Rooney, Mary Anne – HomeworkTwo
MISE - Physical Basis of Chemistry
Second Set of Problems - Due October 29, 2006
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Submit electronically (digital drop box) by Sunday, October 29 – by 6 pm.
Note: When submitting to digital Drop Box label your files with your name first and
then the label - HmwkTwo. Please put you name in the ‘Header’ along with the
already-inserted page #. There is a useful equation “template” document
posted below this assignment in Bb. Give it a look.
The following information may be useful when solving the problems:
Assume that all gases behave ideally, i.e., PV = nRT ;
where R = 0.0821 L•atm•mole-1•K-1.
1 mole = 6.022 x 1023 “things”.
The atomic weight of an element listed in the periodic table is the mass of one
mole of atoms in grams.
The molecular weight - determined from molecular formula as the sum of the
atomic weights of all of the constituent atoms - is the mass of one mole of
molecules in grams.
1 atmosphere (atm) = 760 mm Hg = 76 cm Hg = 101.3 kilopascals.
453.6 grams = 1 lb ;
1000 cm3 = 1 L ;
2.54 cm = 1 inch ;
{32ºF = 0ºC & 212ºF = 100ºC} … In case you have to figure out the linear
relationship to convert ºF <----> ºC.
K = ºC + 273
Volume of a cylinder = π•(radius)2•(height)
1. A child is holding two balloons – “A” and “B” – each containing a different gas on a
day when the local weatherman reports the barometric pressure to be 29 inches of mercury
and a temperature of 78ºF. If balloon A contains 6.00 L of nitrous oxide (N2O(g)) and
balloon B contains 8.00 L of helium (He(g)), determine the moles and grams of each
gas in each balloon.
PV = nRT
PA = 29in Hg
PB = 29in Hg
PAVA = R = PBVB
VA = 6L
VB = 8L
O
nATA
NBTB
TA = 78 F
TB = 78OF
nA = ?
nB = ?
nA = PAVA
RTA
nB = PBVB
RTB
Pressure conversion:
29in x 2.54 cm x 1 atm = 73.66 = .969atm = .97 atm
Rooney, Mary Anne – HomeworkTwo
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1 in
76 cm 76
Temperature conversion:
(78oF – 32OF)/1.8OC = 25OC + 273 = 298OK
nA = .97atm x 6 L
.0821 L mol-1K-1 x 298K
nB = .97atm x 8 L
.0821 L mol-1K-1 x 298K
nA = 5.82
24.47
nB =
nA = .238 moles
nB = .317 moles
7.76
24.47
AWA = 44g/mole x .238 mole  10.47 gA
AWB = 4g/mole x .317 mole  1.27 gB
2. An ideal gas at 1.00 atmosphere was collected in a bulb of unknown volume V. A valve is
opened which allows the gas to expand into a previously evacuated bulb whose volume is
known to be exactly 0.500 L. When things settle down (the gas fills the full volume of both
bulbs), it is noted that the temperature had not changed, and the gas pressure was 530 mm
Hg. What is the volume, V, of the first bulb (in liters)?
PiVi = PfVf
Pressure is 1 atm; temperature and moles are the same.
niTi nfTf
Vf = Vi + .5L
Pf = 530 mmHg x 1 atm = .697 atm
760 mm
PiVi = PfVf  PiVi = Pf(Vi + 0.5L)  PiVi = PfVi + 0.5LPf
PiVi - PfVi = 0.5LPf
Vi(Pi – Pf) = 0.5LPf
Vi(Pi – Pf) = 0.5LPf
Pi – Pf
Pi – Pf
Vi = (0.5LPf)/ Pi – Pf
Vi = 0.5L x .697 atm
1 atm - .697 atm
Vi = .3485 L / .303  1.15 L
3. Balancing a chemical reaction of gases and reasoning with moles …
Gaseous propane (C3H8(g)) is a hydrocarbon that is often used as the fuel for
small gas grills. The burning of propane is actually the reaction with molecular
diatomic oxygen (O2(g)) – a combustion – to produce carbon dioxide (CO2) and
water (steam):
C3H8(g) + O2(g)
------>
CO2(g) +
H2O(g) .
Rooney, Mary Anne – HomeworkTwo
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(a) According to Dalton, atoms must be conserved, i.e., there must be the same number
of atoms of each element on both sides of the ------>. Please place integer coefficients
in front of each chemical species – as needed – to conserve the atoms, i.e., balance
the reaction. How should we interpret these coefficients? Read on.
When balanced, the coefficients tell us the correct “recipe”. [For example, the
balanced reaction:
2 H2 + O2 -----> 2 H2O ; can be read as follows:
Every 2 molecules of H2 will react with one molecule of O2 to yield or produce
2 molecules of H2O. Since the relationship between grams and moles is such
a useful one (recall the first homework), we can always count the molecules in
moles (similar to counting in dozens). Then, we can interpret the coefficients above as
follows: Every 2 moles of H2 will react with one mole of O2 to yield or produce
2 moles of H2O. This will be useful in (c) below.]
C3H8 + O2  CO2 + H2O
C3H8 + O2  3CO2 + H2O
C3H8 + 5O2  3CO2 + 4H2O
(b) A tank of propane is advertised as containing 5 lbs. of propane and the tank - assumed
to be a cylinder - has the listed dimensions of 8” in diameter x 12.5” in height.
The temperature of the propane in the tank is 59ºF. Please determine the following.
•
The mass of propane in the tank in grams.
5 lb x 453.6g = 2268g
1 lb
•
The volume of propane in the tank in liters.
V = Πr2h  V = Π4inches212.5inches  638.3 in3
1mL = 1cm3 and 1000 cm3 = 1 L so:
1in/2.54 cm then 1 in3/16.4 cm3
638.3 in3 x 16.4 cm3 x 1 L
= 10304.12 = 10.3 L
3
3
1 in
1000 cm
1000
•
The temperature of propane in the tank in ºC and in K.
(59oF – 32OF)/1.8 = OC
27/1.8 = OC
15 = OC  15 OC + 273 = 288OK
•
The moles of propane in the tank.
36 C + 8 H = 44 g C3H8 = MW
44g/1 mole C3H8, and 2268g = x moles; so 2268g/44g = 51.5 moles C3H8
•
The number of propane molecules in the tank.
51.5 moles C3H8 x 6.022 x 1023molecules = 3.10 x 1025molecules C3H8
1 mole C3H8
•
The pressure of propane in the tank in atmospheres - assuming ideal gas behavior.
PV = nRt  P = nRT  P = 51.5 mol x .0821 L atm mol-1K-1 x 288 K
V
10.3 L
P = 1217.7/10.3
P = 118.2 atm
(c) If all of the propane in the tank was completely reacted with the exact proportion of
Rooney, Mary Anne – HomeworkTwo
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O2(g) indicated in your balanced reaction, how many moles of O2(g) would be
required? How many molecules of O2(g) is this? From the original balancing of the
equation at the beginning of this question, we know that one mole of propane reacts with
5 moles of oxygen. Based on that:
5 mol O2
=
x
1 mol C3H8
51.5 molC3H8
x = 257.5 molO2
257.5 mol O2 x 6.022 x 1023 = 1.55 x 1026 molecules O2
1mol O2
(d) How many moles of CO2(g) and how many moles of H2O(g) should be produced
by the amount of C3H8(g) and O2(g) determined in (c)?
Again, using the original balanced equation:
51.5 molC3H8 x 3 mol CO2 = 154.5 mol CO2
1 mol C3 H8
51.5 mol C3 H8 x 4 mol H2 O = 206 mol H2O
1 mol C3 H8
(e) Extra Credit: Air is a mixture of gases and is popularly proclaimed to contain
20 % oxygen (O2(g)) and 80 % nitrogen (N2(g)). This percentage is a mole
percent - not a mass percent. Thus, every 100 moles of air contains 20 moles
of O2(g). How many moles of air would contain the amount of O2(g) determined
in (c)?
4. An allotrope is one of several forms of the same element with different structure.
Consider the element carbon (C). Several years ago an allotrope of carbon
- neither graphite nor diamond - was discovered and named buckminsterfullerine
or “buckyball”, CX , where “X” is a whole number. A sample of CX(g) at
400.0oC and 680.0 mm Hg is found to have a density of 11.67 g/L. Determine
“X” and hence the formula CX for one of these buckyball molecules.
volume = 1 L
temperature = 400.0 oC
pressure = 680.0 mm Hg
density = 11.67 g/L
n = #mol = g/MW = g/(g/mol) = mol
n = g/MW
PV = (g/MW)RT
MW = gRT  g/V x (RT/P)
PV
g/mol = (g/L) x ((L x atm)/mol x K) x K
atm
Rooney, Mary Anne – HomeworkTwo
Convert pressure to atm  680 mm x 1 atm
= 680 = .89 atm
760 mm 760
MWCX = 11.67 g/L x (.0821 L x atm)/(mol x K) x 673oK
.89 atm
MWCX = 724.5 g/mol
(12 g/mol)/(12g/mol) = (724.5 g/mol)/(12g/mol)
MWC = 60 atoms of carbon so Cx = C60 = “buckyball”
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