Unit 4 – The Language of Chemistry: Part Deux. Amadeo Avogadro II.B.2(f) – Describe Avogadro’s...
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Transcript of Unit 4 – The Language of Chemistry: Part Deux. Amadeo Avogadro II.B.2(f) – Describe Avogadro’s...
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Unit 4 – The Language of Chemistry: Part Deux
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Amadeo Avogadro
II.B.2(f) – Describe Avogadro’s hypothesis and use it to solve stoichiometric problems
III.A.2(a) – Explain the meaning of mole and Avogadro’s number
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Avogadro’s Hypothesis
Equal amounts of gases at the same temperature contain equal numbers of molecules
Leads to definition of the “mole” Mole Def: the number equal to the number of
atoms in 12.01 grams of carbon
We will return to this idea later
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Atomic Mass Units (amu)
Since the mass of one atom is so tiny, it is more practical to use relative atomic masses
Carbon has been arbitrarily assigned a mass of exactly 12 atomic mass units
One atomic mass unit is defined as exactly 1/12 the mass of a carbon-12 atom
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Avogadro’s Number
Experimentally determined that there are 6.022 x 1023 atoms in exactly one mole
Remember this number!
6.022 x 1023
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Molar Mass
The mass of one mole of a pure substance
Units: grams/mol
Equal to the mass of 6.022 x 1023 atoms of a pure substance
Molar mass of an element is numerically equal to the atomic mass of the element in atomic mass units
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Converting Mass, Moles, & Atoms/Molecules/Ions
III.A.2(b) -Interconvert between mass, moles, and number of particles
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Gram/Mole/Atom Conversions
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Example: What is the mass in grams of 3.50 mol of copper?
Cu g 222Cu mol 1
Cu g 63.55Cu mol 50.3
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More Examples
What is the mass in grams of 2.25 mol of Fe? What is the mass of 0.375 mol of K? What is the mass of 16.3 mol of Ni?
OR How many moles are in 5.00 g of calcium? How many moles are in 3.60 x 10-10 g Au?
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More Complex Examples
How many moles of Ag are in 3.01 x 1023 atoms of Ag?
How many atoms of Na are in 36.0 grams of Na?
How many oxygen atoms are in 180.18 g of glucose?
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Determining Chemical Formulas
III.A.1(e) - Calculate the percent composition of a substance, given its formula or masses of each component element in a sample
III.A.1(f) - Determine the empirical formulas and molecular formulas of compounds, given percent composition data or mass composition data
III.A.2(c) - Distinguish between formula mass, empirical mass, molecular mass, gram molecular mass, and gram formula mass
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Empirical Formulas
III.A.1(f) - Determine the empirical formulas and molecular formulas of compounds, given percent composition data or mass composition data
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Calculating an Empirical Formula
Empirical Formula – consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole-number ratio of the different atoms in the compound
Always assume 100.0 g sample Change percents to grams Calculate moles of each element Divide moles by smallest mole amount to
determine ratio Multiply to get whole numbers (if necessary)
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Determine the Empirical Formula of Aspirin
What is the empirical formula of aspirin? It has 4.48 % H, 60.00 % C, and 35.52 % O by mass.
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Example #1
An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen. Calculate the empirical formula for this compound.
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Example #2
When a 0.3546 g sample of vanadium metal is heated in air, it reacts with oxygen to achieve a final mass of 0.6330 g. Calculate the empirical formula of this vanadium oxide.
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Example #3
A sample of lead arsenate, an insecticide used against the potato beetle, contains 1.3813 g of lead, 0.00672 g of hydrogen, 0.4995 g arsenic, and 0.4267 g of oxygen. Calculate the empirical formula for lead arsenate
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MolecularFormulas
III.A.1(f) - Determine the empirical formulas and molecular formulas of compounds, given percent composition data or mass composition data
III.A.2(c) - Distinguish between formula mass, empirical mass, molecular mass, gram molecular mass, and gram formula mass
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Calculating a Molecular Formula
Molecular Formula – the actual formula of a molecular compound
Example: C2H4 – ethylene
C3H6 – cyclopropane
Both have a 2H:1C ratio
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Determining Molecular Formula
Must know formula mass to calculate molecular formula
Divide experimental formula mass by empirical mass
Multiply subscripts by quotient
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Example #4
Find the molecular formula of a compound with an empirical formula of CH and a formula mass of 78.110 amu
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Answer
C – 12.011
H – 1.008
Total = 13.019
78.110 ÷ 13.019 ≈ 6
6(CH) = C6H6
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Example #5
A white powder is analyzed and found to have an empirical formula of P2O5. The compound has a molar mass of 283.88g. What is the compounds molecular formula?
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Finding Molecular Formula from Empirical Formula
An unknown compound is found in tree sap. It has been shown to be composed of 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. It was also discovered that 5.00 moles of the material has a mass of 900 grams.
What is the molecular formula of this compound?
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Combustion Analysis
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Example #6
What is the empirical formula of a hydrocarbon that produces 2.703 g CO2 and 1.108 g H2O when combusted?
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Example #7
What is the empirical formula of a substance containing carbon, hydrogen, and oxygen if 1.000 g of substance produces 1.467 g CO2 and 0.6003 g H2O upon combustion?
The molar mass of the substance is 120 g/mol. What is the molecular formula?
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Example #8
What is the molecular formula of a substance containing carbon, hydrogen, and oxygen if it has a molar mass of 234 g/mol and 0.360 g of substance produces 0.406 g CO2 and 0.250 g H2O upon combustion?