Answer Key
2014 Ch 1a, Problem Set One
Side 1 of 8
Problem One – Stoichiometry – 25 Points
a. 2N 2H 4 + N2O4 → 3N 2 + 4H 2O
(CH3 )2 N2H2 + 2N2O4 →3N2 + 4H2O +2CO 2
5 points total (2.5 points each)
-0.5 for each wrong coefficient
(1)
(2)
b. The mixture is 50/50 by mass hydrazine and dimethylhydrazine. Therefore if the total
mass is 2521 kg, each component is 1260.5 kg (since each fraction is known to infinite
precision). Assuming that all of the reactants are used up, we can use this mass, the molar
masses of each reactant, and the stoichiometric ratios from the above balanced equation
to determine the mass of nitrogen tetroxide
Hydrazine:
⎛ 1mol N 2 H 4 ⎞ ⎛ 1mol N 2O4 ⎞
1260.5 ×103g N 2 H 4 ⎜
⎟ ⎜
⎟ = 19664.6 mol N2O4
32.045g
N
H
2mol
N
H
⎝
2 4 ⎠ ⎝
2 4 ⎠
2 points each equation
1 pt for adding together
Dimethylhydrazine
⎛ 1mol ( CH3 )2 N2 H2 ⎞
1260.5 ×103g ( CH3 )2 N2 H2 ⎜
⎜ 60.098g ( CH ) N H ⎟⎟
3 2
2 2 ⎠
⎝
⎛
⎞
2mol N2O4
⎜⎜
⎟⎟ = 41946.8mol N2O4
⎝ 1mol ( CH3 )2 N2 H2 ⎠
So, 61611.3 total moles N2O4 are used. Using the molar mass of N2O4 (=92.011 g/mol) ,
the mass can be found to be 5668.9 kg, or with the correct number of sig figs…
2 points for conversion to kg
3
Final Answer: 5.669 × 10 kg N2O4 used
2 points for answer, -1 for incorrect number of
sig figs, -2 pts if not in requested units
The stoichiometric ratio of carbon dioxide is the same for nitrogen tetroxide in the
decomposition of dimethylhydrazine, so the same molar amount will be produced
⎛ 1mol ( CH3 )2 N2 H2 ⎞
1260.5 ×103g ( CH3 )2 N2 H2 ⎜
⎜ 60.098g ( CH ) N H ⎟⎟
3 2
2 2 ⎠
⎝
1846.1 kg CO2
Final Answer: 1.846 × 103 kg CO2 evolved
⎛
⎞ ⎛ 43.9987g CO2 ⎞
2mol CO2
⎜⎜
⎟⎟ ⎜
⎟ =
1mol
CH
N
H
1mol
CO
(
)
3
2
2
⎝
2
⎠
⎝
2
⎠
2 points for conversion
2 points, -1 for incorrect number of sig figs
-2 pts if not in requested units
Answer Key
2014 Ch 1a, Problem Set One
Side 2 of 8
c. First determine the molar amount of water produced by this reaction.
⎛ 1mol N 2 H 4 ⎞ ⎛ 2mol H 2O ⎞
1260.5 ×103g N 2 H 4 ⎜
⎟ ⎜
⎟ = 78658.3 mol H2O
⎝ 32.045g N 2 H 4 ⎠ ⎝ 1mol N 2 H 4 ⎠
1 point for each equation
⎛ 1mol ( CH3 )2 N2 H2 ⎞
1260.5 ×103g ( CH3 )2 N2 H2 ⎜
⎜ 60.098g ( CH ) N H ⎟⎟
3 2
2 2 ⎠
⎝
⎛
⎞
4mol H2O
⎜⎜
⎟⎟ = 83893.5 mol H2O
1mol
CH
N
H
(
)
3 2
2 2 ⎠
⎝
1 point for adding together
Or 162551.9 mol H2O total. There are molar mass of water is 18.0152 and the density is
given as 0.92 g/cm 3 . Putting this all together into a cylinder 1.8 km in diameter:
h=
mol H 2O × MW
2
π d2 ρ
( )
1.251x10-4 cm
= 162551.9 moles H2O ×
2 points for correct
equation
(18.0152g/mol ) /(1.8/2×1000×100)2cm2 =
π × ( 0.92g/cm3 )
Final Answer = 1.3 × 10-4 cm
2 points for answer, -1 for sig figs
-2 pts if not in requested units
Answer Key
2014 Ch 1a, Problem Set One
Side 3 of 8
Problem Two – Stoichiometry – 25 Points
a. The molecular weight may be calculated from the osmotic pressure experiment.
ρghV
3 points
π = ρgh = cRT and πV = nRT so n =
RT
Where π is the osmotic pressure, ρ is the density of the solution, g is the acceleration due
to gravity, h is the height of the liquid in the thistle tube, c is the solution concentration, n
is the number of moles, V is the volume of the solution, and RT are the universal gas
constant and temperature, respectively. Rearranging this equation and converting the
experimental numbers to their proper units we see that
2 points
⎛ g ⎞
⎛ m ⎞
0.842⎜ 3 ⎟ × 9.807⎜ 2 ⎟ × 0.9231m
⎝ cm ⎠
⎝ s ⎠
n=
× 250cm3=7.6900×10-4 moles
⎛ g ⎞
J
1000⎜⎜ ⎟⎟ × 8.31447
× 298.15 K
mol * K
⎝ kg ⎠
The molar mass can be found from the mass of the solid that was dissolved:
2 points
0.132 g /7.6900×10-4 moles = 171.71 g/mol
2 points, -1 for sig figs.
Final Answer = 172 g/mol We will accept this answer in g/molec (2.85×10−22 g/molec), but the
convention for molecular weight is g/mol
b. We want to balance the equation C x H y O z + O2 → CO2 + H 2 O .
To do this, first balance the carbon and hydrogen atoms (as they each appear in only one
compound on either side of the equation. When you do this you get the unbalanced equation:
Y
C x H y Oz + O2 → XCO2 + H 2 O .
2
Next the oxygen atoms need to be balanced by finding the correct coefficient for O2 on
the left side of the equation. As written above, there are 2+Z oxygen atoms on the left, and
2X+Y/2 oxygen atoms on the right. So, to balance the equation, we need to put a coefficient
1
Y
of (2 X + − Z ) . The balanced equation is:
2
2
%
1"
Y
Y
6 points total
Cx H yOz + $ 2X + − Z ' O2 → XCO2 + H 2O
(2 pt per coefficient)
&
2#
2
2
Or, multiplied out so there are no fractional coefficients,
4Cx H yOz + ( 4X +Y − 2Z ) O2 → 4XCO2 + 2YH 2O
c. The mass of CO2 and water collected is 320.6 g and 87.6 g, respectively. By molar
masses, this is 7.2848 moles of carbon dioxide and 4.8664 moles of water. Similarly,
136.2 g of initial white solid would be 0.79318 moles from our calculation of the
molecular weight in the first part.
Answer Key
2014 Ch 1a, Problem Set One
Side 4 of 8
From here, these ratios may be used to calculate x and y. Dividing 7.2848 by 0.79318
indicates that 9.18 moles of carbon dioxide are formed from every mole of white solid.
Since the carbon in CO2 is formed 1:1 from the carbon in the white solid, x = 9.184.
Similarly, dividing 4.8664 by 0.7931 indicates that 6.135 moles of water are formed from
every mole of white solid. Here though, water is formed 2:1 from the white solid so
y = 12.3.
Lastly, we can solve for z from the constraint that:
9 points total (3 points each for the
rationales behind x, y, and z)
12.0107x + 1.0079y + 15.9994z = 171.7 g mol-1
We find that z = 3.06.
Now we have fractional values in our molecular formula: C9.14H12.2O3.05. We do not want
to use fractional subscripts in the molecular formula, so we need to do some rounding.
This seems to indicate that there was some slight experimental error, probably in the
determination of the molecular weight in part a.
Final Answer: C9H12O3
1 point: final answer
Answer Key
2014 Ch 1a, Problem Set One
Side 5 of 8
Problem Three – Boiling Point Elevation – 12 Points
a. The boiling point elevation of water as a function of solute is
ΔT = K bm
For water, Kb = 0.512
2 points total
(1 points each)
K kg
mol
The amount of salt in the solution is
12g ×
1
= 0.20533 mol
g
58.4428 mol
1 point
Since salt dissociates to two separate ionic species, there
are 0.41066 moles of ions. Both must be accounted for!!!
4 points (see note
below)
The mass of water (**, see note below) is:
⎛ 1000mL ⎞⎛ 1.0g ⎞⎛ 1kg ⎞
8L × ⎜
⎟⎟ = 8kg
⎟⎜
⎟⎜⎜
⎝ 1L ⎠⎝ 1mL ⎠⎝ 1000g ⎠
The molality, m, of all ionic species in this solution is therefore
2 points
m = 0.41066 mol/8 kg = 0.0513 mol/kg
Therefore the temperature increase is
ΔT = (0.512 Kmolkg )× 0.0513 mol/kg = 2.628x10-2K or ºC
Temperature Increase: 0.03 Kelvin or °C
2 points for
answer
-1 point off for sig fig…only 1 SF here!
-2 pts if not in °C or K
NOTE: 4 points off for not doubling m to get the total ion molality. Simply doubling the
value at the end to get the right number is not acceptable since it is not representative of
the science involved.
** We have tried to clarify that students should assume room temperature water, but
depending on their interpretation of “8 L of water”, Students may have used the density of
water at 0°C (1.00 g/mL), 25°C (0.995 g/mL), or 100°C (0.948 g/mL). Using any of these
values is acceptable, and will not impact the final answer to 1 sig fig. **
b. Final Answer: No. This difference is too small to affect the boiling time.
1 point for answer
Answer Key
2014 Ch 1a, Problem Set One
Side 6 of 8
Problem Four – Gas Laws – 13 Points
First, get all of the pressures into the same units, say atmospheres…
⎛ 10 5 Pa ⎞⎛ 1 atm ⎞
Outside pressure: 1016 x 10 bar × ⎜⎜
⎟⎟⎜
⎟ = 1.0027 atm
⎝ 1 bar ⎠⎝ 101325Pa ⎠
1 point each (atm not
⎛ 1 atm ⎞
required here, simply
Gauge pressure: 33 psi × ⎜⎜
⎟⎟ = 2.2455 atm
consistent units)
14.696psi
⎝
⎠
-3
Tire gauges do not read absolute pressure, psi, they read pressure difference between the
tire and the ambient pressure (referred to as psig). Therefore, the absolute pressure (i.e.
relative to a pure vacuum) inside of the tire is the sum or 3.2482 atm…we’ll call this P1.
3 points for conversion to absolute pressure.
No deductions for labels of psi vs psig
Now we have to convert the temperatures to absolute temperature. The formula for
conversion from Fahrenheit to Celsius is:
5 !
(
F − 32 )= !C
9
2 points
So the temperature at the start of the trip, T1, is 13.8889 C = 287.0389 K. At the end of
the trip the temperature, T2, is 73.8889 C = 347.0389 K.
To get the pressure at the end of the trip we invoke the law of Gay-Lussac:
P2 T 2
= ,
P1 T1
or
P2 =
T2
P1
T1
1 point for equation, 1 point for
computation
So P2 is found to be 3.92717 atm. This is the absolute pressure! To get the gauge
pressure, we have to subtract the ambient atmospheric pressure and convert to psi (really
psig, but either unit is acceptable since relative pressures are understood):
2 points for conversion back to gauge
pressure
⎛ 14.696 psi ⎞
(3.92717 atm – 1.0027atm) × ⎜
⎟ = 42.978 psig
⎝ 1 atm ⎠
Final Answer: 43 psig
2 points, -1 for sig figs…
Again, no penalty for labeling psi instead of psig
But, -2 pts. if units are not psi / psig
Answer Key
2014 Ch 1a, Problem Set One
Side 7 of 8
Problem Five – Henry’s Law – 25 Points
a. First convert the depth to meters:
2 points
⎛ 12in ⎞⎛ 0.0254m ⎞
60 ft × ⎜
⎟⎜
⎟ = 18.29m
⎝ 1ft ⎠⎝ 1in ⎠
The pressure at that depth can be found by P=ρgh (density, gravity and height of water
above diver)…
2 point
(
g
)(
P = 1.025 cm3 9.807
m
s2
)
⎛ 100cm ⎞
⎜
⎟
⎝ m ⎠
3
⎛ kg ⎞ ⎛ 1atm ⎞ ⎛ 1Pa ⎞
× 18.29m
⎜
⎟ ⎜
⎟ ⎜
2 ⎟
⎝ 1000g ⎠ ⎝ 101325Pa ⎠ ⎝ 1kg/ms ⎠
=1.81438 atm
That is just the pressure of the water. The total pressure includes the air pressure at sea
level too, or the addition of 1.0 atm. Which gives 2.81438 atm.
1 points
Corrected for sig figs…
2 points, -1 for incorrect sig figs
-2 pts if not in requested units
Final Answer: 2.8 atm
b. We are told the solubility at 1.0 atm. Using the equation
4 points
⎛ mg ⎞
⎜ 24
⎟
⎛ P2 ⎞
solubility@P 2 P2
L
⎜
⎟ × 2.8 atm = 67.2 mg/L
⎜
⎟
= ,or sol@P2 = ⎜ ⎟sol@P1 =
solubility@P1 P1
P
1
.
0
atm
⎜
⎟
⎝ 1 ⎠
⎜
⎟
⎝
⎠
If the density of blood is assumed to be 1.0 g/mL then one L of blood is 1000 g. So at 60
Assumption: 3 points
ft, the maximum amount of nitrogen in the blood is:
2 points
67.5432
mg
L
⎛ L blood ⎞⎛ g ⎞
⎜⎜
⎟⎟⎜⎜
⎟⎟ = 67.5432 ppm (based on mass)
1000g
1000mg
⎝
⎠⎝
⎠
Final Answer: 68 ppm
2 point, -1 for incorrect sig figs
-2 pts if not in requested units
c. Any "extra" nitrogen in the blood can be related to the difference in the solubility of
nitrogen on the surface and at depth.
Extra N2 = 67.5432 mg
− 24 mgL = 43.5432 mg/L
L
To find out the volume of the diver's blood:
3 point
Answer Key
2014 Ch 1a, Problem Set One
Side 8 of 8
⎛ 7 kg blood ⎞⎛ 1.0mL ⎞⎛ 1000 g ⎞⎛ L ⎞
⎟⎟⎜⎜
⎟⎟⎜⎜
⎟⎟⎜
⎟ = 5.6 L blood
100
kg
body
g
kg
1000
mL
⎠
⎝
⎠⎝
⎠⎝
⎠⎝
80 kg⎜⎜
Since the diver has 5.6 L of blood, then there is (43.5432 mg/L)(5.6 L) = 0.2438 g of N2 that
cannot dissolve in the bloodstream at standard pressure. If a person were to rise suddenly out of
the water, the volume of this dissolved gas can be found by the ideal gas law:
PV = nRT
or
V=
nRT
P
1 point
5
We need to convert 98.6ºF to K. 98.6 F = 273.15 + (98.6 − 32) = 310.15 K
9
€
Plugging the proper values into the above equation we find this volume to be:
( 0.0820574
V=
Latm
molK
) × 310.15K
(0.2438g N2) = 0.2215 L
⎛ 28.0134gN 2 ⎞
⎜
⎟ ×1atm
mol
⎝
⎠
With sig figs and change of units…
Final Answer = 2.2 × 102 mL N2
2 point, -1 for incorrect sig figs
-2 pts if not in requested units
1 point