the mole
DESCRIPTION
The Mole. Not the type of mole we are talking about. Measuring Matter. You often measure the amount of something by one of three different methods: Count Mass Volume. Conversions Try the practice problems on page 289!!!. 12 apples = 1 dozen apples Count 12 apples 1dozen apples - PowerPoint PPT PresentationTRANSCRIPT
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The Mole
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Not the type of mole we are talking about
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Measuring Matter• You often measure the amount
of something by one of three different methods:
1. Count2. Mass3. Volume
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Conversions
• 12 apples = 1 dozen apples Count12 apples1dozen apples
1 dozen apples = .20 bushel Volume1 dozen apples.20 bushel
1 dozen apples = 2.0 kg Mass1 dozen apples2.0 kg
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What is a Mole?
Just as 12 eggs is a dozen ( a specific number of particles), a mole is 6.02 x 10 23 particles.
This number of representative particles is called Avogadro’s number.
*That’s Avogadro, notAvocado!!!!
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What is meant by “representative particles”?
Representative particles are whatever you are talking about: atoms, molecules, or formula units (ions).
The representative particle of most elements is the atom. Seven elements exist as diatomic molecules, and, as such, its representative particle is the molecule.
The seven: H2, N2, O2, F2, Cl2, Br2, I2.
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OK, so now we know that a mole is 6.02 x1023 representative particles. So how many atoms are in one mole of an compound, or how many moles are in 6.02 x 1023 atoms?
Example:
How many moles are in 6.02 x 1023 atoms of silver?
Solution:
1. Determine the conversion factor.
1 mole = 6.02 x 1023 representative particles
1 mol representative particles
6.02 x 1023 particles
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To find the number of atoms in a mole of a compound, you must determine the number of atoms in a representative formula of that compound.
Example:
How many oxygen atoms are in a mole of CO2?
6.02 x 1023 molecules CO2
1 mole CO2
2 oxygen atoms
1 molecule CO2
x1 mole CO2 x
= 12.04 x 1023 atoms O2; or, 1.204 x 1024 atoms O2.
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Molar mass• The mass of one mole is called “molar mass”• E.g. 1 mol Li = 6.94 g Li• This is expressed as 6.94 g/mol• What are the following molar masses?
S SO2
Cu3(BO3)2
32.06 g/mol 64.06 g/mol308.27 g/mol
Calculate molar masses (to 2 decimal places) CaCl2 (NH4)2CO3
O2
Pb3(PO4)2
C6H12O6
Cu x 3 = 63.55 x 3 = 190.65B x 2 = 10.81 x 2 = 21.62O x 6 = 16.00 x 6 = 96.00
308.27
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Molar mass• The mass of one mole is called “molar mass”• E.g. 1 mol Li = 6.94 g Li• This is expressed as 6.94 g/mol• What are the following molar masses?
S SO2
Cu3(BO3)2
32.06 g/mol 64.06 g/mol308.27 g/mol
Calculate molar masses (to 2 decimal places) CaCl2 (NH4)2CO3
O2
Pb3(PO4)2
C6H12O6
110.98 g/mol (Ca x 1, Cl x 2)96.11 g/mol (N x 2, H x 8, C x 1, O x 3)32.00 g/mol (O x 2)
811.54 g/mol (Pb x 3, P x 2, O x 8)180.18 g/mol (C x 6, H x 12, O x 6)
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Example:
How many grams are in 7.20 mol of N2O3?
N2 = 28.0 gO3 = 48.0 g
1 mole N2O3 = 76.0 g7.20 mol N2O3 x 76.0 g N2O3
1 mol N2O3
= 547.2 g N2O3
= 5.47 x 102 g N2O3
You Try It!
1. How many grams in .720 mol Be?
2. How many moles in 2.40 g N2?
6.48 g
0.086 mol
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Now let’s see if the fog has lifted….
Easy peasy!!
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Example
• If I have 2.00 moles of C13H18O2, how many moles of each atom would I have?
C mol 26.0OHC mol 1
C mol 13OHC mol 2.0021813
21813
H mol 36.0OHC mol 1
H mol 18OHC mol 2.0021813
21813
O mol 4.00OHC mol 1
O mol 2OHC mol 2.0021813
21813
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Formula Mass• mass of a molecule, ion, or formula unit• sum of mass of all atoms in the chemical formula• in amu• Ex: H2O
• 18.01528 amu• formula mass = molecular mass for molecular
compound
amu 2.01588atom H 1
amu 1.00794atoms H 2
amu 9994.51atom O 1
amu 15.9994atoms O 1
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Example
• Find formula mass of potassium chlorate.
• KClO3
• 1(39.0983) + 1(35.4527) + 3(15.9994)
• 122.549 amu
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Molar Masses
• mass of one mole of pure substance• numerically equal to formula mass • units: g/mol
• Find molar mass of barium nitrate.• Ba(NO3)2
• 1(137.327) + 2(14.00674) + 6(15.9994)• 261.337 g/mol
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Molar Mass in Conversions
• can be used as a conversion factor• between grams and moles
• What is the mass in grams of 2.50 mol of oxygen gas?
22
2 O g 80.0 79.997O mol 1
g 2(15.9994)O mol 2.50
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Example
• Ibuprofen, C13H18O2, is the active ingredient in many pain relievers.
• Find molar mass:
(13 x 12.011) + (18 x 1.00794) + (2 x 15.9994) =206.29 g/mol
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Example
• If the tablets in a bottle contain a total of 33 g of ibuprofen, how many moles are in one bottle?
• How many molecules of ibuprofen are in the bottle?
2181321813
2181321813 OHC mol 16.0
OHC g 206.29OHC mol 1OHC g 33
2181322
21813
2181323
21813 OHC molec. 10 x 9.6OHC mol 1
OHC molec. 10 x 6.022OHC mol 16.0
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Example
• What is the number of moles of carbon in that bottle?
• What is the total mass in grams of carbon in the bottle?
C mol 1.2OHC mol 1
C mol 13OHC mol 16.021813
21813
C g 25C mol 1
C g 12.011C mol 2.1
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The Amount of A Mole of Gas
We have seen that one-mole amounts of liquids or solids have different volumes than other solids or liquids. What about the volume of gas?
Moles of gases have very predictable volumes.
Changing temperature or pressure of a gas can vary the volume. That is why the volume of a gas is measured at a standard temperature and pressure (STP).
Standard temperature is 0°C. Standard pressure is 101.2 kPa or 1 atmosphere (atm)
At STP, 1 mol of any gas occupies a volume of 22.4 L.
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22.4 L is known as the molar volume of a gas which means it contains 6.02 x 1023 representative particles of that gas.
Example:
Determine the volume, in liters, of 0.600 mol of SO2 gas at STP.
0.600 mol SO2 x 22.4 L SO2
1 mol SO2
= 13.4 L SO2
Assuming STP, how many moles are in 67.2 L SO2?
67.2 L SO2 x 1 mole SO2
22.4 L SO2
= 3 mol SO2
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Review:
1. What volume, in liters, will 0.680 mol of a certain gas occupy?
2. How many moles is 1.33 x 104 mL of O2 at STP?
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Mass/Density of a Gas
Would 22.4 L of one gas also have the same mass as 22.4 L of another gas at STP?
Probably not. A mole of one gas have a mass equal to its gfm. Different gases usually have different gfm’s.
Measuring the volume of a gas is preferred to measuring mass. Knowing the volume can also help find the density of the gas.
Density is found by dividing the mass of a gas by its volume. Because volume can change with a change in temperature, density is measured at STP.
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Density (at STP) =
molar massmolar volume =
g/mol22.4 L/mol
=gL
Example:
What is the density of oxygen gas at STP? (in grams per liter.)
D = molar massmolar volume
= 32 g/mol22.4 L/mol
= 1.43 g/L
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Percent Composition
To keep your lawn healthy, wealthy, and wise you need to use fertilizer. You can’t just use any ol’ fertilizer. You need to use one that has the right mixture of elements or compounds depending on what you need to do.
You need to know the relative amount of each nutrient. This is the same in the laboratory. When you make a new compound, you need to determine its formula by finding the relative amounts of elements in the compound.
The relative amounts are expressed as the percent composition, the percent by mass of each element in a compound.
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There are as many percent values as there are elements in the compound.
The percentages must add up to 100%.
% mass of element X = grams of element X grams of compound x 100%
Example:
An 8.20-g piece of magnesium combines with a 5.40-g sample of oxygen completely to form a compound. What is the percent composition of this compound?
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1. Find mass of compound.
13.60 g = mass of compound (8.20 g + 5.40 g)
2. Find % of each element
% Mg = mass Mgmass cmpd. =
8.20 g13.6 g
= 60.3%
% O = mass Omass cmpd. =
5.40 g 13.6 g
= 39.7 %
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Once you determine the percent composition of a compound you can determine the number of grams of an element in a specific amount of compound.
Example:
Calculate the mass of carbon in 82.0 g of propane, C3H8.
1. Determine % composition of C3H8.
C3H8 = 44.0 g
C:36 g44 g = 81.8 %
H: 8 g44 g = 18.2 % H
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2. Use conversion factor based on percent by mass of carbon in ethane.
a. 81.8 % C means that for every 100 g C3H8, 81.8 g will be C.
82 g C3H8 x 81.8 g C100 g C3H8
= 67.1 g C
Check to see if 67.1/82 = 81.8 %
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Problems:
Calculate the amount of hydrogen in:
a. 350 g C2H6.
b. 20.2 g NaHSO4
c. 124 g Ca(C2H3O2) 2
d. 378 g HCN
e. 100 g H2O
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Empirical Formulas
Once you make a new compound in the laboratory, you can determine the percent composition information. Once you know the percent composition, you can determine the empirical formula of the compound.
The empirical formula gives the lowest whole number ratio of the atoms of the elements in a compound.
CO2 is an empirical formula because it is the lowest whole-number ratio. N2H4 (an explosive) has an empirical formula of NH2.
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What is the empirical formula of a compound that is 25.9% N and 74.1% O?
1. Remember % composition means that in 100 g of that compound each % equals that many grams.
2. Change g to moles
25.9 g N x1 mol N14.0 g N = 1.85 mol N
74.1 g O x1 mol O
16.0 g O= 4.63 mol O
N1.85O4.63 = mole ratio
Not empirical formula because needs to be whole-numbers.
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3. Divide both molar values by smallest value to give you a “1” for element with smallest value.
1.85 mol N 1.85
= 1 mol N4.63 mol O 1.85
= 2.50 mol O
Is N1O2.5 correct? No, not a whole number.
Just multiply both values by a number to make a whole number.
1 x 2 = 2 mol N; 2.50 x 2 = 5 mol O
Now you have whole numbers.
empirical formula: N2O5
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If grams are already given to you, just convert to moles.
Example:
Analysis of a compound indicates it contains 2.08 g K, 1.40 g Cr, and 1.74 g O. Find its empirical formula.
2.08 g K x 1 mol K39.1 g K
= .053 mol K
1.40 g Cr x 1 mol Cr 52 g Cr
= .027 mol Cr
1.74 g O x 1 mol O16.00 g O
= .109 mol O
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Then divide by smallest mole number to get mole ratio. Multiply if you need to.
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Molecular Formula
Determining the empirical formula does not always tell you the actual molecular formula.
An example is methanol and glucose.
CH2O = methanol (empirical and molecular)
C6H12O6 = glucose (molecular)
Same empirical; different molecular formula.
Well, then, how do you know if you have empirical or molecular?
You need to know the molar mass of the compound.
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The molecular formula is some multiple of the empirical formula based on their masses.
(simplest formula)x = molecular formula,
where “x” = whole-number multiple
(simplest-formula mass)x = molecular-formula mass
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Example:
The simplest formula of a compound containing phosphorus and oxygen was found to be P2O5. The molar mass of this compound is 283.889 g/mol. What is the molecular formula of this compound?
(simplest-formula mass)x = molecular-formula mass
x =molecular-formula mass simplest-formula mass
= 283.889 g/mol 141.945 g/mol
= 1.999
(P2O5)2 = molecular formula
= P4O10