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Example Problems
The temperature of 2.5 L of a gas initially at STP
is raised to 250 oC at constant volume. Calculate the
final pressure of the gas in atm.
PV
=R
nT
so
Pi V i
ni T i
=
Pf V f
nf T f
or
Pi P f
=
(since V and n are constant)
Ti T f
Example Problems
The temperature of 2.5 L of a gas initially at STP
is raised to 250 oC at constant volume. Calculate the
final pressure of the gas in atm.
P i×T f
Pf =
Ti
=
1 atm×523 K
K
298
= 1.75atm
Example Problems
At STP, 0.280 L of a gas weighs 0.400g. Calculate the
molar mass of the gas.
mRT
M=
PV
=
0.400g×0.0821 Latm/molK ×298 K
1 atm×0.280l
=35.0 g/mol
Example Problems
A compound has the empirical formula SF4. At 20 oC,
0.100g of the gaseous compound occupies a volume of
22.1 mL and exerts a pressure of 1.02 atm. What is
the molecular formula of the compound?
Empirical formula mass = 108.059 g/mol
M = mRT/PV
= 0.100 x 0.0821 x 293/ 1.02 x 0.0221
= 107 g/mol
ratio molar mass to empirical formula mass:
107 : 108
1:1
molecular formula: SF4
Example Problems
Methane, the principal component of natural gas, is
used for heating and cooking. The combustion process
is:
CH4
+
2O2
CO2
+
2H2O
If 15.0 moles of CH4 are reacted what is the volume
of CO2 in liters produced at 23.0 oC and 0.985 atm?
Mols CO2 produced = 15.0 mol CH4 x 1 mol CO2/1 mol CH4
= 15.0 mols
V = nRT/P
= 15.0 mol x 0.0821 latm/molK x 296 K/ 0.985 atm
= 370 L
Example Problems
A piece of sodium metal reacts completely with water
as follows:
2Na
+
2H2O
2NaOH
+
H2
The H2 gas is collected over water at 25.0 oC. The
volume of the gas is 246 mL measured at 1.00 atm.
Calculate the number of grams of Na used in the
reaction (Vapor pressure of water at 25 oC = 0.0313
atm).
PT = PH2 +
PH2 = PT = 1.00
= 0.97
PH2O
PH2O
– 0.0313
atm
Example Problems
A piece of sodium metal reacts completely with water
as follows:
2Na
+
2H2O
2NaOH
+
H2
The H2 gas is collected over water at 25.0 oC. The
volume of the gas is 246 mL measured at 1.00 atm.
Calculate the number of grams of Na used in the
reaction (Vapor pressure of water at 25 oC = 0.0313
atm).
nH2 = PV/RT
= 0.97 atm x 0.246 L / 0.0821 Latm/molK x 298 K
= 0.0097 mol
mols Na = 0.0097 mol H2 x 2 mol Na/ 1 mol H2
= 0.020 mol
mass Na = 0.020 mol x 22.99 g/mol = 0.45g
Example Problems
A mixture of Ar and N2 gases has a density of 1.413
g/L at STP. What is the mole fraction of the gas?
Recalling the equation M = dRT/P that can be used to
calculate a molar mass from a density:
M = 1.413 g/L x 0.0821 Latm/molK x 298 K / 1 atm
= 34.6 g/mol
This is intermediate between the molar masses of N2,
28.013 g/mol, and Ar, 39.948 g/mol and is the
weighted average of the two ... in exactly the same
way that the average atomic mass is the weighted
average of isotopes.
Example Problems
A mixture of Ar and N2 gases has a density of 1.413
g/L at STP. What is the mole fraction of the gas?
34.6 = XN2 x 28.013 + (1-XN2) x 39.948
XN2 = (39.948 – 34.6)/(39.948 – 28.013)
= 0.448
therefore, XAr = 1 – 0.45 = 0.55
Example Problems
A 36.4 L volume of methane gas is heated from 25
to 88 oC at constant pressure. What is the final
volume of the gas?
P i×T f
Pf =
Ti
1atm×523K
=
K
298
=1.75atm
oC
Example Problems
Iodine, I2, reacts with aqueous thiosulfate ion in neutral solution
according to the equation:
I2 + S2O32-
S4O62- + I-
How many grams of I2 are present in a solution if 35.20 mL of
0.150 M Na2S2O3 solution is needed to titrate the I2 solution?
Balance the equation:
1.
I2 + 2e2I2S2O32S4O62- + 2e2.
I2 + 2S2O32-
S4O62- + 2I-
Example Problems
Iodine, I2, reacts with aqueous thiosulfate ion in neutral solution
according to the equation:
I2 + 2S2O32-
S4O62- + 2I-
How many grams of I2 are present in a solution if 35.20 mL of
0.150 M Na2S2O3 solution is needed to titrate the I2 solution?
Mols Na2S2O3 = 0.03520 L x 0.150 mol/L
= 0.00528 mol
mols I2 = mols Na2S2O3 x 1 mol I2/ 2 mols Na2S2O3
= 0.00264 mol
mass I2 = 0.00264 mol x 253.809 g/mol
= 0.670 g
Example Problems
Iron in a sample can be determined by titrating with cerium
ion:
Fe2+ + Ce4+
Fe3+ + Ce3+
What is the mass percentage of iron in a sample if titration of
1.2284 g of the sample requires 57.91 mL of 0.1018 M Ce4+?
Mols Ce4+ = 0.05791 L x 0.1018 mol/L
= 0.005895 mol
mols Fe2+ = 0.005895 mol Ce4+ x 1 mol Fe2+/1 mol Ce4+
= 0.005895 mol
mass Fe = 0.005895 mol x 55.85 g/mol
= 0.3292 g
mass %Fe = (0.3292g/1.2284g) x 100 = 26.80%
Example Problems
If 1.87 g of acetic acid, C2H4O2, reacts with 2.31 g of isopentyl
alcohol, C5H12O to give 2.96 g of isopentyl acetate, C7H14O2,
what is the percent yield?
Balanced equation:
C2H4O2 + C5H12O
C7H14O2 + H2O
limiting reagent approach
1. mols acetic acid = 1.87 g / 60.05 g/mol
= 0.0311 mol
mols C7H14O2 = 0.0311 mol
2. mols C5H12O = 2.31 g / 88.15 g/mol
= 0.0262 mol
mols C7H14O2 = 0.0262 mol
So, C5H12O is the limiting reagent.
Example Problems
If 1.87 g of acetic acid, C2H4O2, reacts with 2.31 g of isopentyl
alcohol, C5H12O to give 2.96 g of isopentyl acetate, C7H14O2,
what is the percent yield?
Mass C7H14O2 = 0.0262 mol x 130.19 g/mol
= 3.41 g (theoretical yield)
Percent yield = (actual yield/theoretical yield) x 100
= (2.96/3.41) x 100
= 78.9%
Example Problems
An average cup of coffee contains about 125 mg of caffeine,
C8H10N4O2. How many moles of caffeine are in a cup? How
many molecules of caffeine?
Mols caffeine = 125 mg x (1 g/1000 mg) / 194.19 g/mol
= 6.44 x 10-4 mol
Molecules caffeine = 6.44 x 10-4 mol x 6.022 x 1023 molecule/mol
= 3.88 x 1020 molecules
Example Problems
A 1.268 g sample of a metal carbonate MCO3 was reacted with
100.00 mL of 0.1083 M H2SO4 yielding CO2 gas and an
aqueous solution of the metal sulfate. The CO2 was removed
by boiling and the remaining H2SO4 was titrated with 0.1241
M NaOH, requiring 71.02 mL to reach the endpoint. What is the
metal?
2NaOH + H2SO4
Na2SO4 + 2H2O
starting mols H2SO4 = 0.10000 L x 0.1083 mol/L
= 0.01083 mol
mols H2SO4 left = mols NaOH x 1 mol H2SO4 / 2 mol NaOH
= 0.07102 L x 0.1241 mol/L x ½
= 0.004407 mol
mols H2SO4 used = 0.01083 – 0.004407 = 0.006423 mol
Example Problems
A 1.268 g sample of a metal carbonate MCO3 was reacted with
100.00 mL of 0.1083 M H2SO4 yielding CO2 gas and an
aqueous solution of the metal sulfate. The CO2 was removed
by boiling and the remaining H2SO4 was titrated with 0.1241
M NaOH, requiring 71.02 mL to reach the endpoint. What is the
metal?
MCO3 + H2SO4
MSO4 + H2O
mols H2SO4 used = mols MCO3
= 0.006423 mol
molar mass MCO3 = 1.268 g/0.006423 mol
= 197.4 g/mol
mass M in 1 mol = mass MCO3 – mass CO3
= 197.4 – 60.01
= 137.4 g/mol
Barium
Example Problems
The fictional isotope Xium-60 has an atomic mass of 60.15
units while Xium –70 has an atomic mass of 70.16 units. The
average mass of Xium is 69.41 units. What is the fractional
abundance of Xium 70?
Average atomic mass = fraction A x atomic mass A + fraction
B x atomic mass B
69.41 amu = fraction 60 x 60.15 amu + fraction 70 x 70.16
= (1 - fraction 70) x 60.15 + fraction 70 x 70.16
fraction 70 = 0.9251
Example Problems
Calculate the number of neutrons in an atom of 239Pu.
Atomic number of plutonium = 94
mass number = 239 = number of neutrons + number of
protons
number of neutrons = mass number – number of protons
= 239 – 94
= 145
Example Problems
Permanganate, MnO4-, reacts with nitrite, NO2-, to produce
manganese (IV) oxide and nitrate. Balance the equation.
1. half reactions:
MnO4- + 3eMnO2
NO2NO3- + 2e2. balance oxygens:
MnO4- + 3eMnO2 + 2H2O
NO2- + H2O
NO3- + 2e3. balance hydrogens:
MnO4- + 4H+ + 3eMnO2 + 2H2O
NO2- + H2O
NO3- + 2H+ + 2e-
Example Problems
Permanganate, MnO4-, reacts with nitrite, NO2-, to produce
manganese (IV) oxide and nitrate. Balance the equation.
4. add half reactions:
2MnO4- + 8H+ + 6e2MnO2 + 4H2O
3NO2- + 3H2O
3NO3- + 6H+ + 6e2MnO4- + 2H+ + 3NO2-
2MnO2 + 3NO3- + H2O
5. adjust for basic solution if necessary:
2MnO4- + 2H+ + 2OH- + 3NO22MnO2 + 3NO3- + H2O +
2OH2MnO4- + H2O + 3NO2-
2MnO2 + 3NO3- + 2OH-
Example Problems
Acetylsalicylic acid (C9H8O4) commonly known as "aspirin" is an
acid that can ionize to produce one H+ ion. A typical aspirin
tablet, however, contains only a small amount of the acid. In an
experiment to determine its composition, an aspirin tablet was
crushed and dissolved in water. It took 12.25 mL of 0.1466 M
NaOH to neutralize the solution. Calculate the number of grams
of aspirin in the tablet.
HA + NaOH
NaA + H2O
mols NaOH = 0.01225 L x 0.1466 mol/L
= 0.001796 mol
mols HA = mols NaOH x 1 mol HA / 1 mol NaOH
= 0.001796 mol
mass HA = 0.001796 mol x 180.16 g/mol
= 0.3235 g
Example Problems
Titration with solutions of potassium bromate, KBrO3, can be
used to determine the concentration of As(III). What is the
molar concentration of As(III) in a solution if 22.35 mL of 0.100
M KBrO3 is needed to titrate 50.00 mL of the As(III) solution?
The (unbalanced) equation is:
H3AsO3 + BrO3-
Br- + H3AsO4
1. half reactions:
H3AsO3
H3AsO4 + 2eBrO3- +6eBr2. balance oxygens:
H3AsO3 + H2O
H3AsO4 + 2eBrO3- + 6eBr- + 3H2O
Example Problems
Titration with solutions of potassium bromate, KBrO3, can be
used to determine the concentration of As(III). What is the
molar concentration of As(III) in a solution if 22.35 mL of 0.100
M KBrO3 is needed to titrate 50.00 mL of the As(III) solution?
The (unbalanced) equation is:
H3AsO3 + BrO3-
Br- + H3AsO4
3. balance hydrogens:
H3AsO3 + H2O
H3AsO4 + 2H+ + 2eBrO3- + 6H+ + 6eBr- + 3H2O
4. add half reactions:
3H3AsO3 + 3H2O
3H3AsO4 + 6H+ + 6eBrO3- + 6H+ + 6eBr- + 3H2O
3H3AsO3 + BrO3-
3H3AsO4 + Br-
Example Problems
Titration with solutions of potassium bromate, KBrO3, can be
used to determine the concentration of As(III). What is the
molar concentration of As(III) in a solution if 22.35 mL of 0.100
M KBrO3 is needed to titrate 50.00 mL of the As(III) solution?
The (unbalanced) equation is:
H3AsO3 + BrO3-
Br- + H3AsO4
Mols BrO3- = 0.02235 L x 0.100 mol/L
= 0.00224 mol
mols H3AsO3 = mols BrO3- x 3 mol H3AsO3 / 1 mol BrO3= 0.00671 mol
[H3AsO3] = 0.00671 mol / 0.05000 L
= 0.134 mol/L