6-1 Ideal Gas Law Constant Lab Kristen Brown
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Determination of Ideal Gas Law Constant
S
tudent: Kristen Brown
Date: 11/30/23
Activity 1
If your concentration, moles, or R calculations are incorrect or your calculation work does not actually result in
the desired final units, or units are missing/incomplete, or cancelled incorrectly, you will lose points. Be sure
that every number you include has its associated unit.
Email me if you are unsure or need assistance.
Data Table 1
Trial 1
(1 mL
H
2
O
2
)
Trial 2
(1 mL
H
2
O
2
)
Trial 3
(2 mL
H
2
O
2
)
Trial 4
(2 mL
H
2
O
2
)
Trial 5
(3 mL
H
2
O
2
)
Trial 6
(3 mL
H
2
O
2
)
Trial 7
(4 mL
H
2
O
2
)
Trial 8
(4 mL
H
2
O
2
)
Air temperature
(°C)
25°C
25°C
25°C
25°C
25°C
25°C
25°C
25°C
Volume H
2
O
2
liquid (mL)
1.0mL
1.0mL
2.0mL
2.0mL
3.0mL
3.0mL
4.0mL
4.0mL
Initial Volume
Gas (mL)
0.0mL
0.2mL
0.0mL
0.3mL
0.0mL
1.0mL
0.0mL
1.2mL
Final Volume
Gas (mL)
9.2mL
11.3mL
18.5mL
19.4mL
32.5mL
34.8mL
40.0mL
41.9mL
ΔV (mL)
9.2mL-
0.0mL =
9.2mL
11.3mL-
0.2mL =
11.1mL
18.5mL
-
0.0mL =
18.5mL
19.4mL
-
0.0mL =
19.1mL
32.5mL
-
0.0mL =
32.5mL
34.8mL
-
1.0mL =
33.8mL
40.0mL
-
0.0mL =
40.0mL
41.9mL
-
1.2mL =
40.7mL
©2016 2 Carolina Biological Supply Company
2021
Activity 2
Data Table 2
Show work for determining Concentration of H
2
O
2
in the box below. Concentration should be mol/L.
You
need to begin your calculation using the provided 3%m/v H
2
O
2
concentration
.
Concentration
H
2
O
2
(mol/L)
30.0g H
2
O
2
/ 1000mL
30.0g H
2
O
2
x 1.0mol H
2
O
2
/ 34.01g H
2
O
2
= 0.88mols H
2
O
2
/ L
Trial 1
1 mL H
2
O
2
Trial 2
1 mL H
2
O
2
Trial 3
2 mL H
2
O
2
Trial 4
2 mL H
2
O
2
Trial 5
3 mL H
2
O
2
Trial 6
3 mL H
2
O
2
Trial 7
4 mL H
2
O
2
Trial 8
4 mL H
2
O
2
Mole
s
H
2
O
2
0.0009mol
H
2
O
2
0.0009mol
H
2
O
2
0.0018mol
H
2
O
2
0.0018mol
H
2
O
2
0.0026mo
l
H
2
O
2
0.0026mo
l
H
2
O
2
0.0035mol
H
2
O
2
0.0035mol
H
2
O
2
Mole
s O
2
*
0.00045mo
l
O
2
0.00045mol
O
2
0.0009mol
O
2
0.0009mol
O
2
0.0013mo
l
O
2
0.0013mo
l
O
2
0.00175mo
l
O
2
0.00175mo
l
O
2
ΔV
(L)
1L =
1000mL
9.2mL
/1000 =
0.0092L
1L =
1000mL
11.1mL
/1000 =
0.0111L
1L =
1000mL
18.5mL
/1000 =
0.0185L
1L =
1000mL
19.1mL
/1000 =
0.0191L
1L =
1000mL
32.5mL
/1000 =
0.0325L
1L =
1000mL
33.8mL
/1000 =
0.0338L
1L =
1000mL
40.0mL
/1000 =
0.04L
1L =
1000mL
40.7mL
/1000 =
0.0407L
*Hint: Use reaction stoichiometry/molar ratio to solve for moles of O
2
.
Show work for determining moles of H
2
O
2
for Trial 1
here:
1.0mL H
2
O
2
x 1L/1000mL x 0.88mols H
2
O
2
/1L = 0.00088mols H
2
O
2
Show work for determining moles of O
2
for Trial 1
here:
0.0009mols H
2
O
2
x 1.0mol O
2
/2.0mols H
2
O
2
= 0.00045mols O
2
Show work for determining moles of H
2
O
2
for Trial 3
here:
2.0mL H
2
O
2
x 1L/1000mL x 0.88mols H
2
O
2
/1L = 0.00176mols H
2
O
2
Show work for determining moles of O
2
for Trial 3
here:
0.0018mols H
2
O
2
x 1.0mols O
2
/2.0mol H
2
O
2
= 0.0009mols O
2
Show work for determining moles of H
2
O
2
for Trial 5
here:
3.0mL H
2
O
2
x 1L/1000mL x 0.88mols H
2
O
2
/1L = 0.00264mols H
2
O
2
Show work for determining moles of O
2
for Trial 5
here:
0.0026mols H
2
O
2
x 1.0mol O
2
/2.0mols H
2
O
2
= 0.0013mols O
2
© 2016 Carolina Biological Supply Company
2021
Show work for determining moles of H
2
O
2
for Trial 7
here:
4.0mL H
2
O
2
x 1L/1000mL x 0.88mols H
2
O
2
/1L = 0.00352mols H
2
O
2
Show work for determining moles of O
2
for Trial 7
here:
0.0035mols H
2
O
2
x 1.0mol O2/2.0mols H
2
O
2
= 0.00175mols O
2
Insert a copy of your graph. Your graph must include the
equation of the line displayed, your name and date
in
the title.
You cannot complete the rest of this lab without this graph and the equation of the line. See the Week 6
Announcement for an example. Don’t forget to set the y-intercept to 0.
Graph:
0
0
0
0
0
0
0
0
0
0
0
0.01
0.01
0.02
0.02
0.03
0.03
0.04
0.04
0.05
f(x) = 23.45 x
Volume of O2 vs Moles O2
Kristen Brown 12/2/2023
Moles O2
Vol. O2 (L)
Show work,
including all units
, for the calculations/conversions for Air Temperature, Gas Constant R, and
Percent Error in the boxes below. Calculations must include complete units to be correct. This includes the
correct units for the slope.
Air Temperature (K)
Show work in the space
provided.
25°C + 273.15 = 298.15K
Air Pressure (atm)
Show work in the space
provided.
1 atm
Equation of the Line
Y = 23.451x
© 2016 Carolina Biological Supply Company
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2021
Gas Constant R
Show work in the space
provided.
23.451 * l/mols * (1 atm)
/298.15 k
=
0.0787 L(atm)/mols(K)
Percent Error
Show work in the space
provided.
Use 0.0821 L*atm/mol*K in
your calculation
(|0.0821 L
∗
atm/mol
∗
K−0.0787 L
∗
atm/mols
∗
K|
/0.0821 L
∗
atm/mol
∗
K)
* 100
=
4.14% error
Question 1:
Would the volume of oxygen that is generated be affected if a smaller mass of yeast were used?
Why or why not?
Yes, if less yeast is used, there would be a smaller amount of catalase present to perform as the catalyst to the
decomposition of H
2
O
2
. This would directly result in a smaller amount of O
2
being produced.
Question 2:
Identify at least two potential sources of error in the experiment. Are any assumptions made that
would add to the experimental error?
Some potential sources of error would be the scale having trouble reading such low numbers. Another would be
how much gas is lost between the time of adding H
2
O
2
and putting the cap back on the bottle. Lastly, another
potential error could be if any residual liquid is still in the container from the previous trials.
© 2016 Carolina Biological Supply Company
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