chemistry chapter 11 the mole
TRANSCRIPT
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Chapter 11
The Mole
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Sec. 11.1
Measuring Matter
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The Mole (mol):
SI base unit used to measure the amount of a substance.
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1 mole = 6.02 x 10²³ representative particles such as atoms, molecules, formula units, electrons, or ions.
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One mole of anything has 6.02 x 10²³ number of representative particles!
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Avogadro’s Number:
6.02 x 10²³
602 000 000 000 000 000 000 000
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Because Avogadro’s number is so enormous, it is used to count extremely small particles!
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Converting moles to particles:
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If one mole of copper contains 6.02 x 10²³ copper atoms, then how many copper atoms does two moles of copper contain?
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Answer:
2 x (6.02 x 10²³) = ? Copper atoms
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Complete practice problems p. 311
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Converting particles to
moles:
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Question:
If one mole of iron contains 6.02 x 10²³, how many moles of iron contain 2.50 x 1020 contain?
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Answer:
2.50 x 10 ²º/ 6.02 x 10²³
= ?
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Complete practice problems p. 312
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Sec 11.2Mass and the
Mole
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Would a dozen limes have the same mass as a dozen eggs?
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No, because they have different size and composition.
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Would the mass of one mole of copper differ from the mass of one mole of carbon?
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Yes, because carbon atoms differ from copper atoms.
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Molar Mass: The mass in grams of one mole of any pure substance.
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The molar mass of any element is numerically equal to its atomic mass and has the units g/mol
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What is the atomic mass of manganese?
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54.94amu
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What is the molar mass of manganese?
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54.94 g/mol
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What is the molar mass of:
Carbon= Zinc= Calcium=Iron=
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Using molar mass:
Mole=mass (g)/molar mass
(g/mol)
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Calculate the mass in grams of 0.0450 moles of chromium.
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By rearranging the previous equation:
Mass=mole x molar mass
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0.0450 mol Cr x 52.00g/mol Cr= 2.34g Cr
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How many moles of calcium are there in 525g calcium?
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mole= mass/molar mass
525g/40.08g/mol=
13.1mol Ca
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Complete practice problems p.316
(11 and 12).
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Conversion from mass to atoms:
How many atoms of gold (Au) are in a pure gold nugget having a mass of 25.0g?
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First, find the number of moles.
mol= mass/molar mass
mol= 25.0g/196.97g/mol=
0.127mol Au
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Next, find the number of particles in 0.127
mol of Au using Avogadro’s #
0.127 x 6.02 x 10²³
= 7.65 x 10²² atoms Au
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complete practice problems p. 318 (13).
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Converting atoms to mass:
A party balloon contains 5.50 x 10²² atoms of helium gas. What is the mass in grams of the helium?
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First, find the number of moles:
Mol=5.5 x 10²²/6.02 x 10²³=0.0914 mol He
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Next, find the mass in grams:
mass=mol x molar mass
0.0914mol x 4.00g/mol= 0.366g He
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Complete practice problems p. 318 (14).
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Draw the diagram of conversions on p. 319
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Sec 11.3Moles of
Compounds
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Freon CCl2F2
Freon contains one atom of carbon, two atoms of chlorine and two atoms of fluorine.
The ratio is 1:2:2
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Suppose you have one mole of freon:
The you would have Avogadro’s number of freon molecules.
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One mole of freon would have one mole of carbon atoms, two moles of chlorine atoms and two moles of fluorine atoms.
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Determine the moles of aluminum ions
(Al+3)in 1.25 moles of Al2O3
1 mole Al2O3 contains 2 mol Al+3 ions, therefore 1.25 mol contains
1.25 x 2 = 2.5 mol Al+3
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Complete practice
problems p. 321.
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The Molar Mass of Compounds:
The sum of the masses of every particle that makes up the compound.
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Determine the molar mass of K2CrO4
(2 x 39.10) + 52.00 + (4 x 16.00) = 194.20g/mol
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Complete practice problems p. 322
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Converting moles of a compound to mass:
First, find the molar mass.
Next, multiply molar mass x moles to get the mass.
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What is the mass of 2.5 moles of allyl sulfide (C3H5)2S?
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Molar mass = (6x12.01) + (10x1.008) + (1x32.07)= 114.21g/mol
Mass = 2.5 x 114.21= 286g
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Complete practice problems p. 323.
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Converting the mass of a compound to moles:First find the molar mass.
Next, find the mol by dividing mass/molar mass
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Calculate the number of moles in calcium hydroxide Ca(OH)2 in 325g
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Molar mass= (1x40.08) + (2x16.00) + (2x1.008)= 74.096g/mol
mol = 325/74.096= 4.39mol
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Complete practice problems p. 324 (30).
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Converting the mass of a compound to number of
particles:
First find the molar mass.
Next, divide mass by molar mass to find mol.
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Then, multiply by Avogadro’s number to find the number of particles.
Use the ratio of each particle from the formula to find the number of each particle.
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A sample of aluminum chloride AlCl3 has a mass of 35.6g.
a. How many Al+3 ?b. How many Cl1-?c. What is the mass in
grams of one formula unit of aluminum chloride?
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Molar mass = (1x26.98) + (3x35.45) = 133.33g/mol
Mass/molar mass=mol
35.6/133.33= 0.267mol
Mol x Avogadro’s #= number of particles.
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0.267 x 6.02 x 10²³= 1.61 x 10²³ formula units AlCl3
Number of Al+3= 1 x 1.61 x 10²³= 1.61 x 10²³ Al+3
Number of Cl1-= 3 x 1.61 x 10²³ = 4.83 x 10²³ Cl1-
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To calculate the mass of one formula unit, divide the molar mass by Avogadro’s #
133.33/6.02 x 10²³ = 2.21 x 10 ²²־ g AlCl3/formula unit
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Complete practice problems p. 326
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Copy diagram on p. 327.
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Sec 11.4
Empirical and Molecular Formulas
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Percent Composition:
The percent by mass of each element in a compound.
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If a 100g sample of a compound contains 55g of element x and 45g of element y,
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Use the equation:
percent by mass= mass of element/mass of compound x100
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For element x:55g/100gx100=55%
For element y:45g/100gx100=45%
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Percent composition can be determined from the chemical formula.
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H2OFirst find the molar mass of water:
(2x1.01)+(1x16)=18.02g/mol
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Use the molar mass as the mass of the compound water.
Next, divide the molar mass of hydrogen (element) over the molar mass of water (compound) x100
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2.02g/18.02gx100==11.2% HDo the same with oxygen:
16g/18.02gx100=88.80% O
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Percent compositions add up to 100
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Complete practice problems p. 331.
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Empirical Formula for a Compound:
The formula with the smallest whole- number mole ratio of the elements.
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The empirical formula for hydrogen peroxide is HO
The molecular formula is H2O2
In both formulas, the ratio of O to H is 1:1
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Finding empirical formula:
First, find the number of moles for each element: mol=mass/molar mass
Then change ratios to whole numbers by dividing by the lowest number of moles.
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The percent composition of an oxide of sulfur is 40.05% S and 59.95% O
100g of the oxide contains 40.05 g S and 59.95g O
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Find the mol for each:
40.05g/32.07g/mol=
1.249 mol S59.95g/16g/mol=3.747 mol O
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Change to whole numbers by dividing both by 1.249
1.249/1.249=1 mol3.747/1.249=3 molThe simplest whole number ratio of S to O is 1:3, the empirical formula is SO3
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Complete practice problems p. 333
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Molecular formula:The actual number of atoms of each element in one molecule or formula unit of a substance.
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To find the molecular formula:
First, find the number of moles for each element.
Next, find the simplest ratio by dividing number of moles of each element by smallest #
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The simplest ratio is the empirical formula.
Find the molar mass of the empirical formula.
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Divide the experimentally determined molar mass by the molar mass of the empirical formula. This is n
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Multiply the subscripts in the empirical formula by n to determine the molecular formula.
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Molecular formula =
(empirical formula) n
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Complete practice problems p. 335 and 337
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Write the diagram on p. 337
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