Example: Grams to Moles How many formula units are in 25.5 g NaCl Na Cl 1 22.99 g/mole 35.45 g/mole  = + 58.44 g/mole 1 moles NaCl formula units = 58.44 g 25.5 g NaCl 1 mol NaCl 6.02 x 1023 FU’s 58.44 g NaCl 1 mol NaCl 2.63 x 1023 = ____ FU’s NaCl

Gases & Moles Amounts

Molar Volume The molar volume of a gas is measured at STP (standard temperature and pressure) 1 mole of any gas = 22.4 L

Molar Volume as a Conversion Factor The molar volume at STP has about the same volume as 3 basketballs can be used to form 2 conversion factors: 22.4 L and 1 mole 1 mole 22.4 L

Let’s Try it Out! Example When converting between volume and moles, STP must be a condition to use molar volume Example: An experiment requires .0580 moles of NO. What volume would you need at STP? 1 mole NO = 22.4 L .O580 mol NO 22.4 L NO = _______ L NO 1.2992 1 mol NO

Suppose you need 4.22 g of Cl2. What volume at STP would you use? Try Another! Example: Suppose you need 4.22 g of Cl2. What volume at STP would you use? 1 moles Cl2 = 70.90 g Cl2 1 mol = 22.4 L at STP 4.22 g Cl2 1 L Cl2 mol Cl2 22.4 70.90 g Cl2 1 mol Cl2 = _________ L Cl2 1.33

Percent Composition Defined as the percent by mass of each element in a compound Steps to Finding Percent Composition Add up the mass of each element within the compound to get the mass of the compound. Divide each element’s mass by the mass of the compound. 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. CaCO3 Molar 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 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%

Heating of A Hydrate Animation Calculating the experimental % composition of water in a hydrate. http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/flashfiles/stoichiometry/empirical.html

Empirical Formula A 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 Assume 100 grams of the sample of compound. Switch the percent sign to grams Convert each element’s mass into moles. 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. 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? 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 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 = 3 1.250 mole 1.250 mol S1O3  SO3

Example: What is the empirical formula for 43.64% P and 56.36% O? 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 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 ≠ 3 1.409 mole 1.409 mol 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 formula N2O4 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 What is PRESSURE? 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

Pressure and Moles (# of Molecules) If pressure is molecular collisions with the container… As number of molecules increases, there will be more molecules to collide with the wall Collisions between molecules and the wall increase Pressure increases As # of moles increase, pressure increases Think about blowing up a balloon!

# of Gas Particles vs. Pressure

Pressure & Volume As volume increases, pressure decreases. If pressure is molecular collisions with the container… As volume increases, molecules can travel farther before hitting the wall Collisions between molecules & the wall decrease Pressure decreases As volume increases, pressure decreases. Think about how your lungs work! http://www.youtube.com/watch?v=q6-oyxnkZC0

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, molecular motion increases Collisions between molecules & the wall increase Pressure increases As temperature increases, 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, molecular motion increases molecules will move farther away from each other Volume increases As temperature increases, volume increases Think of liquid nitrogen and the balloon. http://www.youtube.com/watch?v=QEpxrGWep4E

Pressure In Versus Out A container will expand or contract until the pressure inside equals atmospheric pressure outside 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. The internal pressure of the bag at low altitude is high At high altitude there is lower pressure Lower pressure Higher pressure Lower pressure The internal pressure is higher than the external pressure. The bag will expand in order to reduce the internal pressure.

When Expansion Isn’t Possible Rigid containers cannot expand Example: An aerosol can is left in a car trunk in the summer. What happens? The temperature inside the can begins to rise. As temperature increases, pressure increases. Lower pressure Higher pressure Can Explodes! The internal pressure is higher than the external pressure. The can is rigid—it cannot expand, it explodes!

Air Pressure Crushing Cans http://www.csun.edu/scied/4-discrpeant-event/the_can_crush/index.htm

Air Pressure Crushing “Cans” http://www.youtube.com/watch?v=Zz95_VvTxZM Another cool video http://www.youtube.com/watch?v=JsoE4F2Pb20

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

Assumptions of the KMT All gases are made of atoms or molecules that are in constant, rapid, random motion 1 The temperature of a gas is proportional to the average kinetic energy of the particles 2 Gas particles are not attracted nor repelled from one another *** 3 All gas particle collisions are perfectly elastic (no kinetic energy is lost to other forms) 4 The volume of gas particles is so small compared to the space between the particles, that the volume of the particle itself is insignificant*** 5

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 Gas particles do have attractions and repulsions towards one another Assumption #5 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 particle size (mass) increases, the particles move slower it takes them more time to find the hole or to go across the room As mass of the particles increases, rate of effusion and diffusion is lowered.

Rate of Diffusion & Particle Mass H2 Watch as larger particles take longer to get to your nose CO2

Section 3.6—Gas Laws How can we calculate Pressure, Volume and Temperature of our airbag?

Pressure Units 1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi Several units are used when describing pressure Unit Symbol atmospheres atm Pascals, kiloPascals Pa, kPa millimeters of mercury mm Hg pounds per square inch 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 Convert 654 mm Hg to atm Convert 879 mm Hg to kPa 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 Examples 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: P= Pressure V= Volume T=Temperature n= moles(molecules) 1= initial amount 2= final amount

Boyle’s Law Pressure Increases as Volume Decreases

Boyles’ Law Volume & Presssure are INVERSELY proportional when temperature and moles are held constant P = pressure V = volume 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 ***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

Boyles Law: Graph

Charles’ Law Volume Increases as Temperature Increases

Charles’ Law Volume & Temperature are DIRECTLY proportional when pressure and moles are held constant. V = Volume T = Temperature 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 25C to 50C? Temperature needs to be in Kelvin! V1 = 10.5 L T1 = 25C V2 = ? L T2 = 50C 25C + 273 = 298 K 50C + 273 = 323 K

Charles’ Law ***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 25C to 50C? V1 = 10.5 L T1 = 25C V2 = ? L T2 = 50C = 298 K = 323 K V2 = 11.4 L

Charles Law: Graph

Gay-Lussac’s Law Temperature decreases as Pressure decreases

Gay-Lussac’s Law Pressure & temperature are DIRECTLY proportional when moles and volume are held constant P = Pressure T = Temperature The two pressure units must match and temperature must be in Kelvin! A sample of hydrogen gas at 47C exerts a pressure of .329 atm. The gas is heated to 77C at constant volume and moles. What will the new pressure be? Example: Temperature needs to be in Kelvin! P1 = .329 atm T1 = 47C P2 = ? atm T2 = 77C 47C + 273 = 320 K 77C + 273 = 350 K

Gay-Lussac’ Law Example: A sample of hydrogen gas at 47C exerts a pressure of .329 atm. The gas is heated to 77C at constant volume and moles. What will the new pressure be? P1 = .329 atm T1 = 47C P2 = ? atm T2 = 77C = 320 K = 350 K P2 = .360 atm

Gay Lussac Law: Graph

Avogadro’s Law Example: Moles and Volume are directly proportional when temp. & pressure are held constant V = Volume n = # of moles of gas The two volume units must match! 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?

Avogadro’s Law The two volume units must match! 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? n1 = 0.15 moles V1 = 2.5 L n2 = 0.55 moles V2 = ? L V2 = 9.2 L

Combined Gas Law Example: P = Pressure V = Volume n = # of moles T = Temperature 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 = Pressure V = Volume T = 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 STP is standard temperature (273 K) and pressure (1 atm) V2 = 14.1 L

Why you really only need 1 of these 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:

Transforming the Combined Law Watch as variables are held constant and the combined gas law “becomes” the other 3 laws Hold pressure and temperature constant Avogadro’s Law Hold moles and temperature constant Boyles’ Law Hold pressure and moles constant Charles’ 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? PT = 2.3 atm + 1.7 atm 4.0 atm

Modified Dalton’s Law When a gas is Collected over water, the total pressure of the mixture collected is a combination of water vapor and the gas you are collecting! 2KClO3 (s) 2KCl (s) + 3O2 (g) PT = PO + PH O 2

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? 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 = Pressure V = Volume n = moles R = Gas Law Constant T = Temperature There are three possibilities for “R”! 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 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 The Ideal Gas Law does not compare situations—it describes a gas in one situation. 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 Chosen to match the kPa in the “P” above V2 = 13 L

The Ideal Gas Law Example 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 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