h advanced chemistry unit 1
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Matter and Measurement / Atoms, Molecules, and Ions/Stoichiometry: Calculations with Chemical Formulas and Equations. H Advanced Chemistry Unit 1. Objective #6 Isotopes / Calculating Average Atomic Mass / Mass Spectrometer. - PowerPoint PPT PresentationTRANSCRIPT
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Matter and Measurement / Atoms, Molecules, and Ions/Stoichiometry: Calculations with
Chemical Formulas and Equations
H Advanced ChemistryUnit 1
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Objective #6 Isotopes / Calculating Average Atomic Mass / Mass Spectrometer
*average atomic masses can be determined from the masses of the various isotopes that naturally occur for an element and their percent of abundance in nature
*examples:*operation of mass spectrometer:*instrument that provides the atomic mass
and percent of abundance of the isotopes that make up a sample of an element
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Operation of Mass Spectrometer
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*gaseous sample is introduced into instrument and the particles are bombarded with high energy electrons
*positive ions are produced and these are passed through a magnetic field
*the amount of deflection is measured; the more massive the particle the less the deflection
*a graph of the intensity of the detector signal vs. the particle mass is called a mass spectrum
*it is from the data provided from this graph that one obtains the atomic mass and percent abundance necessary to calculate the average atomic mass for an element
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Example of Mass Spectrograph for Hg
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Most Common Isotopes of Mercury Based on Mass Spectrograph
Isotope Percent Abundance*mercury-196 .15 %*mercury-198 9.97 %*mercury-199 16.87 %*mercury-200 23.1 %*mercury-201 13.18 %*mercury-202 29.86 %*mercury-204 6.87 %
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Objective #12 Description of Combustion Analysis
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Objective #12 Combustion Analysis
*Example I: An unknown compound is composed of carbon, nitrogen, and hydrogen. When .1156 grams of this compound is reacted with oxygen and burned, the following components are produced: .1638 g of carbon dioxide and .1676 g of water. Calculate the percent composition and empirical formula for this unknown compound.
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Step I All of the carbon that was in the original sample is now contained in the carbon dioxide. Convert from grams of carbon dioxide to grams of carbon.
.1638 g CO2 X 12.0 g C/44.0 g CO2 = .04467 g C
Step II Divide the resulting mass of carbon by the original mass of the compound and multiply by 100% to obtain the percentage of carbon in the sample
% C = .04467 g/.1156 g X 100 = 38.64% C
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Step III All of the hydrogen that was in the original sample is now contained in the water. Convert from grams of water to grams of hydrogen.
.1676 g H2O X 2.0 g H / 18.0 g H2O = .01862 g
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Step IV Divide the resulting mass of hydrogen by the original mass of the compound and multiply by 100% to obtain the percentage of the hydrogen in the sample
%H = .01862 g/.1156 g X 100 = 16.11% HStep V The mass and percentage of the final
component can be determined by mass and percentage difference.
.1156 g - .04467 g - .01862 g = .05231 g N100% - 38.64% - 16.11% = 45.25% N
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Step VI Use the percentages or masses of components to compute the empirical formula of the compound.
.04467 g C X 1 mole C/12.0 g C = .003723 moles C
.01862 g H X 1 mole H/1.0 g H = .01862 moles H
.05231 g N X 1 mole N/14.0 g N = .003736 moles N
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.003723 moles C/.003723 = 1
.01862 moles H/.003723 = 5
.003736 moles N/.003723 = 1CH5N
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Example II Caproic acid, which is responsible for the foul odor of dirty socks, is composed of C, H, and O atoms. Combustion analysis of a .225 g sample of this compound produces .512 g of carbon dioxide and .209 g of water. What is the emprical formula of caproic acid? Determine the molecular formula if the molecular mass is 116 g.
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.512 g CO2 X 12.0 g C/44.0 g CO2 = .140 g C
.209 g H2O X 2.0 g H/18.0 g H2O = .0232 g HBy difference .0618 g O.140 g C X 1 mole C/12.0 g C = .0117 g C.0232 g H X 1 mole H/1.0 g H = .0232 mole H.0618 g O X 1 mole O/16.0 g O =.00386mole O
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.0117 mole C/.00386 = 1 mole C
.0232 mole H/.00386 = 6 mole H
.00386 mole O / .00386 = 1 mole OC3H6O empirical formula116 g/58 g = 2C6H12O2 molecular formula
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Objectives #13-14 Stochiometry
1. Calculate the mass of water produced from the decomposition of 5.00 g of ammonium dichromate.
5.00 g X 1mole A.D. / 252.0 g A.D. X 4 mole H2O / 1 mole A.D. X 18.0 g H2O = 14.4 g
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2. Calculate the number of water molecules produced from 10. g of A.D.
10. g A.D. X 1 mole A.D. / 252.0 g A.D. X
4 mole H2O / 1 mole A.D.X 6.02 X 1023 molecules / 1 mole H2O = 9.56 X 1022 molecules
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3. Calculate the amount of A.D. required to produce 5500 J of energy if the change in energy of the reaction is 2750 J.
5500 J X 1 mole A.D. / 2750 J X 252.0 g A.D. / 1 mole A.D. = 5.0 X 102
g
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4. Calculate the number of chromium atoms produced from the decomposition of 10.0 g of A.D.
10.0 g A.D. X 1 mole A.D. / 252.0 g A.D. X 1 mole Cr2O3 / 1 mole A.D. X
6.02 X 1023 f. units Cr2O3 / 1 mole Cr2O3 X 2 Cr atoms / 1 f. unit Cr2O3 =
4.78 X 1022 atoms
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Example II: The reaction below produced 4.50 g of aluminum sulfate in the laboratory. Calculate the theoretical yield and the percent yield given 5.00 g of sulfuric acid reacted with 5.00 g of aluminum hydroxide.
5.00 g H2SO4 X 1 mole H2SO4 / 98.1 g X 1 mole Al2(SO4)3 / 3 mole H2SO4 X 342.3 g Al2(SO4)3 / 1 mole = 5.82 g
5.00 g Al(OH)3 X 1 mole Al(OH)3 / 78.0 g X 1 mole Al2(SO4)3 / 2 mole Al(OH)3 X 342.0 g Al2(SO4)3 =
11.0 g %Yield = 4.50 g / 5.82 g X 100 = 77%
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Use the above information to determine the amount of the excess reactant leftover.
5.00 g H2SO4 X 1 mole H2SO4 / 98.1 g X 2 mole Al(OH)3 / 3 mole H2SO4 X 78.0 g Al(OH)3 = 2.65 g (amount used)
5.00 g – 2.65 g = 2.35 g leftover
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Objectives #15-16 Molarity, Molality, Solution Stoichiometry and Titrations
Example I: Determine the molarity of a solution containing 2.37 moles of potassium nitrate in enough water to give 650 ml of solution.
M = 2.37 moles / .650 L = 3.65 M
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Example II: Determine the molarity of a solution containing 25.0 g of sodium hydroxide dissolved in enough water to give 2.50 L of solution.
M= 25.0 g NaOH / 40.0 g / 2.50 L = .250 M
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Example III: How many grams of sucrose are present in 125 ml of a 1.07 M sucrose solution?
Moles = .125 ml X 1.07 M = .134 moles
.134 moles sucrose X 342.0 g / 1 mole= 45.8 g
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*MolalityExample I: Calculate the molality of a
solution containing 5.0 g of sodium chloride dissolved in 250 g of water
5.0 g NaCl X 1mole NaCl / 58.5 g / .250 kg H2O
= .34 m
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Example II: Calculate the molality of a solution containing 5.0 moles of NaCl dissolved in 250 g of water.
5.0 moles NaCl / .250 kg H2O = 20. m
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*Dilution*moles before dilution = moles after
dilution*formula: M1V1 = M2V2
Example I: What is the molarity of a solution prepared by diluting 12.0 ml of .405 M NaCl to a final volume of
80.0 ml?12.O ml X .405 M / 80.0 ml = .0608 M
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Example II: What volume of .25 M HCl solution must be diluted to prepare
1.00 L of .040 M HCl?.040 M X 1.00 L / .25 M = .16 L
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*Precipitation Reactions and Solution Stoichiometry*key concept: to find moles of substance in
solution multiply moles by volumeExample I: How many grams of precipitate can be
produced from the reaction of 1.00 L of .375 M silver nitrate solution with an excess of sodium chloride solution?
.375 M X 1.00 L silver nitrate = .375 moles
.375 moles AgNO3 X 1 mole AgCl / 1 mole AgNO3 X 143.4 g AgCl / 1 mole AgCl = 53.8 g
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Example II: How many grams of precipitate can be produced from the reaction of 2.50 L of a .200 M calcium chloride solution with an excess of phosphoric acid?
.200 M calcium chloride X 2.50 L = .500 moles CaCl2
.500 moles CaCl2 X 1 mole Ca3(PO4)2 / 3 moles CaCl2 X 310.3 g Ca3(PO4)2 X 1 mole = 51.7 g
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Example III: What volume in ml of a .300 M silver nitrate solution is needed to react with 40.0 ml of a .200 M potassium phosphate?
.200 M potassium phosphate X .0400 L = .00800 moles
.00800 moles K3PO4 X 3 mole AgNO3 / 1 mole K3PO4 = .0240 moles AgNO3
moles / M = L .0240 moles AgNO3 / .300 M = 80.0 ml
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*Titration Problems:Example I: What is the molarity of a 37.5 ml
sample of a sulfuric acid solution that will completely react with 23.7 ml of a .100 M sodium hdyroxide solution?
.100 M NaOH X .0237 L = .00237 moles NaOH
.00237 moles NaOH X 1 mole H2SO4 / 2 mole NaOH = .00119 moles H2SO4
M = .00119 moles H2SO4 / .0375 L = .0317 M
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Example II: What volume in liters of a 1.00 M hydrogen sulfide solution is needed to react completely with .500 L of a 4.00 M nitric acid solution?
4.00 moles HNO3 X .500 L = 2.00 moles HNO3
2.00 moles HNO3 X 3 moles H2S / 2 moles HNO3 = 3.00 moles H2S
L = moles / M 3.00 moles H2S / 1.00 L = 3.00 L