Chapter 6 Chemical Quantities How you measure how much? You can measure mass, or volume, or you can...
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Transcript of Chapter 6 Chemical Quantities How you measure how much? You can measure mass, or volume, or you can...
Chapter 6
Chemical Quantities
How you measure how much?
• You can measure mass,
• or volume,
• or you can count pieces.
• We measure mass in grams.
• We measure volume in liters.
• We count pieces in MOLES.
Moles
• Defined as the number of carbon atoms in exactly 12 grams of carbon-12.
• 1 mole is 6.02 x 1023 particles.
• Treat it like a very large dozen
• 6.02 x 1023 is called Avagadro’s number.
Representative particles
• The smallest pieces of a substance.
• For a molecular compound it is a molecule.
• For an ionic compound it is a formula unit.
• For an element it is an atom.
Types of questions
• How many oxygen atoms in the following?– CaCO3
– Al2(SO4)3
• How many ions in the following?– CaCl2– NaOH
– Al2(SO4)3
Types of questions
• How many molecules of CO2 are the in
4.56 moles of CO2 ?
• How many moles of water is 5.87 x 1022 molecules?
• How many atoms of carbon are there in 1.23 moles of C6H12O6 ?
• How many moles is 7.78 x 1024 formula units of MgCl2?
Measuring Moles
• Remember relative atomic mass?
• The amu was one twelfth the mass of a carbon 12 atom.
• Since the mole is the number of atoms in 12 grams of carbon-12,
• the decimal number on the periodic table is also the mass of 1 mole of those atoms in grams.
Gram Atomic Mass
• The mass of 1 mole of an element in grams.
• 12.01 grams of carbon has the same number of pieces as 1.008 grams of hydrogen and 55.85 grams of iron.
• We can right this as 12.01 g C = 1 mole
• We can count things by weighing them.
Examples
• How much would 2.34 moles of carbon weigh?
• How many moles of magnesium in 24.31 g of Mg?
• How many atoms of lithium in 1.00 g of Li?
• How much would 3.45 x 1022 atoms of U weigh?
What about compounds?
• in 1 mole of H2O molecules there are two moles
of H atoms and 1 mole of O atoms• To find the mass of one mole of a compound
– determine the moles of the elements they have– Find out how much they would weigh– add them up
What about compounds?
• What is the mass of one mole of CH4?
• 1 mole of C = 12.01 g
• 4 mole of H x 1.01 g = 4.04g
• 1 mole CH4 = 12.01 + 4.04 = 16.05g
• The Gram Molecular mass of CH4 is 16.05g
• The mass of one mole of a molecular compound.
Gram Formula Mass
• The mass of one mole of an ionic compound.
• Calculated the same way.
• What is the GFM of Fe2O3?
• 2 moles of Fe x 55.85 g = 111.70 g
• 3 moles of O x 16.00 g = 48.00 g
• The GFM = 111.70 g + 48.00 g = 159.70g
Molar Mass
• The generic term for the mass of one mole.
• The same as gram molecular mass, gram formula mass, and gram atomic mass.
Examples
• Calculate the molar mass of the following and tell me what type it is.
• Na2S
• N2O4
• C
• Ca(NO3)2
• C6H12O6
• (NH4)3PO4
Using Molar Mass
Finding moles of compounds
Counting pieces by weighing
Molar Mass
• The number of grams of 1 mole of atoms, ions, or molecules.
• We can make conversion factors from these.
• To change grams of a compound to moles of a compound.
For example
• How many moles is 5.69 g of NaOH?
For example
• How many moles is 5.69 g of NaOH?
5 69. g
For example
• How many moles is 5.69 g of NaOH?
5 69. g mole
g
need to change grams to moles
For example
• How many moles is 5.69 g of NaOH?
5 69. g mole
g
need to change grams to moles for NaOH
For example
• How many moles is 5.69 g of NaOH?
5 69. g mole
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
For example
• How many moles is 5.69 g of NaOH?
5 69. g mole
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
For example
• 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
For example
• 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 1 mole NaOH = 40.00 g
Examples
• How many moles is 4.56 g of CO2 ?
• How many grams is 9.87 moles of H2O?
• How many molecules in 6.8 g of CH4?
• 49 molecules of C6H12O6 weighs how
much?
Gases and the Mole
Gases
• Many of the chemicals we deal with are gases.
• They are difficult to weigh.• Need to know how many moles of gas
we have.• Two things effect the volume of a gas• Temperature and pressure• Compare at the same temp. and
pressure.
Standard Temperature and Pressure
• 0ºC and 1 atm pressure
• abbreviated STP
• At STP 1 mole of gas occupies 22.4 L
• Called the molar volume
• Avagadro’s Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles.
Examples
• What is the volume of 4.59 mole of CO2
gas at STP?
• How many moles is 5.67 L of O2 at STP?
• What is the volume of 8.8g of CH4 gas at
STP?
Density of a gas
• D = m /V• for a gas the units will be g / L• We can determine the density of any gas at
STP if we know its formula.• To find the density we need the mass and the
volume.• If you assume you have 1 mole than the mass
is the molar mass (PT)• At STP the volume is 22.4 L.
Examples
• Find the density of CO2 at STP.
• Find the density of CH4 at STP.
The other way• Given the density, we can find the molar
mass of the gas.• Again, pretend you have a mole at STP, so V
= 22.4 L.• m = D x V• m is the mass of 1 mole, since you have 22.4
L of the stuff.• What is the molar mass of a gas with a
density of 1.964 g/L?• 2.86 g/L?
All the things we can change
We have learned how to
• change moles to grams
• moles to atoms
• moles to formula units
• moles to molecules
• moles to liters
• molecules to atoms
• formula units to atoms
• formula units to ions
Moles
Mass
Moles
MassPT
Moles
MassVolume PT
Moles
MassVolume PT22.4 L
Moles
MassVolume
Representative Particles
PT22.4 L
6.02 x 1023
Moles
MassVolume
Representative Particles
PT22.4 L
Moles
MassVolume
Representative Particles
6.02 x 1023
PT
Atoms
22.4 L
Moles
MassVolume
Representative Particles
6.02 x 1023
PT
Atoms Ions
22.4 L
Percent Composition
• Like all percents
• Part x 100 % whole
• Find the mass of each component,
• divide by the total mass.
Example
• Calculate the percent composition of a compound that is 29.0 g of Ag with 4.30 g of S.
Getting it from the formula
• If we know the formula, assume you have 1 mole.
• Then you know the pieces and the whole.
Examples
• Calculate the percent composittion of C2H4?
• Aluminum carbonate.
Empirical Formula
From percentage to formula
The Empirical Formula
• The lowest whole number ratio of elements in a compound.
• The molecular formula the actual ration of elements in a compound.
• The two can be the same. • CH2 empirical formula
• C2H4 molecular formula
• C3H6 molecular formula
• H2O both
Calculating Empirical
• Just find the lowest whole number ratio
• C6H12O6
• CH4N• It is not just the ratio of atoms, it is also the
ratio of moles of atoms.
• In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen.
• In one molecule of CO2 there is 1 atom of C and 2 atoms of O.
Calculating Empirical
• Means we can get ratio from percent composition.
• Assume you have a 100 g.
• The percentages become grams.
• Can turn grams to moles.
• Find lowest whole number ratio by dividing by the smallest.
Example
• Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.
• Assume 100 g so• 38.67 g C x 1mol C = 3.220 mole C
12.01 gC • 16.22 g H x 1mol H = 16.09 mole H
1.01 gH• 45.11 g N x 1mol N = 3.219 mole N
14.01 gN
Example
• The ratio is 3.220 mol C = 1 mol C 3.219 molN 1 mol N
• The ratio is 16.09 mol H = 5 mol H 3.219 molN 1 mol N
• C1H5N1
• A compound is 43.64 % P and 56.36 % O. What is the empirical formula?
• Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?
Empirical to molecular
• Since the empirical formula is the lowest ratio the actual molecule would weigh more.
• By a whole number multiple.
• Divide the actual molar mass by the the mass of one mole of the empirical formula.
• Caffeine has a molar mass of 194 g. what is its molecular mass?
Example
• A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mas is known (from gas density) is known to be 98.96 g. What is its molecular formula?
Chapter 8
Stoichiometry• Greek for “measuring elements”
• The calculations of quantities in chemical reactions based on a balanced equation.
• We can interpret balanced chemical equations several ways.
In terms of Particles• Atom - Element
• Molecule – Molecular compound (non- metals)
– or diatomic (O2 etc.)
• Formula unit – Ionic Compounds (Metal and non-metal)
2H2 + O2 2H2O
• Two molecules of hydrogen and one molecule of oxygen form two molecules of water.
• 2 Al2O3 Al + 3O2
2 formula units Al2O3 form 4 atoms Al
and 3 molecules O2
2Na + 2H2O 2NaOH + H2
2Na + 2H2O 2NaOH + H2
Look at it differently• 2H2 + O2 2H2O
• 2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen molecules of water.
• 2 x (6.02 x 1023) molecules of hydrogen and 1 x (6.02 x 1023) molecules of oxygen form 2 x (6.02 x 1023) molecules of water.
• 2 moles of hydrogen and 1 mole of oxygen form 2 moles of water.
In terms of Moles
• 2 Al2O3 Al + 3O2
• The coefficients tell us how many moles of each kind
36.04 g reactants36.04 g reactants
In terms of mass• The law of conservation of mass applies
• We can check using moles
• 2H2 + O2 2H2O
2 moles H2
2.02 g H2
1 moles H2
= 4.04 g H2
1 moles O2
32.00 g O2
1 moles O2
= 32.00 g O2
In terms of mass
• 2H2 + O2 2H2O
2 moles H2O18.02 g H2O
1 mole H2O= 36.04 g H2O
2H2 + O2 2H2O
36.04 g (H2 + O2) = 36.04 g H2O
Your turn• Show that the following equation follows
the Law of conservation of mass.
• 2 Al2O3 Al + 3O2
Mole to mole conversions• 2 Al2O3 Al + 3O2
• every time we use 2 moles of Al2O3 we make 3 moles of O2
2 moles Al2O3
3 mole O2
or2 moles Al2O3
3 mole O2
Mole to Mole conversions• How many moles of O2 are produced when 3.34 moles of Al2O3
decompose?
• 2 Al2O3 Al + 3O2
3.34 moles Al2O3 2 moles Al2O3
3 mole O2 = 5.01 moles O2
Your Turn
• 2C2H2 + 5 O2 4CO2 + 2 H2O
• If 3.84 moles of C2H2 are burned, how
many moles of O2 are needed?
• How many moles of C2H2 are needed to
produce 8.95 mole of H2O?
• If 2.47 moles of C2H2 are burned, how
many moles of CO2 are formed?
Mole to Mole Conversions
• 2C2H2 + 5 O2 4CO2 + 2 H2O
• How many moles of C2H2 are needed to
produce 8.95 mole of H2O?
• If 2.47 moles of C2H2 are burned, how
many moles of CO2 are formed?
We can’t measure moles!!• What can we do?
• We can convert grams to moles.
• Periodic Table
• Then use moles to change chemicals
• Balanced equation
• Then turn the moles back to grams.
• Periodic table
Periodic Table
MolesA
MolesB
Massg B
Periodic Table
Balanced Equation
Massg A
•Decide where to start based on the units you are given
•and stop based on what unit you are asked for
Conversions
• 2C2H2 + 5 O2 4CO2 + 2 H2O
• How many moles of C2H2 are needed to
produce 8.95 g of H2O?
• If 2.47 moles of C2H2 are burned, how
many g of CO2 are formed?
For example...• If 10.1 g of Fe are added to a solution of
Copper (II) Sulfate, how much solid copper would form?
• Fe + CuSO4 Fe2(SO4)3 + Cu
• 2Fe + 3CuSO4 Fe2(SO4)3 + 3Cu 10.1 g Fe
55.85 g Fe1 mol Fe
2 mol Fe3 mol Cu
1 mol Cu63.55 g Cu
= 17.3 g Cu
2Fe + 3CuSO4 Fe2(SO4)3 + 3Cu
0.181 mol Fe2 mol Fe
3 mol Cu= 0.272 mol Cu
0.272 mol Cu1 mol Cu63.55 g Cu
= 17.3 g Cu
Could have done it
10.1 g Fe55.85 g Fe1 mol Fe
2 mol Fe3 mol Cu
1 mol Cu63.55 g Cu
= 17.3 g Cu
More Examples• To make silicon for computer chips they use
this reaction
• SiCl4 + 2Mg 2MgCl2 + Si
• How many moles of Mg are needed to make 9.3 g of Si?
• 3.74 mol of Mg would make how many moles of Si?
• How many grams of MgCl2 are produced along
with 9.3 g of silicon?
For Example• The U. S. Space Shuttle boosters use this
reaction
• 3 Al(s) + 3 NH4ClO4
Al2O3 + AlCl3 + 3 NO + 6H2O
• How much Al must be used to react with 652 g of NH4ClO4 ?
• How much water is produced?
• How much AlCl3?
How do you get good at this?
Gases and Reactions
We can also change• Liters of a gas to moles
• At STP
• 0ºC and 1 atmosphere pressure
• At STP 22.4 L of a gas = 1 mole
• If 6.45 moles of water are decomposed, how many liters of oxygen will be produced at STP?
For Example• If 6.45 grams of water are decomposed,
how many liters of oxygen will be produced at STP?
• H2O H2 + O2
• 2H2O 2H2 + O2
6.45 g H2O 18.02 g H2O1 mol H2O
2 mol H2O1 mol O2
1 mol O2
22.4 L O2
Your Turn
• How many liters of CO2 at STP will be
produced from the complete combustion of 23.2 g C4H10 ?
• What volume of oxygen will be required?
Example• How many liters of CH4 at STP are required to
completely react with 17.5 L of O2 ?
• CH4 + 2O2 CO2 + 2H2O
17.5 L O2 22.4 L O2 1 mol O2
2 mol O2 1 mol CH4
1 mol CH4 22.4 L CH4
= 8.75 L CH4
22.4 L O2 1 mol O2
1 mol CH4 22.4 L CH4
Avagadro told us• Equal volumes of gas, at the same
temperature and pressure contain the same number of particles.
• Moles are numbers of particles
• You can treat reactions as if they happen liters at a time, as long as you keep the temperature and pressure the same.
Example
• How many liters of CO2 at STP are
produced by completely burning 17.5 L of CH4 ?
• CH4 + 2O2 CO2 + 2H2O
17.5 L CH4 1 L CH4 1 L CO2
= 17.5 L CO2
Particles
• We can also change between particles and moles.
• 6.02 x 1023
– Molecules – Atoms– Formula units
Limiting Reagent• If you are given one dozen loaves of
bread, a gallon of mustard and three pieces of salami, how many salami sandwiches can you make?
• The limiting reagent is the reactant you run out of first.
• The excess reagent is the one you have left over.
• The limiting reagent determines how much product you can make
How do you find out?• Do two stoichiometry problems.
• The one that makes the least product is the limiting reagent.
• For example
• Copper reacts with sulfur to form copper ( I ) sulfide. If 10.6 g of copper reacts with 3.83 g S how much product will be formed?
• If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed?
• 2Cu + S Cu2S
10.6 g Cu 63.55g Cu 1 mol Cu
2 mol Cu 1 mol Cu2S
1 mol Cu2S
159.16 g Cu2S
= 13.3 g Cu2S
3.83 g S 32.06g S 1 mol S
1 mol S 1 mol Cu2S
1 mol Cu2S
159.16 g Cu2S
= 19.0 g Cu2S
= 13.3 g Cu2S
Cu is Limiting Reagent
How much excess reagent?
• Use the limiting reagent to find out how much excess reagent you used
• Subtract that from the amount of excess you started with
Your turn
• Mg(s) +2 HCl(g) MgCl2(s) +H2(g)
• If 10.1 mol of magnesium and 4.87 mol of HCl gas are reacted, how many moles of gas will be produced?
• How much excess reagent remains?
Your Turn II• If 10.3 g of aluminum are reacted with 51.7
g of CuSO4 how much copper will be produced?
• How much excess reagent will remain?
Yield • The amount of product made in a chemical
reaction.• There are three types• Actual yield- what you get in the lab when
the chemicals are mixed• Theoretical yield- what the balanced
equation tells you you should make.• Percent yield = Actual x 100 %
Theoretical
Example• 6.78 g of copper is produced when 3.92 g
of Al are reacted with excess copper (II) sulfate.
• 2Al + 3 CuSO4 Al2(SO4)3 + 3Cu
• What is the actual yield?
• What is the theoretical yield?
• What is the percent yield?
• If you had started with 9.73 g of Al, how much copper would you expect?
Details• Percent yield tells us how “efficient” a
reaction is.
• Percent yield can not be bigger than 100 %.
• How would you get good at this?