Transcript Unit 1 Matter and Meaurement/Atoms, Molecules, and Ions/Stoichiometry: Calculations with Chemical Formulas and Equations

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
*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:
Naturally occurring chlorine is 75.78%
chlorine-35 with a mass of 34.969 amu
and 24.22% chlorine-37 with a mass of
36.966 amu. Determine the average
atomic mass for chlorine.
AAM = (.7578) (34.969 amu) +
(.2422) (36.966 amu)
= 26.50 amu + 8.953 amu
= 35.45 amu
Three isotopes of silicon occur in nature:
Silicon-28 92.23% 27.97693 amu
Silicon-29 4.68% 28.97649 amu
Silicon-30 3.09% 29.97377 amu
Calculate the average atomic mass for
silicon:
AAM = (.9223) (27.97693 amu) + (.0468)
(28.97649 amu) + (.0309) (29.97377 amu)
= 25.80 amu + 1.356 amu + .9262 amu =
28.08 amu
Objective #6 Isotopes / Calculating
Average Atomic Mass / Mass Spectrometer
*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
Operation of Mass Spectrometer
(video clip)
*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
Example of Mass Spectrograph for Hg
Most Common Isotopes of Mercury Based on Mass
Spectrograph
Isotope
*mercury-196
*mercury-198
*mercury-199
*mercury-200
*mercury-201
*mercury-202
*mercury-204
Percent Abundance
.15 %
9.97 %
16.87 %
23.1 %
13.18 %
29.86 %
6.87 %
Objective #11 Description of Combustion
Analysis
Objective #11 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.
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
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
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% H
Step 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 N
100% - 38.64% - 16.11% = 45.25% N
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
.003723 moles C/.003723 = 1
.01862 moles H/.003723 = 5
.003736 moles N/.003723 = 1
CH5N
Determine the molecular formula if the
molecular mass is 62.0 g
Emp mass = 31.0 g
62.0 g / 31.0 g = 2
C2H10N2
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 empirical formula of
caproic acid? Determine the molecular
formula if the molecular mass is 116 g.
.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 H
By difference .0618 g O
.140 g C X 1 mole C/12.0 g C = .0117 mole 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
.0117 mole C/.00386 = 3 mole C
.0232 mole H/.00386 = 6 mole H
.00386 mole O / .00386 = 1 mole O
C3H6O empirical formula
116 g/58 g = 2
C6H12O2 molecular formula
Objectives #12-13 Stochiometry
(NH4)2Cr2O7 --› Cr2O3+ N2 +4H2O +Δ
1. Calculate the mass of water produced
from the decomposition of 5.00 g of
ammonium dichromate.
5.00 g A.D. X 1mole A.D. / 252.0 g A.D. X
4 mole H2O / 1 mole A.D. X
18.0 g H2O / 1 mole H2O = 1.44 g H2O
(NH4)2Cr2O7 --› Cr2O3+ N2 +4H2O +Δ
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 =
9.6 X 1022 molecules H2O
(NH4)2Cr2O7 --› Cr2O3+ N2 +4H2O +Δ
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
(NH4)2Cr2O7 --› Cr2O3+ N2 +4H2O +Δ
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
3H2SO4
+ 2Al(OH)3 --›Al2(SO4)3 +
6H2O
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 Al2(SO4)3 = 5.82 g Al2(SO4)3
5.00 g Al(OH)3 X 1 mole Al(OH)3 / 78.0 g Al(OH)3 X
1 mole Al2(SO4)3 / 2 mole Al(OH)3 X 342.3 g
Al2(SO4)3 / 1 mole Al2(SO4)3 = 11.0 g Al2(SO4)3
%Yield = 4.50 g / 5.82 g X 100 = 77%
3H2SO4
+ 2Al(OH)3 --›Al2(SO4)3 +
6H2O
Use the above information to determine
the amount of the excess reactant
leftover.
5.00 g H2SO4 X 1 mole H2SO4 / 98.1 g
H2SO4 X 2 mole Al(OH)3 / 3 mole
H2SO4 X 78.0 g Al(OH)3 / 1mole
Al(OH)3 = 2.65 g (amount used)
5.00 g – 2.65 g = 2.35 g leftover
Objectives #14-15 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
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
Example III: How many grams of sucrose
C12H22O11 are present in 125 ml of a 1.07
M sucrose solution?
Moles = .125 L X 1.07 M = .134 moles
.134 moles sucrose X 342.0 g / 1 mole
= 45.8 g
*Molality
Example 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
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
*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
M1V1 = M2V2
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
AgNO3 + NaCl
--›
AgCl + NaNO3
*Precipitation Reactions and Solution
Stoichiometry
*key concept: to find moles of substance in
solution multiply molarity by volume
Example 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
3CaCl2
+ 2H3PO4 --› 6HCl
Ca3(PO4)2
+
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 /
1 mole Ca3(PO4)2 = 51.7 g
3AgNO3
+
K3PO4 --›
3KNO3
Ag3PO4 +
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 K3PO4
.00800 moles K3PO4 X 3 mole AgNO3 / 1
mole K3PO4 = .0240 moles AgNO3
M = moles / L
moles / M = L
.0240 moles AgNO3 / .300 M = 80.0 ml
H2SO4 + 2NaOH --› 2H2O + Na2SO4
*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
2HNO3 + 3H2S
--› 2NO + 3S + 4H2O
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 M HNO3 X .500 L = 2.00 moles HNO3
2.00 moles HNO3 X 3 moles H2S / 2 moles
HNO3 = 3.00 moles H2S
M = moles / L
L = moles / M
3.00 moles H2S / 1.00 M = 3.00 L