Moles How can we count how many atoms or molecules are in a piece of matter if we can’t see them?...
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Transcript of Moles How can we count how many atoms or molecules are in a piece of matter if we can’t see them?...
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Moles
How can we count how many atoms or molecules are in a piece of matter if we
can’t see them?
How can we count how many atoms or molecules are in a piece of matter if they
have different masses?
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Moles
What can we measure in the laboratory that will help us?
What is the “common currency”?
MOLES!
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Moles Unit
We will incorporate the following skills and knowledge:
• metric system (esp. mass)• scientific/exponent notation• conversions and dimensional analysis• sig figs
the most important mathematical tool in chemistry – the mole!
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Mole
1 mole = the amount of pure substance that contains as many particles (atoms, molecules, or fundamental units) as there are atoms in exactly 12 grams of carbon-12
(agreed upon by chemists and physicists in 1960/61)
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Mole
= 6.02 x 1023 particles
= Avogadro’s number
• Mole comes from molekül (German) = molecule
• Molecule comes from molecula (New Latin), meaning very small specimen
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Background
Lorenzo Romano Amedeo Carlo Avogadro (1776-1856)
- Italian physics professor
1811 - Avogadro's hypothesis - now a law
"Equal volumes of gases under the same conditions have equal numbers of molecules.“
“universal container”
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Background
Late 1800s – chemists – developed scale of relative atomic masses of gases, based on 1/16 of the average atomic mass of oxygen
1920s – physicists – developed relative atomic masses based on 1/16 of the oxygen-16 atom
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Background
1959-1961 – chemists and physicists - agreed to switch to carbon-12 as the standard, setting its atomic mass at 12
(pragmatic reasons – carbon-12 was the standard in mass spectroscopy, and close to chemistry’s oxygen standards)
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Moles in conversion factors
number of “particles” > moles
6.02 x 1023 particles
1 mole
mass (g) > moles
average atomic mass in g
1 mole
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Moles in conversion factors
number of “particles” > mass (g)
both of the above conversion factors, using moles as the intermediate, or “common currency”
number of atoms in a compound
# atoms
1 compound
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Math and the Mole
(# particles/mole)
1. How many fingers are there on 1 person?
2. How many fingers are there on 1 dozen people?
3. How many fingers are there on 3 dozen people?
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Math and the Mole
(# particles/mole)
4. How many fingers are there on 1 mole of people?
5. How many fingers are there on 3.12 moles of people?
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Math and the Mole
(# particles/mole)
6. How many F atoms in 3.12 moles of F?
7. How many moles of F do you have if you have 2.45 x 1022 atoms of F?
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Math and the Mole (continued)
Hint:
If problem includes numbers of atoms or other representative particles and moles, use Avogadro’s number:
6.02 x 1023 particles
If problem includes grams and moles use the periodic table to find molar mass.
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Math and the Mole (continued)(mass/mole)
8.How many grams of F are in 3.89 moles of F?
9.How many moles of F atoms in 45.6 g of F?
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Math and the Mole (continued)
(# particles/mass)
10. How many F atoms are in 65.8 g F?
11. What is the mass, in grams, of 7.62 x 1024 F atoms?
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Math and the Mole (continued)
(# atoms/compound)
12. How many F atoms are in 3.84 moles of MoF6 molecules?
A: 1.39 x 1025 F atoms
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Molar Mass
Molar mass = mass, in grams, of 1 mole of a substance (6.02 x 1023 particles)
expressed in g/mol
Molar mass is numerically equal to average atomic mass in amus (atomic mass units).
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Molar Mass
So, molar mass of oxygen (O) = mass of 1 mole of O atoms= mass of 6.02 x 1023 atoms= 16.00 g/mol
molar mass of lead (Pb)= mass of 1 mole of Pb atoms= mass of 6.02 x 1023 atoms= 207.2 g/mol
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How do amu’s compare to grams?
e.g. 1 aluminum atom weighs 26.98 amu.
By definition,
1 amu = 1/12 mass of the nucleus of 1 C-12 atom
= 1.66 x 10-24 g
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How do amu’s compare to grams?
mass of 1 Al atom in g is:
26.98 amu x 1.66 x 10-24 g = 4.48 x 10-23 g (per atom)
1 amu
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How do amu’s compare to grams?
If I have a sample of Al with a mass of 26.98 g, how many atoms do I have?
26.98 g Al x 1 atom Al = 6.02 x 1023 atoms Al
4.48 x 10-23g
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How do amu’s compare to grams?
Or how much does 6.02 x 1023 atoms Al weigh?
4.48 x 10-23 g x 6.02 x 1023 atoms Al = 27.0 g
Al atom
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How do amu’s compare to grams?
Mass of 1 mole of any element is the “molar mass” of that element.
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How do we determine the molar mass of compounds?
Add up the molar masses of all elements in the compound, taking into account the number of moles of each element.
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How do we determine the molar mass of compounds?
e.g. What is the molar mass of Na3PO4?
3 moles of Na @ 22.99 g/mol = 3 mol x 22.99 g mol
1 mole of P @ 30.97 g/mol = 1 mol x 30.97 g mol
4 moles of O @ 16.00 g/mol = 4 mol x 16.00 g mol
Add these together: = 163.94 g = molar mass of Na3PO4
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How do we use molar mass of a compound in a conversion
problem?
The same way we use molar mass of an atom…
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How do we use molar mass of a compound in a conversion
problem?
(#atoms/compound and molar mass of compounds)
13. How many Na atoms are in 252 g of Na3PO4?
(A: 2.78 x 1024 Na atoms)
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Where we’ve been(today’s lab quiz; Quiz 6 – next Th/F)
• # particles-mole and mole-# particles conversions
• mass-mole and mole-mass conversions
• # atoms-molecule/formula unit conversion
• molar mass of compounds
• combination of two or more of above conversions
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Where we’ve been(today’s lab quiz; Quiz 6 – next Th/F)
• Use the sample problems in your notes, Moles WS #1, your bookwork and the warmups to help you prepare for Quiz 6.
• Come in for ICP (T/W lunch) for more practice.
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Where we are going next – applications of mole conversions
• % composition by mass from a chemical formula (today) (Moles WS #2)
• Find the empirical chemical formula from % composition data (Moles WS #3 & 4)
• Find the molecular formula from % composition and empirical formula (Moles WS #4)
• Practice problems for all moles material (Moles WS #5)
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Where we are going next – applications of mole conversions
• Calculate molarity of a solution (Moles WS #6 & 7)
• Find the mass of a solute or volume of a solution using molarity (Moles WS #6 &7)
• Find the mass percent of a solution (Moles WS #7)
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Calculations relating to Chemical Formulas
1. Determine the molar mass of a compound:- Find molar masses of all elements.- Multiply each mass taking into account the number of atoms of that element present in the formula.- Add all masses together.
e.g. What is the molar mass of Pb3(PO4)2?
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Calculations relating to Chemical Formulas
1. Determine the molar mass of a compound:- Find molar masses of all elements.- Multiply each mass taking into account the number of atoms of that element present in the formula.- Add all masses together.
e.g. What is the molar mass of Pb3(PO4)2?811.5 g/mol
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Calculations relating to Chemical Formulas
2. Determine percent composition of all elements present in a compound:
- Find the molar masses of each element in the
compound.
- Find the total molar mass.
Calculate:
% composition = molar mass of element x 100%
total molar mass of compound
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Calculations relating to Chemical Formulas
e.g.
What is the % composition of NaCl? CaCl2?
(assume % composition by mass)
Apply % composition:
e.g. What mass of Na is present in 200.0 g NaCl?
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Calculations relating to Chemical Formulas
e.g. What is the % composition of NaCl? CaCl2? (assume % composition by mass)
NaCl: 39.34% Na, 60.66% ClCaCl2: 36.11% Ca, 63.89% Cl
Apply % composition: e.g. What mass of Na is present in 200.0 g NaCl?
200.0 g x 0.3934 = 78.68 g Na
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3. Convert % composition (mass percent) empirical formula
a. Assume a 100 g sample – use same numbers as grams rather than %.
b. Perform mass mole conversions.
c. Divide each result (mole) by the smallest result present (mole ratio).
d. Look for whole number ratio.
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3. Convert % composition (mass percent) empirical formula
Rhyme to remember order of steps to convert % composition empirical formula:
Percent to mass
Mass to mole
Divide by small
Times till whole
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Practice
1) One of the components of fresh
alkaline batteries is a black powdery
compound, of 63% manganese
and 37% oxygen.
What is the compound’s empirical
formula?
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Practice
2) While analyzing a dead alkaline battery,
Antonio finds a compound of 70.0% manganese and 30.0% oxygen.
What is its empirical formula?
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Empirical vs. Molecular Formula
Compound Empirical formula
Empirical molar mass
Molecular molar mass
Molecular formula
Formaldehyde CH2O 30.03 g 30.03 g CH2O
Acetic acid
CH2O 30.03 g 60.06 g C2H4O2
Glucose CH2O 30.03 g 180.18 g C6H12O6
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To find the molecular formula of glucose,
divide the molecular molar mass by the empirical molar mass, round to the nearest whole number.
(molecular) molar mass = 180.18 g = ~ 6
empirical molar mass 30.03 g
then multiply subscripts by 6 => C6H12O6
to get the molecular formula
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Practice Problems – Determining molecular formula
Hydrazine is 87.42% N and 12.58% H.
The (molecular) molar mass of hydrazine is 32.0 g/mol.
a) What is its empirical formula?
b) What is its molecular formula?
(Hint: find the molar mass of the empirical formula)
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Practice Problems – Determining molecular formula
Hydrazine is 87.42% N and 12.58% H.
The (molecular) molar mass of hydrazine is 32.0 g/mol.
a) What is its empirical formula? (NH2)
b) What is its molecular formula? (N2H4)
(Hint: find the molar mass of the empirical formula)
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Describing solution concentration – Percent by Mass
Mass % of component
= mass of component in solution x 100%
total mass of solution
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Describing solution concentration – Percent by Mass
e.g.
In order to maintain a sodium chloride (NaCl) concentration similar to ocean water, an aquarium must contain 3.6 g NaCl per 100.0 g of water. What is the percent by mass of NaCl in the solution?
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Describing solution concentration – Percent by Mass
3.6 g NaCl x 100% = 3.6 g NaCl x 100%
100.0 g H2O + 3.6 g NaCl 103.6 g total
= 3.5%
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Describing solution concentration – Percent by Mass
e.g.
A solution contains 2.7 g of CuSO4 in 75 mL of solution. Assume the density of the solution is 1.0 g/mL.
What is the mass percent of the solution?
2.7 g (1 mL) x 100% = 3.6%
75 mL 1.0 g
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Molarity(Moles WS #6)
• One way to measure concentration
= the # moles solute dissolved in 1 L of solution
Molarity ( M ) = moles of solute
1 liter of solution
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Molarity
• Divide # moles by # liters.
• If the problem gives you grams,
first convert to moles by dividing by
molar mass.
• Also, don't forget to convert mL to L.
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Example
A saline solution contains 0.90 g of NaCl in 100.0 mL. What is the molarity?
(Molar mass of NaCl is 58.5 g.)
• Begin by converting the solute and
solution into the correct units.
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Example
Begin by converting the solute and solution into the correct units.
Solute:0.90 g (1 mol NaCl) = 0.015 mol NaCl 58.5 g NaCl
Solution: 100.0 mL (1 L) = 0.1000 L
103 mL
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Example
Next, divide the number of moles by the volume of the solution:
• Molarity = 0.015 mol NaCl = 0.15 M NaCl
0.1000 L
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Example
OR, in one step:
0.90 g (1 mol NaCl) (103 mL) = 0.015 M NaCl
100.0 mL 58.5 g NaCl 1 L
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Molarity
• You can also use the molarity equation to calculate moles (and grams) or volumes (measured in L or mL).
If M = mol then M x L = # moles
L
and L = mol
M
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Molarity
• Notice that all three of the equations above are just different algebraic versions of each other.
• Use them to solve the next few problems, or use standard conversion techniques:
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Example of # moles = M x L
How many moles of CaCl2 are in 250. mL of 2.0 M CaCl2?
How many grams is this?
Begin with the value you are given in the problem (250. mL), then use molarity of CaCl2 (2.0 M) as a conversion factor.
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Example of # moles = M x L
250. mL (1 L) (2.0 mol CaCl2)
103 mL 1Lmolarity as conversion factor
= 0.50 mol CaCl2
then 0.50 mol CaCl2 (110.08 g CaCl2) = 55.0 g CaCl2 1 mol CaCl2
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Example of # moles = M x L
OR, all in one step:
250. mL (1 L) (2.0 mol CaCl2) (110.08 g CaCl2) 103 mL 1L 1 mol CaCl2
= 55.0 g CaCl2
molarity as conversion factor
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Example of : # L (or mL) = mol M
• How many mL contain 6.25 g of 2.00 M CaCl2?
• Begin with the value you are given in the problem (6.25 g), then use molarity of CaCl2 (2.00 M) as a conversion factor.
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Example of # L (or mL) = mol M
6.25 g CaCl2 (1 mol CaCl2) = 0.0568 mol CaCl2
110.08 g CaCl2
then
0.0568 mol CaCl2 (1 L ) (103 mL) = 28.4 mL
2.00 mol 1 L
Note upside down molarity
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Making Dilutions (Honors only)
• Chemists often use concentrated (or “stock”) solutions to make dilute solutions.
**The number of moles of solute does not change when a solution is diluted.**
Number of moles before dilution = Number of moles after dilution
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Making Dilutions
• Since moles = Molarity (M) x liters ( V ), then: Mc x Vc = Md x Vd
e.g.
How would you prepare 100. mL of 0.40 M MgSO4 from a solution of 2.0 M MgSO4 ?
Mc=2.0 M Vc= ????
Md = 0.40 M Vd = 100. mL
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Making Dilutions
Rearrange the equation:
Vc = Md x Vd
Mc
Vc = 0.40 M x 100. mL 2.0 M
= 20. mL of the concentrated solution
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Making Dilutions
So, you would measure out 20. mL of the original solution, then add enough water to it to bring the volume to 100. mL.
Important: You can do these dilution problems in either mL or L, but the V1 and V2 must both be in the same units (either both mL or both L; don't mix them up).