Chapter 10 Chapter 10

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Chapter 10 Chapter 10

Chemical Quantities Chemical Quantities

oror

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How you measure how much?



You can measure mass, You can measure mass,



or volume, or volume,



or you can count pieces. or you can count pieces.



We measure mass in grams. We measure mass in grams.



We measure volume in liters. We measure volume in liters.



We count pieces in We count pieces in MOLES. MOLES.

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Moles Moles



Defined as the number of carbon Defined as the number of carbon

atoms in exactly 12 grams of carbon- atoms in exactly 12 grams of carbon- 12. 12.



1 mole is 6.02 x 10 1 mole is 6.02 x 10

²³²³

particles. particles.



Treat it like a very large dozen Treat it like a very large dozen



6.02 x 10 6.02 x 10

²³²³

is called Avogadro's is called Avogadro's number.

number.

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Representative particles Representative particles



The smallest pieces of a substance. The smallest pieces of a substance.



For an element it is an For an element it is an atom atom . . – Unless it is diatomic Unless it is diatomic



For a molecular compound it is a For a molecular compound it is a molecule

molecule . .



For an ionic compound it is a For an ionic compound it is a formula unit

formula unit . .

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Conversion factors Conversion factors



Used to change units. Used to change units.



Three questions Three questions

– What unit do you want to get rid of? What unit do you want to get rid of?

– Where does it go to cancel out? Where does it go to cancel out?

– What can you change it into? What can you change it into?

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Calculation question Calculation question



How many molecules of CO How many molecules of CO

₂₂

are the in are the in 4.56 moles of CO

4.56 moles of CO

₂₂

? ?

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Calculation question Calculation question



How many moles of water is 5.87 x 10 How many moles of water is 5.87 x 10

²²²²

molecules?

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Calculation question Calculation question



How many atoms of carbon are there in How many atoms of carbon are there in 1.23 moles of C

1.23 moles of C

₆₆

H H

₁₂₁₂

O O

₆₆

? ?

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Measuring Moles Measuring Moles



The amu was one twelfth the mass of a The amu was one twelfth the mass of a carbon 12 atom.

carbon 12 atom.



Since the mole is the number of atoms Since the mole is the number of atoms in 12 grams of carbon-12,

in 12 grams of carbon-12,



the decimal number on the periodic the decimal number on the periodic table is

table is

– The mass of the average atom in The mass of the average atom in amu amu

– the mass of 1 mole of those atoms in the mass of 1 mole of those atoms in grams.

grams.

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Gram Atomic Mass Gram Atomic Mass



The mass of 1 mole of an element in The mass of 1 mole of an element in grams.

grams.



12.01 grams of carbon has the same 12.01 grams of carbon has the same number of atoms as 1.01 grams of number of atoms as 1.01 grams of hydrogen and 55.85 grams of iron.

hydrogen and 55.85 grams of iron.



We can write this as We can write this as

12.01 g C = 1 mole 12.01 g C = 1 mole



We can count things by weighing We can count things by weighing them.

them.

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Examples Examples



How much would 2.34 moles of How much would 2.34 moles of carbon weigh?

carbon weigh?

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Examples Examples



How many moles of magnesium in How many moles of magnesium in 4.61 g of Mg?

4.61 g of Mg?

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Examples



How much would 3.45 x 10 How much would 3.45 x 10

²²²²

atoms atoms of U weigh?

of U weigh?

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What about compounds?



in 1 mole of H in 1 mole of H

₂₂

O molecules there are O molecules there are two moles of H atoms and 1 mole of two moles of H atoms and 1 mole of

O atoms O atoms



To find the mass of one mole of a To find the mass of one mole of a compound

compound

– determine the moles of the determine the moles of the elements they have

elements they have

– Find out how much they would Find out how much they would weigh

weigh

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What about compounds?



What is the mass of one mole of CH What is the mass of one mole of CH

₄₄

? ?



1 mole of C = 12.01 g 1 mole of C = 12.01 g



4 mole of H x 1.01 g = 4.04g 4 mole of H x 1.01 g = 4.04g



1 mole CH 1 mole CH

₄₄

= 12.01 + 4.04 = 16.05g = 12.01 + 4.04 = 16.05g

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Molar Mass Molar Mass



The mass of 1 mole The mass of 1 mole



What is the molar mass of Fe What is the molar mass of Fe

₂₂

O O

₃₃

? ?



2 moles of Fe x 55.85 g = 111.70 g 2 moles of Fe x 55.85 g = 111.70 g



3 moles of O x 16.00 g = 48.00 g 3 moles of O x 16.00 g = 48.00 g



The GFM = 111.70 g + 48.00 g = 159.70g The GFM = 111.70 g + 48.00 g = 159.70g

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Using Molar Mass Using Molar Mass

Finding moles of compounds

Counting pieces by weighing

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Molar Mass Molar Mass



The number of grams in 1 mole of The number of grams in 1 mole of atoms, formula units, or molecules.

atoms, formula units, or molecules.



We can make conversion factors We can make conversion factors from these.

from these.



To change grams of a compound to To change grams of a compound to moles of a compound.

moles of a compound.



Or moles to grams Or moles to grams

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For example For example



How many moles is 5.69 g of NaOH? How many moles is 5.69 g of NaOH?

5 69 . g 1 mole 40.00 g



 





need to change grams to moles



for NaOH



1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g



1 mole NaOH = 40.00 g

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For example For example



How many moles is 5.69 g of NaOH? How many moles is 5.69 g of NaOH?

5 69 . g 1 mole

40.00 = 0.142 mol NaOH g



 





need to change grams to moles



for NaOH



1mole Na = 22.99g 1 mol O = 16.00 g

1 mole of H = 1.01 g

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Gases and the Mole

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Gases Gases



Many of the chemicals we deal with are Many of the chemicals we deal with are gases.

gases.



They are difficult to weigh, so we’ll They are difficult to weigh, so we’ll measure volume

measure volume



Need to know how many moles of gas Need to know how many moles of gas we have.

we have.



Two things affect the volume of a gas Two things affect the volume of a gas



Temperature and pressure Temperature and pressure



Compare at the same temp. and Compare at the same temp. and pressure.

pressure.

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Standard Temperature and Standard Temperature and

Pressure Pressure



Avogadro's Hypothesis Avogadro's Hypothesis - at the same - at the same

temperature and pressure equal volumes temperature and pressure equal volumes of gas have the same number of particles.

of gas have the same number of particles.



0ºC and 1 atmosphere pressure 0ºC and 1 atmosphere pressure



Abbreviated atm Abbreviated atm



273 K and 101.3 kPa 273 K and 101.3 kPa



kPa is kiloPascal kPa is kiloPascal

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At At S S tandard tandard T T emperature emperature and and P P ressure ressure



abbreviated STP abbreviated STP



At STP 1 mole of gas occupies 22.4 L At STP 1 mole of gas occupies 22.4 L



Called the Called the molar volume molar volume



Used for conversion factors Used for conversion factors



Moles to Liter and L to mol Moles to Liter and L to mol

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Examples Examples

 What is the volume of 4.59 What is the volume of 4.59 mole of CO

mole of CO

₂₂

gas at STP? gas at STP?

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Density of a gas Density of a gas



D = m /V D = m /V



for a gas the units will be g / L for a gas the units will be g / L



We can determine the density of any gas We can determine the density of any gas at STP if we know its formula.

at STP if we know its formula.



To find the density we need the mass To find the density we need the mass and the volume.

and the volume.



If you assume you have 1 mole than the If you assume you have 1 mole than the mass is the molar mass (PT)

mass is the molar mass (PT)



At STP the volume is 22.4 L. At STP the volume is 22.4 L.

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Examples Examples



Find the density of CO Find the density of CO

₂₂

at STP. at STP.

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Quizdom



Find the density of CH Find the density of CH

₄₄

at STP. at STP.

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The other way The other way



Given the density, we can find the Given the density, we can find the molar mass of the gas.

molar mass of the gas.



Again, pretend you have a mole at Again, pretend you have a mole at STP, so V = 22.4 L.

STP, so V = 22.4 L.



m = D x V m = D x V



m is the mass of 1 mole, since you m is the mass of 1 mole, since you have 22.4 L of the stuff.

have 22.4 L of the stuff.



What is the molar mass of a gas with a What is the molar mass of a gas with a density of 1.964 g/L?

density of 1.964 g/L?

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All the things we can change

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Volume

Atoms Ions

Representative Particles

PT Mass Moles

6.02 x 10

²³

22.4 L

Count

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Percent Composition Percent Composition



Like all percents Like all percents



Part x 100 % Part x 100 % whole

whole



Find the mass of each component, Find the mass of each component,



divide by the total mass. divide by the total mass.

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Example Example



Calculate the percent composition of Calculate the percent composition of a compound that is 29.0 g of Ag with a compound that is 29.0 g of Ag with

4.30 g of S.

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Getting it from the formula Getting it from the formula



If we know the formula, assume you If we know the formula, assume you have 1 mole.

have 1 mole.



Then you know the pieces and the Then you know the pieces and the whole.

whole.

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Examples Examples



Calculate the percent composition of Calculate the percent composition of

C C

₂₂

H H

₄₄

? ?

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Examples Examples



What is the percent composition of What is the percent composition of Aluminum carbonate.

Aluminum carbonate.

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Percent to Mass Percent to Mass



Multiply % by the total mass to find Multiply % by the total mass to find the mass of that component.

the mass of that component.



How much aluminum in 450 g of How much aluminum in 450 g of aluminum carbonate?

aluminum carbonate?

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Empirical Formula Empirical Formula

From percentage to formula

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The Empirical Formula The Empirical Formula



The lowest whole number ratio of The lowest whole number ratio of elements in a compound.

elements in a compound.



The molecular formula the actual ratio The molecular formula the actual ratio of elements in a compound.

of elements in a compound.



The two can be the same. The two can be the same.



CH CH

₂₂

empirical formula empirical formula



C C

₂₂

H H

₄₄

molecular formula molecular formula



C C

₃₃

H H

₆₆

molecular formula molecular formula



H H

₂₂

O both O both

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Finding Empirical Formulas Finding Empirical Formulas



Just find the lowest whole number ratio Just find the lowest whole number ratio



C C

₆₆

H H

₁₂₁₂

O O

₆₆



CH CH

₄₄

N N

₂₂



It is not just the ratio of atoms, it is also It is not just the ratio of atoms, it is also the ratio of moles of atoms.

the ratio of moles of atoms.

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Calculating Empirical Formulas Calculating Empirical Formulas



Means we can get ratio from percent Means we can get ratio from percent composition.

composition.



Assume you have a 100 g. Assume you have a 100 g.



The percentages become grams. The percentages become grams.



Turn grams to moles. Turn grams to moles.



Find lowest whole number ratio by Find lowest whole number ratio by dividing everything by the smallest dividing everything by the smallest

moles.

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Example Example



Calculate the empirical formula of a Calculate the empirical formula of a compound composed of 38.67 % C, compound composed of 38.67 % C,

16.22 % H, and 45.11 %N.



Assume 100 g so Assume 100 g so



38.67 g C x 1mol C = 3.220 mole C 38.67 g C x 1mol C = 3.220 mole C 12.01 gC

12.01 gC



16.22 g H x 1mol H = 16.1 mole H 16.22 g H x 1mol H = 16.1 mole H 1.01 gH

1.01 gH



45.11 g N x 1mol N = 3.220 mole N 45.11 g N x 1mol N = 3.220 mole N

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Example Example



The ratio is 3.220 mol C = 1 mol C The ratio is 3.220 mol C = 1 mol C 3.220 molN 1 mol N 3.220 molN 1 mol N



The ratio is 16.1 mol H = 5 mol H The ratio is 16.1 mol H = 5 mol H 3.220 molN 1 mol N 3.220 molN 1 mol N



C C

₁₁

H H

₅₅

N N

₁₁



Caffeine is 49.48% C, 5.15% H, Caffeine is 49.48% C, 5.15% H,

28.87% N and 16.49% O. What is its 28.87% N and 16.49% O. What is its

empirical formula?

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Empirical to molecular Empirical to molecular



Caffeine is 49.48% C, 5.15% H, 28.87% N Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical

and 16.49% O. What is its empirical formula?

formula?



Since the empirical formula is the lowest Since the empirical formula is the lowest ratio the actual molecule would weigh the ratio the actual molecule would weigh the

same or more.



By a whole number multiple. By a whole number multiple.



Divide the actual molar mass by the the Divide the actual molar mass by the the

mass of one mole of the empirical formula.



You will get a whole number. You will get a whole number.

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Example Example



A compound has an empirical

formula of ClCH

₂

and a molar mass of 98.96 g/mol. What is its molecular formula?



A compound has an empirical

formula of CH

₂

O and a molar mass of 180.0 g/mol. What is its molecular

formula?

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Percent to molecular Percent to molecular



Take the percent x the molar mass Take the percent x the molar mass

– This gives you mass in one mole of the This gives you mass in one mole of the compound

compound



Change this to moles Change this to moles

– You will get whole numbers You will get whole numbers – These are the subscripts These are the subscripts



Caffeine is 49.48% C, 5.15% H, 28.87% N Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. It has a molar mass of 194 and 16.49% O. It has a molar mass of 194