lab-report-analysis-of-a-two-component-alloy
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LAB Report Analysis of a Two-Component Alloy
Chemistry (Michigan State University)
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Chemistry (Michigan State University)
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OBJECTIVE
The experiment was conducted in order for students to know how to apply the Ideal Gas Law to
analyse and determine the composition of a zinc-aluminum alloy by calculation of the mass
percentage of each metal in the alloy.
INTRODUCTION
Most metals, such as zinc, react with strong acids, like HCl, to produce hydrogen gas and metal
salt. The reaction below shows that in every mole of magnesium that reacts one mole of
hydrogen gas is produced.
Zn (s) + 2 H
+
(aq)
Zn
2+
(aq) + H
2
(g)
Aluminum also reacts irreversibly with HCl similar to the example given above. The ionic
equation below shows that it takes 2 moles of aluminum to react with acid to produce 3 moles of
hydrogen gas.
2 Al (s) + 6 H
+
(aq)
2 Al
3+
(aq) + 3H
2
(g)
Based on the chemical equations shown above, the amount of gas produced by one gram of
magnesium is different from the amount produces from one gram of aluminum. Hence, to
determine the amount of hydrogen gas evolved and the percentages of Mg and Al in the given
alloy, we need to have the initial weight of each metal.
The number of moles of hydrogen gas produced can be determined using the Ideal Gas equation:
PV = nRT
n = PV
RT
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Our goal to determine the percentages of metals present in the alloy can be calculated using the following equation:
Mass of Zinc + Mass of Aluminum = Mass of Alloy
mol of H
2 from Zn + mol of H
2
from Al = total mol of H
2
collected
Hence, using the Ideal Gas equation, we get:
mol of H
2 from Zn + mol of H
2
from Al = PV
RT
Where:
P:
Partial pressure of hydrogen gas
V:
Volume of the hydrogen gas
T:
Temperature of the hydrogen
n:
moles of hydrogen gas collected
R:
universal gas constant
PROCEDURE
Before the commencement of the experiment, all the materials needed such as Zn-Al alloy (good
for three trials), empty capsule, graduated cylinder, bottle container, beaker, corks, rubber tubing,
iron stand with clamp, and thermometer was collected. Then, a set-up according the the
following diagram was made:
Figure 1: Experimental Set-up
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It was made sure that there were no leaks in between the set-up to prevent the escape of the
hydrogen gas.
The mass of the Zn-Al alloy (3 replicates) were recorded as well as the mass of the bottle and the
water before and after the reaction took place. About 50 – 75 mL of dilute (6M) hydrochloric
acid was placed in the generator flask. The encapsulated alloy was dropped into the generator
flask, allowing the acid to dissolve the capsule and react with the allow giving off hydrogen gas.
The system was allowed to equilibrate for about 10 minutes before measuring the difference of
water level between the bottle and the beaker. The temperature of the water from the big bottle
was measured and the remaining water was weighed. The temperature of the gas above the water
was also measured. The experiment was done for three replicates using the same acid solution
for every trial. The water and acid solution used in the experiment was properly discarded.
RESULTS AND CALCULATIONS
Table 1 below shows all the analytical data gathered from the experiment.
Table 1. Raw Data Gathered from the Experiment
Trial 1
Trial 2
Trial 3
Mass alloy reacted (g)
2.5099
1.3676
1.3271
Mass Large Bottle filled with water (g)
4652
4720
4636
Mass Large Bottle after experiment (g)
2007
3277
3242
Atmospheric pressure (mm Hg)
742
742
742
Volume of water displaced (L)
2.65
1.45
1.4
Difference in water heights in bottles (cm)
11.74
7.13
7.23
Type "1" if level in starting bottle is higher at end of experiment. (Subtract from Dalton’s Law) Type "2" if level in displaced water bottle is higher. (Add to Dalton’s Law)
2
1
1
Temperature of gas in bottle (
o
C)
24.3
24.7
23.4
Temperature of water in bottle (
o
C)
24.4
24.9
23.3
Vapor pressure of water (mm Hg)
22.4
23.8
21.1
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TRIAL 1
Calculate the pressure of dry hydrogen gas
The pressure of dry hydrogen can be calculated using Dalton’s Law which states that the total
pressure is equal to the sum of the partial pressure of the individual gases in the mixture.
P
bar
= P
H2O
+ P
H2 ± P
ΔH
Note that in the first trial, the level in displaced water bottle is higher than the starting bottle.
Hence, P
ΔH
will be added in the equation.
P
ΔH
= Height difference
(
mm
)
13.6
= 117.4
mm
13.6
=
8.6 mmHg
P
bar
= 742 mmHg
P
H2O
= 22.4 mmHg
P
bar
= P
H2O
+ P
H2 +
P
ΔH
P
H2
= P
bar
- P
H2O - P
ΔH = (742 – 22.4 – 8.6) mmHg = 711 mmHg or
0.9355 atm (Pressure of dry
hydrogen)
Calculate the number of moles of hydrogen gas produced
n = PV
RT
= (
(
0.9355
atm
) (
2.65
L
)
(
0.08206
L.atm
mol. K
)
(
297.45
K
)
)
= 0.1017 moles of H
2
gas
Calculate the % Zn and % Al in the Alloy
Mass Zn + Mass Al = Mass Alloy
Mass Al = Mass Alloy – Mass Zn
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Mass Zn (
1
mol Zn
63.38
g
)(
1
mol H
2
1
molZn
)
+ (Mass Alloy – Mass Zn)
(
1
mol Al
26.98
g
)(
3
mol H
2
2
molZn
)
=
0.1017 mol H
2
Mass Zn (
1
mol Zn
63.38
g
)(
1
mol H
2
1
molZn
)
+ (2.5099 g – Mass Zn)
(
1
mol Al
26.98
g
)(
3
mol H
2
2
molZn
)
= 0.1017
mol H
2
Mass Zn = 0.95 grams
% Zn = 0.95
grams Zn
2.5099
grams Alloy
x 100% = 37.82% Zn
% Al =
(
2.5099
grams
−
0.95
grams
)
2.5099
grams Alloy
x 100% = 62.18% Al
Zn and Al ratio = 1:2
Calculate the number of moles of hydrogen gas produced from Zn and Al
Moles of H
2
from Zn = 0.95 grams Zn (
1
mol Zn
63.38
g
)(
1
mol H
2
1
molZn
)
= 0.0149 moles H
2
Moles of H
2
from Al = (2.5099 grams – 0.95 grams)
(
1
mol Al
26.98
g
)(
3
mol H
2
2
molZn
)
= 0.0867 moles H
2
Parameters for the second and third trials were calculated in a similar manner.
The table shown below summarizes the calculated parameters for the three trials conducted.
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Table 2: Calculated Parameters for Three Trials
DISCUSSION
The experiment was conducted in triplicate in order to evaluate the precision of the method
based on the values of the results. It can be observed that the percent components of alloys in the
three trials conducted is highly similar with 36.53 mean value for % Zn and a very acceptable
standard deviation of 1.13; 63.46 mean value for % Al and a good standard deviation of 1.12.
It’s good to note that although trial 1 used a weight of alloy twice the amount of trials 2 and 3,
the percent composition of the alloys for all trials are significantly close to each other. This
means that the procedure utilized can be successfully used for the determination of the
percentage components of two-component alloy whatever the amount of alloy used.
CONCLUSION
Trial 1
Trial 2
Trial 3
Pressure of dry H
2
(mmHg)
710.97
723.44
726.22
Pressure of dry H
2
(atm)
0.9355
0.9519
0.9555
Temperature of gas in bottle (
o
C)
24.30
24.70
23.40
Molar Gas Constant (R) L*atm/(mol*K)
0.08206
0.08206
0.08206
Moles of H
2
produced from Zn (mol)
0.0149
0.00778
0.00748
Moles of H
2
produced from Al (mol)
0.0867
0.04862
0.0474
Total moles of H
2
produced (mol)
0.1017
0.05636
0.05488
Mass of Zinc (g)
0.9492
0.4931
0.4742
% Zn
37.82
36.05
35.73
Average % Zn
36.53
Standard Deviation for % Zn
1.13
Mass of Al (g)
1.5606
0.8745
0.8529
% Al
62.18
63.94
64.27
Average % Al
63.46
Standard Deviation for % Al
1.12
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The experimental procedure utilized is a good yet simple way to determine the percent metal
composition of alloys. However, this may not be recommended when the type of the metals used
is unknown and yet to be determined because of the type of calculations necessary. Also one has
to effectively estimate the amount of the acid to be used for the reaction process because if the
acid used is not sufficient then you will not be able to accurately calculate for the required
parameters to come up with the final percent composition values.
POST LAB QUESTIONS
1.
Why was it not necessary to discard and refill the generator flask with fresh 6 M
HCl for subsequent reactions? Show a calculation to demonstrate your reasoning.
The volume of 6 M HCl acid used in the experiment was 75 mL and the number of moles of H
+
present is equal to 0.450 mole H
+
(n = M x V). The total amount of alloy used for three trials is
5.2046 grams. The following are the chemical reactions that took place in the generator flask.
Zn (s) + 2 H
+
(aq)
Zn
2+
(aq) + H
2
(g)
2 Al (s) + 6 H
+
(aq)
2 Al
3+
(aq) + 3H
2
(g)
Total amount of Zn used = 1.9165 g
Total amount of Al used = 3.288 g
Total moles of H
+
needed to dissolve Zn = 1.9165 g Zn 2
mol H
+
¿
1
mol Zn
¿
(
1
mol Zn
63.38
g
)
¿
= 0.06047 moles
Total moles of H
+
needed to dissolve Al = 3.288 g Al 6
mol H
+
¿
2
mol Al
¿
(
1
mol Al
26.98
g
)
¿
= 0.3656 moles
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The total number of moles of H
+
needed to react to every component of the alloy for three trials
is 0.42607 moles, hence the HCl used was enough for the whole set-up. 2.
In an experiment similar to yours, students were given an unknown sample that
produced H
2
S or NH
3
gas. What problem(s), if any, might arise if the same procedures as
this experiment were followed?
When this experiment will be repeated using other samples that produce H
2
S or NH
3
gas,
the outcome will not be the same. The polarities of the gases dictate the outcome of the
experiment. The non-polar H
2
gas produced by reacting a metal with HCl does not readily form
bonds with water molecules which causes the displacement to happen. However, when H
2
S or
NH
3
gases are produced, they will react with water because of their polar nature. Hence, no water
displacement happens for this case.
3.
Why is it not necessary to consider the amount of air that was in the generator flask
at the start of the reaction?
The amount of air present in the generator flask was not considered because both the
generator flask and the bottle were saturated with air or water molecules and the only force that
will cause any change is the pressure that will be influenced by the production of the hydrogen
gas. 4.
If there was a leak in the rubber tubing that allowed approximately 150 mL of H
2
gas to escape, how would it affect the relative amounts (% values) that you have calculated
as the results? Would the percent zinc in the sample appear to be high, low, or remain the
same? Show a calculation to justify your answer.
In the event that 150 mL of hydrogen escapes from the set-up, the % values will change,
specifically, the % Zn will increase and the % Al will decrease due to the ratio of their reaction
coefficients.
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Sample calculations (data from trial 1):
Number of moles of hydrogen gas n = PV
RT
= (
(
0.9355
atm
) (
2.65
−
0.15
L
)
(
0.08206
L.atm
mol. K
)
(
297.45
K
)
)
= 0.09581
moles
Mass Zn (
1
mol Zn
63.38
g
)(
1
mol H
2
1
molZn
)
+ (Mass Alloy – Mass Zn)
(
1
mol Al
26.98
g
)(
3
mol H
2
2
molZn
)
=
0.09581 moles
Mass Zn (
1
mol Zn
63.38
g
)(
1
mol H
2
1
molZn
)
+ (2.5099 g – Mass Zn)
(
1
mol Al
26.98
g
)(
3
mol H
2
2
molZn
)
= 0.09581
moles
Mass Zn = 1.097 g
Mass Al = 2.5099 g – 1.097 g = 1.4129 g % Zn = 1.097
g
2.5099
g
x
100
=
43.71%
(Trial 1 %Zn is 37.82%) hence, there is an increase in % Zn
% Al = 1.4129
g
2.5099
g
x
100
=
56.29%
(Trial 1 %Al is 62.18%) hence, there is a decrease in % Al
5.
What assumptions do we make regarding the temperatures of the generator and
collection flask? Are they valid?
It is assumed that the reaction in the experiment took place at a constant temperature and
pressure both in generator flask and collection flask. The assumptions are valid in ideal gas law
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because when there is a slight change of temperature to the experimental set-up, the value of the
parameters calculated will not be the same. Also, the net energy for the reaction will be zero
because the energy released from the generator flask via the production of hydrogen gas will be
utilized in the collection flask which influences the displacement of water.
6.
How would the calculated number of moles of H
2
collected change if the generator
flask were not given time to fully equilibrate?
The generator flask should be allowed to equilibrate for several minutes to make sure that
the reaction is totally complete and no more formation of hydrogen gas is going on. If the
generator flask will not be allowed to equilibrate, a possible error in the measurement of the
volume of the water displaced and hence the volume of the hydrogen gas formed. Incorrect
measurement of the volume of hydrogen gas formed will directly affect the calculation for the
number of moles.
7.
The molecular mass of the gas can be determined using an application of Ideal Gas
Law:
PV = nRT
and
n = m
MM
so
PV = m
MM
RT
An unknown gas with a mass of 1.545 ± 0.001 g at a temperature of 298.5 ± 0.1 K
and a pressure of 715 ± 1 mm Hg is found to occupy 0.260 ± 0.001 L. Determine the molar
mass and absolute error of this unknown. Assume that the gas constant, R, has no error.
PV = m
MM
RT (0.94079 atm)(0.260 L) = 1.545
g
MM
(
0.08206
L.atm
mol.K
)
(298.5 K) MM
=
1.545
g
(
0.94079
atm
)(
0.260
L
)
(
0.08206
L.atm
mol.K
)
(298.5 K) = 154.72 g/mol
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Absolute Error
154.72
= 0.001
1.545
+ 0.1
298.5
+ 1
715
+ 0.001
0.260
Absolute error = 0.963 ≈ 1
Hence, 154.72 ± 1 g/mol
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CI
Br
For each pair of substrates below, choose the one that will react faster in a substitution reaction, assuming that:
1. the rate law for the substitution reaction has a first-order dependence on the concentration of the alkyl halide, and
2. doesn't depend on the concentration of the other reactant at all.
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O KINETICS AND EQUILIBRIUM
Calculating an equilibrium constant from a heterogeneous...
Nickel carbonyl decomposes to form nickel and carbon monoxide, like this:
Ni (CO) (9) Ni(s)+4 CO(g)
!
1
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compound amount
Ni (CO)4 0.115 g
K = 0
Calculate the value of the equilibrium constant K for this reaction. Round your answer to 2 significant digits.
Explanation
Q
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Ni
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At a certain temperature, a chemist finds that a 3.5 L reaction vessel containing a mixture of nickel carbonyl, nickel, and carbon monoxide at equilibrium has
the following composition:
C
N
17.954
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Orbitals which result from solving the Schrodinger Wave Equation can be represented by electron "cloud" pictures.
The shapes of the orbitals are distinguished by their nodes, places where the electron density equals zero.
Erwin
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Based upon the number of spherical and angular nodes, label the orbitals below as 1s, 2p, etc.
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. Experimental data mass (g) and volume (mL) of a substance was gathering, recorded, and plotted in Figure 1.
a.
Correctly la bel the graph axis (include units).
b. The equation for the trendline (y = mx + b) is displayed on the graph. Which variable, in the equation,
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Chapter 8: Chemical Composition
ž AA A
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47. Calculate the percent by mass of the element listed first in the formulas for each of the following
compounds.
a. methane, CH4
b. sodium nitrate, NaNO3
c. carbon monoxide, CO
d. nitrogen dioxide, NO2
e. 1-octanol, Cg H18 O
f. calcium phosphate, Caz (PO4)2
g. 3-phenylphenol, C12 H10O
h. aluminum acetate, Al(C,H,02),
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