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Calculations Calculations with Elements with Elements and Compounds and Compounds The Mole The Mole Concept Concept Rev. 051111 050212 Pisgah High School Pisgah High School M. Jones M. Jones
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Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

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Page 1: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Calculations with Calculations with Elements and Elements and

CompoundsCompounds

The Mole ConceptThe Mole Concept

Rev. 051111050212

Pisgah High SchoolPisgah High SchoolM. JonesM. Jones

Page 2: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.
Page 3: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Moles?Moles?

Page 4: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

No. Not real moles, No. Not real moles, or Whack-a-moles, or Whack-a-moles, or even the moles on or even the moles on your back.your back.

Chemistry molesChemistry moles

Page 5: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

What is a mole?What is a mole?

It’s a way of measuring It’s a way of measuring atoms and molecules.atoms and molecules.

The “mole concept” is a lot The “mole concept” is a lot like the “dozen concept.”like the “dozen concept.”

Page 6: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

A dozen is 12.A dozen is 12.It doesn’t make any It doesn’t make any difference what you difference what you are counting.are counting.

A dozen is always 12.A dozen is always 12.

Page 7: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

A dozen is 12.A dozen is 12.It doesn’t make any It doesn’t make any difference what you difference what you are counting.are counting.

A dozen is always 12.A dozen is always 12.

A dozen donuts.A dozen donuts.

Page 8: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

A dozen is 12.A dozen is 12.It doesn’t make any It doesn’t make any difference what you difference what you are counting.are counting.

A dozen is always 12.A dozen is always 12.

A dozen donuts.A dozen donuts.

A dozen pencils.A dozen pencils.

Page 9: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

A dozen is 12.A dozen is 12.It doesn’t make any It doesn’t make any difference what you difference what you are counting.are counting.

A dozen is always 12.A dozen is always 12.

A dozen donuts.A dozen donuts.

A dozen pencils.A dozen pencils.

A dozen Volkswagons.A dozen Volkswagons.

Page 10: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.
Page 11: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

A dozen is always

12.

Page 12: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Like we can count by Like we can count by dozens, or multiples of 12, dozens, or multiples of 12,

we can also count by we can also count by multiples of a much larger multiples of a much larger

number, which we call number, which we call Avogadro’s number.Avogadro’s number.

Page 13: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Avogadro’s number is much Avogadro’s number is much larger…larger…

Avogadro’s number is Avogadro’s number is 6.022 x 106.022 x 102323..

So how did chemists come up So how did chemists come up with such a huge number?with such a huge number?

Page 14: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

It began with It began with Amedeo Avogadro Amedeo Avogadro about a 150 years about a 150 years ago.ago.

He said that for equal volumes of He said that for equal volumes of gases at constant temperature and gases at constant temperature and pressure, the masses of the gases are pressure, the masses of the gases are proportional to their atomic weights.proportional to their atomic weights.

Page 15: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

10.0 liters of CO10.0 liters of CO22 weighs 19.6 grams weighs 19.6 grams

10.0 liters of N10.0 liters of N22 weighs 12.5 grams weighs 12.5 grams

10.0 liters of Xe weighs 58.6 grams10.0 liters of Xe weighs 58.6 grams

44.0 g44.0 g

28.0 g28.0 g==1.571.57

Avogadro’s lawAvogadro’s law

1.571.5719.6 g19.6 g

12.5 g12.5 g==

Ratio of the masses Ratio of the masses of equal volumes of of equal volumes of COCO22 and N and N22

Ratio of the molar Ratio of the molar masses of COmasses of CO22 and N and N22

Page 16: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

10.0 liters of CO10.0 liters of CO22 weighs 19.6 grams weighs 19.6 grams

10.0 liters of N10.0 liters of N22 weighs 12.5 grams weighs 12.5 grams

10.0 liters of Xe weighs 58.6 grams10.0 liters of Xe weighs 58.6 grams

Avogadro’s lawAvogadro’s law

19.6 g19.6 g

58.6 g58.6 g== 0.3350.335

44.0 g44.0 g

131.3 g131.3 g== 0.3350.335

Ratio of the masses Ratio of the masses of equal volumes of of equal volumes of COCO22 and Xe and Xe

Ratio of the molar Ratio of the molar masses of COmasses of CO22 and Xe and Xe

Page 17: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

He said that for He said that for equal volumes of equal volumes of gases at constant gases at constant temperature and temperature and pressure, the masses pressure, the masses of the gases are of the gases are proportional to their proportional to their atomic weights.atomic weights.

See!I told you

so.

Page 18: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

But Avogadro never But Avogadro never determined the value determined the value we know as we know as Avogadro’s number.Avogadro’s number.

I wish I knew what my number

was.

It was called It was called Avogadro’s number Avogadro’s number as a tribute to his as a tribute to his pioneering work.pioneering work.

Page 19: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

After Avogadro’s death, Stanislao After Avogadro’s death, Stanislao Cannizzaro published an explanation Cannizzaro published an explanation of Avogadro’s law. of Avogadro’s law.

Equal volumes of gases at the Equal volumes of gases at the same temperature and same temperature and pressure have equal numbers pressure have equal numbers of particles.of particles.

Page 20: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Cannizzaro was able to convince Cannizzaro was able to convince chemists of the value of chemists of the value of standardized atomic weights.standardized atomic weights.

Loschmidt (1865) was the Loschmidt (1865) was the first to determine the number first to determine the number

of atoms or molecules in a of atoms or molecules in a fixed volume of gas. fixed volume of gas.

Page 21: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Jean Perrin was the first to use the Jean Perrin was the first to use the term term Avogadro’s numberAvogadro’s number in 1909 in 1909 to describe the number of atoms in to describe the number of atoms in the atomic weight of an element.the atomic weight of an element.

He was the first to determine a He was the first to determine a value of Avogadro’s number value of Avogadro’s number

which is close to the present-day which is close to the present-day value of 6.022 x 10value of 6.022 x 102323. .

Page 22: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

We can relate Avogadro’s We can relate Avogadro’s number to the amount of an number to the amount of an element equal to its atomic element equal to its atomic mass measured in grams.mass measured in grams.

Atomic MassAtomic Mass

Page 23: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Atomic MassAtomic Mass

This used to be called the “gram This used to be called the “gram atomic weight”. Atomic weights are atomic weight”. Atomic weights are now called atomic masses.now called atomic masses.

Atomic masses are given in units Atomic masses are given in units of “atomic mass units”, or amu. of “atomic mass units”, or amu.

The amu is also called the dalton.The amu is also called the dalton.

Page 24: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

The “atomic mass unit” (amu) The “atomic mass unit” (amu) is based on the mass of the is based on the mass of the carbon-12 isotope.carbon-12 isotope.

It was decided that an atom of C-It was decided that an atom of C-12 be assigned a mass of exactly 12 be assigned a mass of exactly

12.0000 amu12.0000 amu

Atomic MassAtomic Mass

Page 25: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

This way, hydrogen,the lightest This way, hydrogen,the lightest element (H) would have a mass element (H) would have a mass very close to 1.very close to 1.

The average atomic mass of The average atomic mass of hydrogen is 1.0079 amu.hydrogen is 1.0079 amu.

Atomic MassAtomic Mass

Page 26: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Instead of 12.0000 amu of Instead of 12.0000 amu of carbon-12, which is too small to carbon-12, which is too small to even see, we consider 12.0000 even see, we consider 12.0000 gramsgrams of C-12. of C-12.

How many atoms are in How many atoms are in 12.0000 g of C-12?12.0000 g of C-12?

Atomic MassAtomic Mass

Page 27: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

A C-12 atom has a A C-12 atom has a mass of 1.993 x 10mass of 1.993 x 10-23-23 g g

How many atoms are in 12.000 g of How many atoms are in 12.000 g of carbon-12?carbon-12?

12.000 g C12.000 g C xx1 C atom1 C atom

1.993 x 101.993 x 10-23-23 g C g C

= 6.022 x 10= 6.022 x 1023 23 C atomsC atoms

C atomC atom

Atomic MassAtomic Mass

Page 28: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

A mole is the quantity of an element A mole is the quantity of an element which has a mass equal to the which has a mass equal to the atomic mass expressed in grams. atomic mass expressed in grams.

Atomic MassAtomic Mass

The mole is the “official” metric The mole is the “official” metric unit for the “quantity of matter”.unit for the “quantity of matter”.

Page 29: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

The word “mole” can be The word “mole” can be abbreviated as “mol”.abbreviated as “mol”.

Page 30: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

So, what is a mole?So, what is a mole?It is a word that describes a very It is a word that describes a very large number of atoms.large number of atoms.

6.022 x 106.022 x 1023 23 is also called is also called Avogadro’s Number.Avogadro’s Number.

It comes from the Latin word It comes from the Latin word molesmoles which means “quantity” which means “quantity”

6.022 x 106.022 x 1023 23 atomsatoms

Page 31: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

A mole is …A mole is …

– – any particle.any particle.

Since they are so small, we Since they are so small, we must count atoms indirectly.must count atoms indirectly.

Can you count to a mole?Can you count to a mole?

6.022 x 106.022 x 102323 atoms, molecules, atoms, molecules, electrons or ions …electrons or ions …

Page 32: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Suppose you wanted to Suppose you wanted to know the number of know the number of marbles in a large jar.marbles in a large jar.

use a conversion factor.use a conversion factor.weigh all the marbles,weigh all the marbles,

Weigh a few marbles,Weigh a few marbles,

Page 33: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Like this:Like this:

10 marbles = 86.45 grams10 marbles = 86.45 grams

Avg. marble = 8.645 gramsAvg. marble = 8.645 grams

All the marbles = 2472.47 g All the marbles = 2472.47 g

1 marble1 marble

8.645 g8.645 g286 marbles286 marbles

This is a This is a conversion factorconversion factor

Suppose …Suppose …

2472.47 g 2472.47 g xx = =

Page 34: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

We can count a large We can count a large number of particles by number of particles by

weighing them.weighing them.

Page 35: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

We can count atoms if we We can count atoms if we know the mass of an know the mass of an

Avogadro's number of Avogadro's number of atoms.atoms.

Page 36: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

An Avogadro's number of atoms has a An Avogadro's number of atoms has a mass equal to the molar mass.mass equal to the molar mass.

6.022 x 106.022 x 102323 atoms of sodium atoms of sodiumhave a mass of 23.0 ghave a mass of 23.0 g

6.022 x 106.022 x 102323 atoms of iron atoms of ironhave a mass of 55.8 ghave a mass of 55.8 g

6.022 x 106.022 x 102323 atoms of aluminum atoms of aluminumhave a mass of 27.0 ghave a mass of 27.0 g

Page 37: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Q. How many atoms are in Q. How many atoms are in 50.0 grams of potassium?50.0 grams of potassium?

We know that 1 mole of K has a We know that 1 mole of K has a mass of 39.1 grams; and mass of 39.1 grams; and

1 mole is 6.02 x 101 mole is 6.02 x 102323 atoms. atoms.

50.0 g K 50.0 g K 39.1 g K39.1 g K

6.02 x 106.02 x 102323 atoms K atoms K

7.70 x 107.70 x 102323 atoms K atoms K

x =

Page 38: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

The The molar massmolar mass

of an element is the mass of of an element is the mass of 6.022 x 106.022 x 102323 atoms atoms

oror1 mole1 mole

of that element.of that element.

Page 39: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

FYI: We now use FYI: We now use ““molar mass” instead of.. molar mass” instead of..

gram formula weight.gram formula weight.

gram molecular weight,gram molecular weight,

gram atomic weight,gram atomic weight,

Page 40: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

The molar mass of an element is The molar mass of an element is usually rounded to one decimal place:usually rounded to one decimal place:

sodium (Na) sodium (Na) 23.0 g/mol23.0 g/mol

iron (Fe)iron (Fe) 55.8 g/mol55.8 g/mol

chlorine (Cl)chlorine (Cl) 35.5 g/mol35.5 g/mol

phosphorous (P)phosphorous (P) 31.0 g/mol31.0 g/mol

Page 41: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

What do the subscripts in a What do the subscripts in a formula tell us?formula tell us?

1 mole of 1 mole of FeFe22OO33 contains …contains …

2 moles of iron (Fe),2 moles of iron (Fe), and …and …

3 moles of oxygen (O)3 moles of oxygen (O)

The subscripts give the number of The subscripts give the number of moles of each element in one mole moles of each element in one mole of the compound.of the compound.

Page 42: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

How many moles of each kind of How many moles of each kind of element are in one mole of …element are in one mole of …

3 moles Fe, 2 moles P, 8 moles O3 moles Fe, 2 moles P, 8 moles O

12 moles C, 22 moles H, 11 moles O12 moles C, 22 moles H, 11 moles O

1 mole Ca and 2 moles Cl1 mole Ca and 2 moles ClCaClCaCl22

CC1212HH2222OO1111

FeFe33(PO(PO44))22

Page 43: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Calculating molar masses of Calculating molar masses of compounds …compounds …

Molar mass Molar mass of the of the compoundcompound

Sum of the Sum of the molar masses of molar masses of

the elementsthe elements==

The molar mass of an element its The molar mass of an element its atomic massatomic mass

Page 44: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Calculate the molar mass of water:Calculate the molar mass of water:

(2 (2 xx 1.0) + (1 1.0) + (1 xx 16.0) = 16.0) = 18.0 g/mol18.0 g/mol2 hydrogen atoms2 hydrogen atoms 1 oxygen atom1 oxygen atom

Molar mass of HMolar mass of H22O =O =

Page 45: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

General Equation for General Equation for Molar MassMolar Mass

Molar mass =Molar mass = (atomic masses)(atomic masses)

For each element, multiply it’s For each element, multiply it’s atomic mass by its subscript, atomic mass by its subscript, then add them all together.then add them all together.

Page 46: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

What is the molar mass What is the molar mass of calcium chlorate?of calcium chlorate?

Start with the formula:Start with the formula:

CaCa((ClClOO33))22

1 mol Ca,1 mol Ca, 2 mol Cl,2 mol Cl, 6 mol O6 mol O

(1 x 40.1)(1 x 40.1) ++ (2 x 35.5)(2 x 35.5) ++ (6 x 16.0)(6 x 16.0)

= 207.1 g/mol= 207.1 g/mol

Page 47: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

What are the molar masses of What are the molar masses of each of the following?each of the following?

1. HCl1. HCl

2. Fe2. Fe22OO33

3. H3. H22SOSO44

4. Ca4. Ca33(PO(PO44))22

5. (NH5. (NH44))22COCO33

36.5 g/mol36.5 g/mol

159.6 g/mol159.6 g/mol

98.1 g/mol98.1 g/mol

310.3 g/mol310.3 g/mol

96.0 g/mol96.0 g/mol

Page 48: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Molar ConversionsMolar Conversions

Use conversion factors to convert Use conversion factors to convert between moles, grams, molecules and the between moles, grams, molecules and the volumes of gases at STP.volumes of gases at STP.

1 mol = 1 molar mass1 mol = 1 molar mass

1 mol = 6.022 x 101 mol = 6.022 x 102323 particles particles

1 mol = 22.4 L of any gas at STP1 mol = 22.4 L of any gas at STP

Page 49: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Convert the following:Convert the following:

0.314 moles Fe0.314 moles Fe

3. 5.45 x 103. 5.45 x 102424 molecules of SO molecules of SO33 to moles. to moles.

2. 0.0362 moles of CaCl2. 0.0362 moles of CaCl22 to grams. to grams.

1. 17.5 grams of iron metal to moles.1. 17.5 grams of iron metal to moles.

4.02 g CaCl4.02 g CaCl22

9.05 moles SO9.05 moles SO33

Page 50: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Detailed solutions:Detailed solutions:

3

3molecules23

33molecules

24

22

2 2

SO mol 9.05

SO 10 6.02

SO mol 1 SO 10 45.5

CaCl g 4.02 CaCl mol 1

CaCl g 111.1 CaCl mol 0.0362

Fe mol 0.314 Fe g 55.8

Fe mole 1 Fe g 17.5

Page 51: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Why 22.4 L?Why 22.4 L? What is STP?What is STP?

What has this all got What has this all got to do with a mole?to do with a mole?

1 mole equals 22.4 L of any gas 1 mole equals 22.4 L of any gas at STPat STP

Page 52: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

STP stands forSTP stands for

Standard temperature and pressureStandard temperature and pressure

Page 53: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

STP …STP …Standard temperature Standard temperature

… …the freezing point of waterthe freezing point of water

0 C or 273 K0 C or 273 K

Standard pressureStandard pressure … … atmospheric pressure at sea levelatmospheric pressure at sea level

1.00 atmosphere, 101.3 kPa, 1.00 atmosphere, 101.3 kPa, 760. mm Hg, 760. Torr 760. mm Hg, 760. Torr

or 14.7 lbs/inor 14.7 lbs/in22

Page 54: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

1 mole of any gas at STP has a 1 mole of any gas at STP has a volume of 22.4 Lvolume of 22.4 L

Consider a box with a Consider a box with a volume of 22.4Lvolume of 22.4L

O2

1 atm 0 C

1.00 mol

22.4 L

If the box contains 1.00 If the box contains 1.00 mole of oxygen gas, at a mole of oxygen gas, at a temperature of 0 C, the temperature of 0 C, the pressure will be 1.00 atm.pressure will be 1.00 atm.

Page 55: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

More molar conversions …More molar conversions …

3. How many molecules are in 3. How many molecules are in 250.0 mL of NH250.0 mL of NH3 3 at STP?at STP?

2. Convert 0.456 moles of HCl(g) 2. Convert 0.456 moles of HCl(g) to liters at STP.to liters at STP.

1. Convert 3.00 liters of N1. Convert 3.00 liters of N22 gas gas

at STP to moles.at STP to moles. 0.134 mol N0.134 mol N22

10.2 L HCl10.2 L HCl

6.72 x 106.72 x 102121 molecules NH molecules NH33Click for solutions

Page 56: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Detailed solutions:Detailed solutions:

321

3

323

3

22

2

NH molecules 10 6.72

NH L 22.4

NH molecules10 6.02 NH L 0.250

HCl L 10.2 HCl mol 1

HCl L 22.4 HCl mol 0.456

N mol 0.134 N L 22.4

N mol 1 N2 L 3.00

Click to return

Page 57: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Calculations with density…Calculations with density…

We can find the density of We can find the density of a gas at STP if we know a gas at STP if we know

the the molar massmolar mass..

Recall that density is the Recall that density is the mass per unit volume …mass per unit volume …

D = D = mmVV

Page 58: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

Sample density calculation:Sample density calculation:

Q.Q. What is the density of carbon What is the density of carbon dioxide at STP?dioxide at STP?

g/L 1.96 L/mol 22.4

g/mol 44.0

V

m D

We know that the molar mass of We know that the molar mass of COCO2 2 is 44.0 g/mol and the volume of is 44.0 g/mol and the volume of

1 mole of gas is 22.4L at STP1 mole of gas is 22.4L at STP

Page 59: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

g/L 1.96 L/mol 22.4

g/mol 44.0

V

m D

Mass of Mass of 1 mole 1 mole of COof CO22

Volume Volume of 1 mole of 1 mole of COof CO22

Sample density calculation:Sample density calculation:

Q.Q. What is the density of carbon What is the density of carbon dioxide at STP?dioxide at STP?

Page 60: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

g/L 1.25 L/mol 22.4

g/mol 28.0

V

m D

1. Find the density of nitrogen 1. Find the density of nitrogen gas (Ngas (N22) at STP.) at STP.

Additional problems:Additional problems:

Page 61: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

2. Find the mass of 200.0 mL 2. Find the mass of 200.0 mL of propane (Cof propane (C33HH88) at STP.) at STP.

g .393 L/mol 22.4

g/mol 44.0 L 0.200 m

This is the density of propaneThis is the density of propane

Hint: What is the density of propane?Hint: What is the density of propane?

m = V Dm = V D

Page 62: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

g .393 L/mol 22.4

g/mol 44.0 L 0.200 m

This is the density of propaneThis is the density of propane

D = -----------------------D = ----------------------- mass of one molemass of one mole

volume of one molevolume of one mole

m = V Dm = V D

Page 63: Calculations with Elements and Compounds The Mole Concept Rev. 051111 050212 Pisgah High School M. Jones.

This concludes the This concludes the presentation on moles presentation on moles

and molar mass.and molar mass.