chemical quantities: the mole chem i / ih : chapter 10
DESCRIPTION
CHEMICAL QUANTITIES: THE MOLE chem i / Ih : Chapter 10. MEASURING MASS. A mole is a quantity of things, just as… 1 dozen= 12 things 1 gross = 144 things 1 mole= 6.02 x 10 23 things “Things” usually measured in moles are atoms, molecules, ions, and formula units. - PowerPoint PPT PresentationTRANSCRIPT
MEASURING MASS
A mole is a quantity of things, just as…
1 dozen = 12 things1 gross = 144 things1 mole = 6.02 x 1023
things “Things” usually measured in moles
are atoms, molecules, ions, and formula units
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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
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A MOLE… is an amount, defined as the
number of carbon atoms in exactly 12 grams of carbon-12
1 mole = 6.02 x 1023 of the representative particles
Treat it like a very large dozen 6.02 x 1023 is called:
Avogadro’s number
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Similar Words for an amount: Pair: 1 pair of shoelaces = 2 shoelaces
Dozen: 1 dozen oranges = 12 oranges
Gross: 1 gross of pencils= 144 pencils
Ream: 1 ream of paper= 500 sheets of paper
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What are Representative Particles (“RP”)?
The smallest pieces of a substance:1. For a molecular compound: it is
the molecule.2. For an ionic compound: it is the
formula unit (made of ions)3. For an element: it is the atom
Remember the 7 diatomic elements? (made of molecules)
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Practice Counting Particles How many oxygen atoms in the following?1. CaCO3 3 atoms of oxygen2. Al2(SO4)3 12 (4 x 3) atoms of oxygen
How many ions in the following?1. CaCl2
3 total ions (1 Ca2+ ion and 2 Cl1- ions)2. NaOH
2 total ions (1 Na1+ ion and 1 OH1- ion)3. Al2(SO4)3
5 total ions (2 Al3+ + 3 SO4 ions)
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EXAMPLES: ATOMS MOLES
How many moles of B are in 3.15 x 1023 atoms of B? Conversion: 1 mole B = 6.02
x 1023 atoms B(b/c the atom is the RP of
boron)
1 mole B
3.15 x 1023 atoms of B
6.02 x 1023 atoms B
= 0.523 mole
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EXAMPLES: MOLES ATOMS
How many atoms of Al are in 1.5 mol of Al? Conversion: 1 mole = 6.02 x 1023
atoms
1.5 mol of Al6.02 x 1023 atoms Al
1 mole Al
= 9.03 x 1023 atoms of Al
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EXAMPLES: MOLECULES MOLESHow many atoms of H are there in 3 moles of H2O? (HINT: Are atoms the RP for water?)Conversions:1 mole = 6.02 x 1023 molecules
(b/c molecules are the RP for H2O)
3 moles of H2O 6.02 x 1023 molec H2O
1 mole H2O
2 atoms H
1 H2O molecule
= 3.612 x 1024 atoms H
H2O molecule = 2 atoms of Hydrogen
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MOLAR MASSDef: The mass of a mole of representative particles of a substance.
Each element & compound has a molar mass.
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MOLAR MASS OF AN ELEMENTDetermined simply by looking at the periodic table
Molar mass (g) = Atomic Mass (amu)
Ca20
40.08
* Thus, 1 mol Ca = 40 g
1 atom of Ca weighs 40.08 amu1 mole of Ca atoms weighs 40.08 grams
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MOLAR MASS FOR COMPOUNDS
To calculate the molar mass of a compound, find the number of grams of each element in one mole of the compound
Then add the masses within the compound
Example: H2O
H= 1.01 2 (1.01) + 1 (15.999)= 18.02 g/mol
O= 15.99915
SOME PRACTICE PROBLEMS How many atoms of O are in 3.7 mol of
O? 2.2 X 1024 atoms of oxygen
How many atoms of P are in 2.3 mol of P? 1.4 x 1024 atoms of phosphorus
How many atoms of Ca are there in 2.5 moles of CaCl2? 1.5 x 1024 atoms Ca
How many atoms of O are there in 1.7 moles of SO4? 4.1 x 1024 atoms of oxygen
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Remember!!!! The molar mass of any substance (in
grams) equals 1 mole This applies to ALL substance: elements,
molecular compounds, ionic compounds Use molar mass to convert between mass
and moles Ex: Mass, in grams, of 6 mol of MgCl2 ?
mass of MgCl2 = 6 mol MgCl2 92.21 g MgCl2 1 mol
MgCl2
= 571.26 g MgCl2
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VOLUME AND THE MOLE
Volume varies with changes in temperature & pressure
Gases are predictable, under the same physical conditions
Avogadro’s hypothesis helps explain:equal volume of gases, at the same temp and pressure contains equal number of particles
Ex: helium balloon
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MOLAR VOLUME Gases vary at different
temperatures, makes it hard to measure
To make it easier to measure gases, we use something called “STP” “Standard Temperature and Pressure” Temperature = 0° C Pressure = 1 atm (atmosphere)
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Molar Volume
At STP:1 mole, 6.02 x 1023 atoms, of any gas has a volume of 22.4 L
1 mole gas = 22.4 L gas
Use it to convert between # of moles and vol of a gas @ STP Ex: what is the volume of 1.25 mol of sulfur
gas?
1.25 mol S 22.4 L S = 28.0 L
1 mol S 20
MOLAR MASS FROM DENSITY Different gases have different densities Density of a gas measured in g/L @ a specific
temperature Can use the following formula to solve :
grams = grams X 22.4 Lmole L 1 mole Ex: Density of gaseous compound containing
oxygen and carbon is 1.964 g/ L, what is the molar mass?
grams = 1.964 g X 22.4 L then you solve
mole 1 L 1 mole = 44.o g/mol
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Molarity
Def: the concentration of a solution. How many moles/liter
Can be used to calculate # of moles of a solute
Ex: Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO). How many moles of solute are present in 1.5 L of 0.70 M NaClO?
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Calculating Percent Composition of a Compound Like all percent problems: a part ÷ the
whole1. Find the mass of each of the components
(the elements)2. Next, divide by the total mass of the
compound3. Then X 100 % = percent
Formula:% Composition = Mass of element X 100%
Mass of compound
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Method #1: % Comp When Actual Masses are Given
A compound is formed when 9.03 g of Mg combines completely with 3.48 g of N.
What is the percent composition of the compound?1. First add the 2 mass of the 2 compounds to
reach the total mass 9.03 g Mg + 3.48 g N = 12.51 g Mg3N2
2. Find the % of each compound
% Mg= 9.03 g Mg X 100 = 72.2 % 12.51 g Mg3N2
% N= 3.48 g N X 100 = 27.8 % 12.51 g Mg3N2
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Method 2: % Comp When Only The Formula is Known Can find the percent composition of a
compound using just the molar mass of the compound and the element
% mass=mass of the element 1 mol cmpd X100%
molar mass of the compound Example:
Find the percent of C in CO2
12.01 g C X 100 = 27.30% C
44.01 g CO2 26
Using % Composition Can use % composition as a conversion factor
just like the mole After finding the % comp. of each element in
a cmpd. can assume the total compound = 100g
Example: C= 27.3% 27.3 g C O= 72.7 % 72.7 g O
In 100 g sample of compound there is 27.3 g of C & 72.7 g of O
How much C would be contained in 73 g of CO2?
73 g CO2 27.3 g C = 19.93 g C
100 g CO2
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EMPIRICAL FORMULAS Empirical formulas are the
lowest WHOLE number ratios of elements contained in a compound
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REMEMBER… Molecular formulas tells the actual
number of of each kind of atom present in a molecule of the compound
Ex:H2O2 HOMolecular Empirical Formula Formula
CO2 CO2
Molecular Empirical For CO2 they are the same
Formula Formula
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Formulas for ionic compounds are ALWAYS empirical (the lowest whole number ratio = can not be reduced)
Examples: NaCl MgCl2 Al2(SO4)3 K2CO3
Simplest whole number ratio for NaCl
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A formula is not just the ratio of atoms, it is also the ratio of moles
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
Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio)
Molecular: H2O C6H12O6 C12H22O11
(Correct formula)
Empirical:
(Lowest whole H2O CH2O C12H22O11
number ratio)
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CALCULATING EMPIRICAL We can get a ratio from the percent
composition 1.Assume you have a 100 g sample the
percentage become grams (75.1% = 75.1 grams)
2.Convert grams to moles3.Find lowest whole number ratio by
dividing each number of moles by the smallest value
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Example calculations Calculate the empirical formula of a
compound composed of 38.67 % C, 16.22 % H, and 45.11 %N
Assume 100 g sample, so 38.67 g C x 1 mol C =
12.0 g C 16.22 g H x 1 mol H =
1.0 g H 45.11 g N x 1 mol N =
14.0 g N *Now divide each value by the smallest value
3.22 mole C
3.22 mole N
16.22 mole H
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…Example 1
The ratio is 3.22 mol C = 1 mol C 3.22 mol N 1
mol N The ratio is 16.22 mol H = 5 mol H
3.22 mol N 1 mol N
C1H5N1 which is = CH5N
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MORE PRACTICE
A compound is 43.64 % P and 56.36 % O
What is the empirical formula? PO3
Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O
What is its empirical formula?
C4H5N2O
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EMPIRICAL TO MOLECULAR
Since the empirical formula is the lowest ratio, the actual molecule would weigh more
Divide the actual molar mass by the empirical formula mass – you get a whole number to increase each coefficient in the empirical formula
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