Ch 10 Notes – Ch 10 Notes – Chemical Quantities: Chemical Quantities:

Working With The Mole Working With The Mole Ch 12: Stoichiometry Ch 12: Stoichiometry

..

I. I. 3 ways that chemists measure the 3 ways that chemists measure the quantity of matter: quantity of matter: massmass, , volumevolume, , amount amount – number of, by counting– number of, by counting.. A. The “mole” is the unit we need to use A. The “mole” is the unit we need to use

when we want to quantify a chemical when we want to quantify a chemical change. change.

The mole is the official SI standard base unit The mole is the official SI standard base unit (for the “(for the “amountamount” of a substance). The ” of a substance). The symbol for mole is symbol for mole is molmol..

B. Definition: One mole of any B. Definition: One mole of any substance is substance is the amountthe amount which which contains the same contains the same numbernumber of particles of particles as there are as there are atomsatoms in exactly 12 grams in exactly 12 grams of carbon-12.of carbon-12.

And, since exactly 12 grams of carbon-12 And, since exactly 12 grams of carbon-12 contains contains 6.02 x 106.02 x 102323 atoms atoms, 1 mole of , 1 mole of anything contains 6.02 x 10 anything contains 6.02 x 10 2323 particles. particles. This number is referred to as “This number is referred to as “Avogadro’sAvogadro’s number”.number”.

It doesn’t even matter what the “particle” is: For It doesn’t even matter what the “particle” is: For example, 1 mole of beetles would equal 6.02 example, 1 mole of beetles would equal 6.02 x 10 x 10 2323 beetles. If each beetle has 6 legs, then beetles. If each beetle has 6 legs, then 1 mole of beetles has 6 moles of legs (3.61 x 1 mole of beetles has 6 moles of legs (3.61 x 10 10 2424 legs). legs).

1. How many atoms are present in 1 1. How many atoms are present in 1 mole of any element? mole of any element? 6.02 x 10 6.02 x 10 2323 atomsatoms

How many pretzels are contained in 1 How many pretzels are contained in 1 mole of pretzels? mole of pretzels? 6.02 x 10 6.02 x 10 2323 pretzels pretzels

How many molecules of COHow many molecules of CO22 are in 1 mole of are in 1 mole of COCO22? ? 6.02 x 10 6.02 x 10 2323 molecules molecules

How many front wheels are on 1 tricycle?How many front wheels are on 1 tricycle? 1 front wheel1 front wheel

How many back wheels are on 1 tricycle?How many back wheels are on 1 tricycle? 2 back wheels2 back wheels

How many atoms of carbon (the front wheel) are How many atoms of carbon (the front wheel) are in 1 mole of COin 1 mole of CO22??

In 1 mole of COIn 1 mole of CO22, there is 1 mole of carbon, , there is 1 mole of carbon, which = 6.02 x 10 which = 6.02 x 10 2323 atoms of carbon atoms of carbon

How many atoms of oxygen (the back wheel) How many atoms of oxygen (the back wheel) are in 1 mole of COare in 1 mole of CO22??

In 1 mole of COIn 1 mole of CO22, there are 2 moles of oxygen, , there are 2 moles of oxygen, which = 1.20 x 10 which = 1.20 x 10 2424 atoms of oxygen atoms of oxygen

Mole road map –see Figure 10.12 p 303.Mole road map –see Figure 10.12 p 303. Left – Left – NumberNumber of “given particles” in a substance of “given particles” in a substance

(atoms, ions, molecules, …) (atoms, ions, molecules, …)

Use Avogadro’s number: Use Avogadro’s number: 6.02 x 10 6.02 x 10 2323 atoms atoms = 1 mole of those atoms= 1 mole of those atoms

Center –Center – Volume Volume of a gaseous substance at STP of a gaseous substance at STP

Use the molar volume (the volume of 1 mole of Use the molar volume (the volume of 1 mole of any gas) 22.4 L of gas = 1 mol of gasany gas) 22.4 L of gas = 1 mol of gas

Right –Right – Mass Mass of a Substanceof a Substance

Use the “molar mass” (the mass of 1 mole of any Use the “molar mass” (the mass of 1 mole of any substance) # of grams substance) # of grams XYZXYZ = 1 mol of = 1 mol of XYZXYZ calculated using the periodic tablecalculated using the periodic table

Formula Factor (or Formula Fraction): Formula Factor (or Formula Fraction): Use the formula of a molecule and/or compound to isolate the Use the formula of a molecule and/or compound to isolate the

atom or ion in question. atom or ion in question. NEED TONEED TO FIX THESEFIX THESE TYPOSTYPOS::

52

52

22

2

22

2

ON mole 1

O mole 5or

O mole 5

ON mole 1

O" of moles 2 are there,O of mole 1in " that means O mole 2

O mole 1or

O mole 1

O mole 2

C" of mole 1 is there,CO of mole 1in " that means CO mole 1

C mole 1or

C mole 1

CO mole 1

52

52

22

2

22

2

ON mole 1

O mole 5or

O mole 5

ON mole 1

O" of mole 1 is there,O of mole 1in " that means O mole 1

O mole 1or

O mole 1

O mole 2

C" of mole 1 is there,CO of mole 1in " that means CO mole 1

C mole 1or

C mole 1

CO mole 1

Means that in 1 mole of N2O5 there are 5 moles of O

1. The mole establishes a connection 1. The mole establishes a connection between the between the massmass, the , the amountamount, and (for , and (for gases) the gases) the volumevolume of a substance. of a substance.

Mathematically, you will need to follow the Mathematically, you will need to follow the arrows in the figure above as you “go arrows in the figure above as you “go through” the mole, to get from, say, a # through” the mole, to get from, say, a # atoms to the mass and/or volume of a atoms to the mass and/or volume of a substance.. Careful – Using the mole road substance.. Careful – Using the mole road map for volume is legitimate when working map for volume is legitimate when working only with a only with a gasgas. When working with a . When working with a solid or liquid, you will need to use the solid or liquid, you will need to use the substance’s substance’s densitydensity to arrive at volume. to arrive at volume.

2. While we were learning how to balance 2. While we were learning how to balance equations, you may have wondered what equations, you may have wondered what unit belongs to each coefficient in the unit belongs to each coefficient in the balanced equation. The answer is “balanced equation. The answer is “molemole”. ”.

This equation would be read as This equation would be read as

““2 moles2 moles of Al combine with of Al combine with 3 moles3 moles of Cl of Cl22, ,

in order to produce in order to produce 2 moles2 moles of AlCl of AlCl3.3.

Thus it is the mole which establishes the Thus it is the mole which establishes the ratioratio of one substance to another in a of one substance to another in a balanced chemical equation.balanced chemical equation.

The coefficients are whole numbers and are The coefficients are whole numbers and are interpreted as having infinite sig figs.interpreted as having infinite sig figs.

The mole is the unit we use to relate the The mole is the unit we use to relate the amount of one substance to the amount of amount of one substance to the amount of another during a chemical change.another during a chemical change.

II. Factor label problem solving Using Avogadro‘s II. Factor label problem solving Using Avogadro‘s number and chemical formulas:number and chemical formulas:

A. The numerical information in each problem A. The numerical information in each problem should be identified: should be identified:

““What you are given” with its unitWhat you are given” with its unit

““what you want to know” with its unitwhat you want to know” with its unit

a conversion factor (that relates the two a conversion factor (that relates the two units – written in the form of a ratio or fraction).units – written in the form of a ratio or fraction).

The top part of the factor must equal the The top part of the factor must equal the bottom part. Ex: 12 eggs / 1 dozenbottom part. Ex: 12 eggs / 1 dozen

This factor is flippable – you can use its reciprocal This factor is flippable – you can use its reciprocal whenever you need to: 1 dozen / 12 eggs.whenever you need to: 1 dozen / 12 eggs.

1. Start with what you are given as a numerator 1. Start with what you are given as a numerator (write down both the number and its unit).(write down both the number and its unit).

2. Multiply the given by your conversion factor, 2. Multiply the given by your conversion factor, which is purposely set up so that the which is purposely set up so that the denominator (bottom) unit is identical with, and denominator (bottom) unit is identical with, and thus will cancel with, the unit of the “given”.thus will cancel with, the unit of the “given”.

3. Last, use your calculator to multiply all the top 3. Last, use your calculator to multiply all the top numbers, and divide by all the bottom numbers.numbers, and divide by all the bottom numbers. Memory aid: Memory aid: MMighighTTy y DDuck uck BBottomsottoms MMultiply All ultiply All TTops ops DDivide By All ivide By All BBottoms.ottoms.

B. Guided Practice #1B. Guided Practice #1 Note: 6.02 x 10 Note: 6.02 x 10 2323 has been rounded to 3 sig figs. As with any has been rounded to 3 sig figs. As with any

chemistry calculation, pay attention to sig figs!chemistry calculation, pay attention to sig figs!

How many atoms of silicon are contained in How many atoms of silicon are contained in 123 mole Si123 mole Si??Given: Given: 123 mole Si123 mole Si Want to know: atoms SiWant to know: atoms SiConversion factor: 1 mol Si = 6.02 x 10 Conversion factor: 1 mol Si = 6.02 x 10 2323 atoms Si atoms Si

Calculator sequence:Calculator sequence:123 x 6.02 “special key” 23 ÷ 1 = 123 x 6.02 “special key” 23 ÷ 1 = Where “special key” might be “EE”, “exp”, “x10Where “special key” might be “EE”, “exp”, “x10yy”, mode, ….”, mode, ….

Si atoms 10 x 7.40

Si mol 1

Si atoms 10 x 6.02 x Si mol 123

2523

Please do Your Turn 1 and 2 at your Please do Your Turn 1 and 2 at your seats; then show your teacher.seats; then show your teacher.

Your Turn 1:Your Turn 1:

Given: 3.70 mol NaGiven: 3.70 mol Na Want to know: atoms NaWant to know: atoms Na Conversion factor:Conversion factor:

1 mole Na = 6.02 x 10 1 mole Na = 6.02 x 10 2323 atoms Na atoms Na

Na atoms 10 x 2.23

Na mol 1

Na atoms 10 x 6.02 x Na mol 3.70

2423

Your turn 2:Your turn 2:

Given: 1.23 x 10 Given: 1.23 x 10 2323 atoms He atoms He Want to know: moles HeWant to know: moles He Conversion factor:Conversion factor:

6.02 x 10 6.02 x 10 2323 atoms He = 1 mole He atoms He = 1 mole He

1.23 x 10 23 atoms He x =1.23 x 10 23 atoms He x =

23

23

1002.6

1 10 23.1

Heatomsx

HemolexHeatomsx

2323

1002.6

1 10 23.1

Heatomsx

HemolexHeatomsx

2323

1002.6

1 10 23.1

Heatomsx

HemolexHeatomsx

2323

1002.6

1 10 23.1

Heatomsx

HemolexHeatomsx

2323

1002.6

1 10 23.1

Heatomsx

HemolexHeatomsx

1 mole He

6.02 x 10 23 atoms He

0.204 mol He

III. Interpreting Formulas of Ionic III. Interpreting Formulas of Ionic Compounds Compounds

Follow the examples given. Then have your Follow the examples given. Then have your teacher check your work.teacher check your work.

IV.IV. Interpreting Formulas of Covalent Interpreting Formulas of Covalent CompoundsCompounds

Follow the examples given. Then have your Follow the examples given. Then have your teacher correct your work.teacher correct your work.

Answers to III:Answers to III:

a.a. 2,1 2,1 2 mole K, 1 mole S2 mole K, 1 mole S

b.b. 2,32,3 2 mole Al, 3 mole 02 mole Al, 3 mole 0

c.c. 1,21,2 1 mole Mg, 2 mole C1 mole Mg, 2 mole C22HH33OO22

d.d. 1,31,3 1 mole Al, 3 mole NO1 mole Al, 3 mole NO33

e.e. 3,13,1 3 mole NH4, 1 mole PO3 mole NH4, 1 mole PO44

f.f. 2,32,3 2 mole Mg, 3 mole SO2 mole Mg, 3 mole SO44

g.g. 3,23,2 3 mole Ca, 2 mole PO3 mole Ca, 2 mole PO44

h.h. 1,11,1 1 mole Na, 1 mole HCO1 mole Na, 1 mole HCO33

Answers to IV:Answers to IV:

a.a. 4,104,10 4 mol P, 10 mol O4 mol P, 10 mol O

b.b. 2,12,1 2 mol H, 1 mol O2 mol H, 1 mol O

c.c. 1,71,7 1 mol I, 7 mol F1 mol I, 7 mol F

d.d. 6,16,1 6 mol B, 1 mol Si6 mol B, 1 mol Si

e.e. 2,52,5 2 mol N, 5 mol O2 mol N, 5 mol O

f.f. 12,22,11 12 mol C, 22 mol H, 11 mol O12,22,11 12 mol C, 22 mol H, 11 mol O

g.g. 1,41,4 1 mol C, 4 mol H1 mol C, 4 mol H

h.h. 1,31,3 1 mol N, 3 mol H1 mol N, 3 mol H

Guided Practice 2Guided Practice 2

Correct this typo:Correct this typo:

Want to know =Want to know = atoms ofatoms of Br Br (NOT Br(NOT Br22))

Br atoms 10 Br molecule 1

Br atoms 2 x Br2 molecules 5.0

2

Your Turn 5Your Turn 5 Given: 7.0 formula units CaGiven: 7.0 formula units Ca33(PO(PO44))22

Want to know: # of POWant to know: # of PO44 ions ions

Conversion factor: Conversion factor:

1 formula unit Ca1 formula unit Ca33(PO(PO44))22 = 2 PO = 2 PO44 ions ions

)(POCaunit formula 1

ions PO 2 x )(POCa units formula 7.0

243

4243

14 PO4 ions

Your turn 6:Your turn 6:

Given: 264 carbon atomsGiven: 264 carbon atoms

Want ot know: # molecules CWant ot know: # molecules C1212HH2222OO1111

Formula factor: Formula factor:

12 C atoms = 1 molecule C12 C atoms = 1 molecule C1212HH2222OO1111

atomscarbon 12

OHC molecule 1 x atoms C 264 112212

22 molecules C12H22O11

V.V. Putting It All Together:Putting It All Together:2 Step Problems2 Step Problems

Given a formula, work with # of atoms, # Given a formula, work with # of atoms, # molecules, and # moles. molecules, and # moles.

Or, given a formula, work with # ions, # Or, given a formula, work with # ions, # formula units, and 3 moles.formula units, and 3 moles.

Guided Practice #3Guided Practice #3 How many atoms of phosphorus are in 35.0 How many atoms of phosphorus are in 35.0

moles of Pmoles of P22OO55??

Given: 35.0 moles PGiven: 35.0 moles P22OO55

Want to know: atoms PWant to know: atoms P Convers Factor: 1 mole P = 6.02 x 10 Convers Factor: 1 mole P = 6.02 x 10 2323 atoms P atoms P Formula Factor: 1 mole PFormula Factor: 1 mole P22OO55 = 2 moles P = 2 moles P

P atoms 10 x 4.21

P mol 1

10 x 6.02 x

OP mol 1

P moles 2 x 0P molecules 35.0

2523

5252

atomsP

Typo: moles

VI. “Average atomic mass” and “molar VI. “Average atomic mass” and “molar mass”:mass”:A. The average mass of 1 atom of an A. The average mass of 1 atom of an

element is indicated in the element’s square element is indicated in the element’s square on the on the periodic tableperiodic table..1. On our large classroom periodic 1. On our large classroom periodic tabletables, s, located located

along the BACK WALLalong the BACK WALL, the average atomic mass , the average atomic mass is rounded to 2 decimal places to the right of the is rounded to 2 decimal places to the right of the point, and is found underneath the symbols and/or point, and is found underneath the symbols and/or names of the elements.names of the elements.

2. On your personal periodic table, you can 2. On your personal periodic table, you can identify the position of the average atomic mass by identify the position of the average atomic mass by using the key provided on the table itself.using the key provided on the table itself.

3. The unit for the mass of 1 atom is called the 3. The unit for the mass of 1 atom is called the “atomic mass unit”, and its symbol is “atomic mass unit”, and its symbol is amuamu..

B. The mass of 1 mole of each element is the B. The mass of 1 mole of each element is the same as the average atomic mass; however, its same as the average atomic mass; however, its unit is unit is grams / molegrams / mole. The mass of 1 mole of . The mass of 1 mole of anything is called its “anything is called its “molar massmolar mass”.”. 1. Ex: The average atomic mass for He is 4.00 amu. 1. Ex: The average atomic mass for He is 4.00 amu.

Thus the molar mass of He is Thus the molar mass of He is 4.004.00 g/mol. g/mol. This means: This means: 4.00 grams He = 1 mole He 4.00 grams He = 1 mole He

2. What atom has an average mass of 24.31 amu? 2. What atom has an average mass of 24.31 amu? MagnesiumMagnesium..

3. What element has a molar mass of 22.99 g/mol? 3. What element has a molar mass of 22.99 g/mol? sodiumsodium

For any compound or molecule, the molar mass is For any compound or molecule, the molar mass is the addition sum of the molar masses of each the addition sum of the molar masses of each element in the compound or molecule.element in the compound or molecule.

Examples below:Examples below: molar mass of Nmolar mass of N2 2 = 14.01 + 14.01 = 28.01 g / mol= 14.01 + 14.01 = 28.01 g / mol

molar mass of COmolar mass of CO22

= 12.01 + 16.00 + 16.00 + 16.00 = 44.01 g/mol= 12.01 + 16.00 + 16.00 + 16.00 = 44.01 g/mol molar mass Pmolar mass P22OO33 = 2P + 3O = = 2P + 3O =

30.97 + 30.97 + 16.00 + 16.00 + 16.00 = 109.94 g/mol30.97 + 30.97 + 16.00 + 16.00 + 16.00 = 109.94 g/mol molar mass of Ca(NOmolar mass of Ca(NO33))22

= 1 Ca +2N + 6O =40.08 + 2(14.01) + 6 (16.00) = 1 Ca +2N + 6O =40.08 + 2(14.01) + 6 (16.00)

= 164.10 g/mol= 164.10 g/mol

Use the rules for Use the rules for additionaddition while calculating while calculating molar mass:molar mass:

Use 2 decimal places to the right of the Use 2 decimal places to the right of the point in your sum*** point in your sum*** (least number of (least number of decimal places to the right, based on the decimal places to the right, based on the addends given).addends given).

Molar mass Ba(NOMolar mass Ba(NO33))22

Ba =Ba = 137.34137.34

N2 =N2 = 14.01 x 2 = 14.01 x 2 = 28.02 28.02

O6 = O6 = 16.00 x 6 = 16.00 x 6 = 96.00 96.00

Total =Total = 261.36 g/mol261.36 g/mol

Your Turn 7Your Turn 7

Please calculate the molar mass of Please calculate the molar mass of aluminum acetate, Al(Caluminum acetate, Al(C22HH33OO22))33..

*****Using the molar masses given on the *****Using the molar masses given on the periodic table on the back of your final periodic table on the back of your final exam study guide.exam study guide.

Show your work. Show your work. Check your answer. Ask for help as Check your answer. Ask for help as

needed.needed.

Al(CAl(C22HH33OO22))33

C2 C2 24.02 24.02H3 H3 3.03 3.03O2 + 32.00O2 + 32.00

59.0559.05

26.98 + 3(59.05) = 26.98 + 3(59.05) =

204.13 grams / 204.13 grams / molemole

VII.VII. Factor label problems using the molar mass:Factor label problems using the molar mass:A.A. Guided Practice Problem #4 – What is the Guided Practice Problem #4 – What is the

mass of 125.0 moles of Ca(OH)mass of 125.0 moles of Ca(OH)22??

22

22 Ca(OH) grams 9263

Ca(OH) mol 1

Ca(OH) grams 74.10 x Ca(OH) mol 125.0

Please do Your Turn # 8 and #9. Ask for help as needed. Then, you may begin the HW # 11 – 19.

Your Turn 8Your Turn 8

Znmol 0.3908 Zng 65.38

Znmol 1 Zn x g 25.55

Given: Given: 25.55 g Zn25.55 g Zn Want: Want: mol Znmol Zn

1 mol Zn = 65.38 g Zn1 mol Zn = 65.38 g Zn

Your Turn 9Your Turn 9

Given: Given: 25.55 g Zn(NO25.55 g Zn(NO33))22

Want: Want: mol Znmol Zn DO YOU SEE THE NEED FOR A DO YOU SEE THE NEED FOR A FORMULA FACTOR?FORMULA FACTOR?

(yes…whenever there is a different substance involved in (yes…whenever there is a different substance involved in the given/want).the given/want).

Formula factor: 1 mol Zn = 1 mol Zn(NOFormula factor: 1 mol Zn = 1 mol Zn(NO33))22

Zn mol 0.1349 ) Zn(NOmol 1

Znmol 1 x

) Zn(NOg 189.40

) Zn(NOmol 1 x ) Zn(NOg 25.55

2323

2323

X. Avogadro’s_X. Avogadro’s_ hypothesis hypothesis::Now recognized as a law of nature:Now recognized as a law of nature:

A. Amedeo Avogadro recognized that unlike A. Amedeo Avogadro recognized that unlike liquids and solids, whose molar volumes differ liquids and solids, whose molar volumes differ dramatically, equal volumes of dramatically, equal volumes of any gasany gas, , measured at the same temperature and measured at the same temperature and pressure, will contain pressure, will contain an equal an equal number of number of particles.particles.

B. Thus, 1 mole (6.02 x 10 23 particles) of B. Thus, 1 mole (6.02 x 10 23 particles) of differentdifferent gases, measured under the gases, measured under the samesame physical conditions, will have identical physical conditions, will have identical volumes.volumes.

C. The reason why the identity of the gas C. The reason why the identity of the gas is irrelevant is because a gas is mostly is irrelevant is because a gas is mostly empty spaceempty space; and since the atoms do NOT ; and since the atoms do NOT touch each other -- their radius, size, touch each other -- their radius, size, mass, etc. (all the stuff that determines the mass, etc. (all the stuff that determines the identity of the gas), is not important.identity of the gas), is not important.

Instead, what is important with gases, is Instead, what is important with gases, is their environment (temperature, their environment (temperature, atmospheric pressure), and atmospheric pressure), and numbernumber of of particles or molecules that are in the gas.particles or molecules that are in the gas.

D. The mole today is commonly referred D. The mole today is commonly referred to as “to as “Avogadro’sAvogadro’s Number Number”, ”, as we as we posthumously remember and honor posthumously remember and honor Amedeo Avogadro, not only for his Amedeo Avogadro, not only for his important work with gases, but also important work with gases, but also because he established the difference because he established the difference between atoms and molecules, and between atoms and molecules, and helped determine the formula for water.helped determine the formula for water.

XI. The “molar volume”:XI. The “molar volume”:

1 mole is1 mole is equal to equal to the number of particles the number of particles of any gas that would fit within a 22.4 L of any gas that would fit within a 22.4 L volume at volume at STP STP (standard temperature and (standard temperature and pressure). pressure).

So, the So, the molar volumemolar volume of any gas at STP is of any gas at STP is

22.4 Liters / mole22.4 Liters / mole..

1. Standard temperature = the freezing 1. Standard temperature = the freezing point of water =point of water = 0 0 ooC = 273 K = 32 C = 273 K = 32 ooFF

Standard pressure = the pressure of our Standard pressure = the pressure of our atmosphere on the earth at sea level atmosphere on the earth at sea level

== 1 atm 1 atm = 760 torr = 760 mm Hg = 760 torr = 760 mm Hg

= 101.3 kPa = 101.3 kPa = 14.7 psi= 14.7 psi

Practice Questions:Practice Questions:

A. What is the volume of 1 mole of radon A. What is the volume of 1 mole of radon gas at STP?gas at STP?

22.4 L / mole22.4 L / mole

How many moles of helium are contained How many moles of helium are contained within a balloon having a volume of 22.4 within a balloon having a volume of 22.4 L? L?

1 mole He1 mole He

How many moles of nitrous oxide (N2O) How many moles of nitrous oxide (N2O) gas would be present in a cube whose gas would be present in a cube whose volume is 22.4 L?volume is 22.4 L?

1 mole N1 mole N22OO

What would be the volume of 1 mole of the What would be the volume of 1 mole of the metal, sodium?metal, sodium?

Gotcha!Gotcha!

Sodium isn’t a gas, it’s a solid – You would need to Sodium isn’t a gas, it’s a solid – You would need to use the molar mass and the density of sodium to use the molar mass and the density of sodium to calculate this answer.calculate this answer.

A container is filled with 22.4 L water. A container is filled with 22.4 L water. How many moles of water is this?How many moles of water is this?

Trick question – would need to use density Trick question – would need to use density and molar mass of water to calculate the and molar mass of water to calculate the answer.answer.

Guided Practice – Molar VolumeGuided Practice – Molar Volume

Problem #5:Problem #5:

What is the volume, What is the volume, given 16.9 moles He given 16.9 moles He at STP?at STP?

Insert “He” hereInsert “He” here::

He L 379 He mol 1

He L 22.4 x He mol 16.9

Problem #6:Problem #6:

How many moles of He, How many moles of He, given 0.500 L at STP?given 0.500 L at STP?

He mol 0.0223 He L 22.4

He mol 1 x He L 0.500

Please do your turn # 10, 11, and 12 now.Please do your turn # 10, 11, and 12 now.

Your TurnsYour Turns

10. 10.

Given: 0.247 mol ArGiven: 0.247 mol Ar

Want: Liters ArWant: Liters Ar

Ar L 5.53 Ar mol 1

Ar L 22.4Ar x mol 0.247

11.11.

Given: 335.2 L ArGiven: 335.2 L Ar

Want: moles ArWant: moles Ar

Ar mol 15.0 Ar L 22.4

Ar mol 1Ar x L 335.2

Putting It TogetherPutting It Together

Your Turn 12Your Turn 12

Given: Given: 10.0 mL H10.0 mL H22

Want: Want: atoms Hatoms H

mL to L: mL to L: 1000 mL = 1 L1000 mL = 1 L

Volume to moles: Volume to moles: 22.4 L = 1 mole22.4 L = 1 mole

Formula factor: Formula factor: 1 mole H1 mole H22 = 2 moles H = 2 moles H

H mole 1

H atoms 10 x 6.02 x

H mole 1

H mole 2 x

H L 22.4

H mole 1 x

H mL 1000

H L 1 x H mL 10.0

23

22

2

2

22

TYPO- Change (fix):

10.0 x 2 x 6.02 x 10 23 ÷ 1000 ÷ 22.4

Answer = 5.38 x 10 20 atoms H

XII. Density Based ProblemsXII. Density Based Problems

Calculating the Molar Mass of a Gas at Calculating the Molar Mass of a Gas at STP from its Density;STP from its Density;

And, Calculating the Density of a gas from And, Calculating the Density of a gas from its molar massits molar mass

You must be more careful than you You must be more careful than you normally are, as you will be starting with a normally are, as you will be starting with a “complex” (two-part) unit.“complex” (two-part) unit.

Start with your “given” in the form of a Start with your “given” in the form of a fraction, with a unit in its numerator and a fraction, with a unit in its numerator and a different unit in its denominator.different unit in its denominator.

Then, cross off in a direction opposite from Then, cross off in a direction opposite from what you normally do. Given’s bottom top what you normally do. Given’s bottom top with top of conversion!with top of conversion!

Guided Practice 7, 8Guided Practice 7, 8

Given: density = Given: density = 1.964 g / L1.964 g / L

Want: molar Want: molar mass….. g / molemass….. g / mole

mole / grams 44.0 CO mole 1

COLiter 22.4 x

COLiter 1

CO grams 1.964

2

2

2

2

Given: molar mass = Given: molar mass = 39.9 g Ar / mol Ar39.9 g Ar / mol Ar

Want: density….Want: density….

g / Literg / Liter

Liter / grams 1.78 Ar Liters 22.4

Ar mole 1 x

Ar mole 1

Ar grams 39.9

Please do Your Turn 13 and 14; then, you Please do Your Turn 13 and 14; then, you may begin the HW.may begin the HW.

Your Turn 13 and 14Your Turn 13 and 14

Given = density = Given = density = 1.24 g N1.24 g N22 / Liter N / Liter N22

Want = molar mass of Want = molar mass of NN22… g N… g N22 / mol N / mol N22

mole / grams 28.0 N mole 1

NLiter 22.4 x

NLiter 1

N grams 1.25

2

2

2

2

Given = molar mass = Given = molar mass = 64.1 g / mole64.1 g / mole

Want = density of SOWant = density of SO22

= g / Liter= g / Liter

Liter / grams 2.86 SO L 22.4

SO mole 1 x

SO mole 1

SO grams 64.1

2

2

2

2