ch. 10 chemical quantities
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Ch. 10 Chemical Quantities. 3 Methods of Measuring. Counting Mass Volume. Example 1. If 0.20 bushel is 1 dozen apples, and a dozen apples has a mass of 2.0 kg, what is the mass of .050 bushel of apples?. Example 1. Count: 1 dozen apples = 12 apples Mass: 1 dozen apples = 2.0 kg apples - PowerPoint PPT PresentationTRANSCRIPT
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Ch. 10 Chemical Quantities
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3 Methods of Measuring
• Counting• Mass• Volume
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Example 1
• If 0.20 bushel is 1 dozen apples, and a dozen apples has a mass of 2.0 kg, what is the mass of .050 bushel of apples?
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Example 1
• Count: 1 dozen apples = 12 apples• Mass: 1 dozen apples = 2.0 kg apples• Volume: 1 dozen apples = 0.20 bushels applesConversion Factors:• 1 dozen 2.0 k.g 1 dozen
12 apples 1 dozen 0.20 bushels
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Example 1
• 0.50 bushel x 1 dozen x 2.0 kg = 0.20 bushel 1 dozen
= 5.0 kg
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Avogadro’s Number
• Named after the Italian scientist Amedo Avogadro di Quaregna
• 6.02 x 10 23
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Mole (mol)
• 1 mol = 6.02 x 10 23 representative particles
• Representative particles: atoms, molecules ions, or formula units (ionic compound)
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Mole (mol)
• Moles= representative x 1 mol particles 6.02 x 10 23
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Example 2 (atoms mol)
• How many moles is 2.80 x 10 24 atoms of silicon?
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Example 2
• 2.80 x 10 24 atoms Si x 1 mol Si 6.02 x 10 23 atoms Si
= 4.65 mol Si
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Example 3 (mol molecule)
• How many molecules of water is 0.360 moles?
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Example 3
• 0.360 mol H2O x 6.02 x 10 23 molecules H2O1 mol H2O
=2.17 molecules H2O
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The Mass of a Mole of an Element• The atomic mass of an element expressed
in grams = 1 mol of that element = molar mass
Molar mass S
Molar mass C
Molar mass Hg
Molar mass Fe
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6.02 x 10 23 atoms S
6.02 x 10 23 atoms C
6.02 x 10 23 atoms Hg
6.02 x 10 23 atoms Fe
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Example 4 (mol gram)
• If you have 4.5 mols of sodium, how much does it weigh?
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Example 4
• .45 mol Na x 23 g Na = 10.35 g Na = 1.0 x 10 2 g Na 1 mol Na
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Example 5 (grams atoms)
• If you have 34.3 g of Iron, how many atoms are present?
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Example 5
• 34.3 g Fe x 1 mol Fe x 6.02 x 10 23 atoms 55.8 g Fe 1 mol Fe=3.70 x 10 23 atoms Fe
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The Mass of a Mole of a Compound
• To find the mass of a mole of a compound you must know the formula of the compound
• H2O H= 1 g x 2
O= 16 g 18 g = 1 mole = 6.02 x 10 23
molecules
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Example 6 (gram mol)
• What is the mass of 1 mole of sodium hydrogen carbonate?
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Example 6
• Sodium Hydrogen Carbonate = NaHCO3
• Na=23 g• H=1 g• C=12 g• O=16 g x3• 84 g NaHCO3 = 1 mol NaHCO3
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Mole-Volume Relationship
• Unlike liquids and solids the volumes of moles of gases at the same temperature and pressure will be identical
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Avogadro’s Hypothesis
• States that equal volumes of gases at the same temperature and pressure contain the same number of particles
• Even though the particles of different gases are not the same size, since the gas particles are spread out so far the size difference is negligible
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Standard Temperature and Pressure (STP)
• Volume of a gas changes depending on temperature and pressure
• STP= 0oC (273 K) 101.3 kPa (1 atm)
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Standard Temperature and Pressure (STP)
• At STP, 1 mol = 6.02 X 1023 particles = 22.4 L of ANY gas= molar volume
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Conversion Factors
• AT STP• 1 mol gas 22.4 L gas 22.4 L gas 1 mol gas
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Example 7
• At STP, what volume does 1.25 mol He occupy?
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Example 7
• 1.25 mol He x 22.4 L He = 28.0 L He 1 mol He
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Example 8
• If a tank contains 100. L of O2 gas, how many moles are present?
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Example 8
• 100. L O2 X 1 mol O2 = 4.46 mol O2
22.4 L O2
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Calculating Molar Mass from Density
• The density of a gas at STP is measured in g/L
• This value can be sued to determine the molar mass of gas present
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Example 9
• A gaseous compound of sulfur and oxygen has a density of 3.58 g/L at STP. Calculate the molar mass.
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Example 9
• 1 mol gas x 22.4 L gas X 3.58 g gas = 1 mol gas 1 L gas
Molar Mass= 80.2 g
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Percent Composition
• The relative amounts of the elements in a compound
• These percentages must equal 100
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Percent Composition
• %element = mass of element x 100 mass of compound
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Example 10
• Find the percentage of each element present in Al2 (CO3)3
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Example 10
• Al2(CO3)3
• Al= 27 g x 2 = 54 g / 234 g x 100=23%• C= 12 g x 3 = 36 g/ 234 g x 100= 15%• O = 16 g x 9 = 144 g / 234 g x 100=62% 234 g Al2(CO3)3