Answers for Gas Laws worksheet
I.5. 1 L containers are filled with
gas.
Calculate the densities of the gases
under various conditions.
n = PV / RT
d = m / V = M n / V
[also = MP/RT]
|
M
(g/mol)
|
n (mol)
|
m (g)
|
d (g/L)
|
H2(1.00 atm, 298 K)
|
2.0158
|
0.0409
|
0.0824
|
0.0824
|
| H2(2.00 atm, 298 K) |
2.0158
|
0.0817
|
0.165
|
0.165
|
| H2(1.00 atm, 398 K) |
2.0158
|
0.0306
|
0.0617
|
0.0617
|
| Cl2(1.00 atm, 298 K) |
70.906
|
0.0409
|
2.90
|
2.90
|
| Cl2(2.00 atm, 298 K) |
70.906
|
0.0817
|
5.79
|
5.79
|
| Cl2(1.00 atm, 398 K) |
70.906
|
0.0306
|
2.17
|
2.17
|
Density is increased by increasing pressure or molar mass.
It is decreased by increasing temperature.
II. Liquid is put into a 100.0 mL
flask with a small opening. The flask is heated until the liquid
boils. At boiling the pressure of the gas equals atmospheric pressure,
0.987 atm. The temperature of the gas is the boiling point of the
liquid, 218°C. When no more liquid is observed, the flask is
removed from heat; the mass of the contents is 0.2797 g.
6. Calculate the moles of gas
in the container.
n =PV/RT = (0.987 atm)(0.1000 L) / (0.0821 L atm mol-1 K-1)(218+273)K
=
0.00245 mol
7. What is the molar mass of
the gas?
M = m/n = 0.2797 g / 0.00245 mol = 114 g/mol
The sample turns out to be a hydrocarbon; when the 0.2797 g of sample
is burned in oxygen, 0.8619 g carbon dioxide and 0.3969 g water
are collected.
8. What is the chemical formula
of the unknown gas?
CO2: 0.8619 g / 44.009 g mol-1 = 0.01958
mol CO2 * (1 mol C/mol CO2) = 0.01958 mol C
H2O: 0.3969 g / 18.015 g mol-1 = 0.02203 mol H2O
*
(2 mol H/mol H2O) = 0.04406 mol H
molar ratio H:C = 0.04406 mol : 0.01958 mol = 2.250:
1
To change to whole number ratio, multiply by
4 H:C =9:4
M(C4H9)
= 4(12.011)+ 9(1.0079) = 57 g/mol
That is one half of the molar mass calculated in question 7, so the
chemical formula must actually be C8H18.
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