SME
General Chemistry I
Gases
Use this for “A gas has a density of 1.31 g/L at 25 °C and 1.00 atm. Find its molar mass.”
Diagram
R = 0.08206 L·atm/(mol·K).
PV = nRT, which gives 1 × 1 = 0.04087 × 0.08206 × 298.15 = 1.
Each particle drawn stands for 0.002 mol: about 20 particles.
Temperature in °C t = 25 °C
This is an example. Type your own values to start a new problem.
M (g/mol)
=
d (g/L)
×
R
×
T (K)
÷
P (atm)
Putting n = m ÷ M and m = dV into the picture’s PV = nRT and solving for M gives M = dRT ÷ P.
t
Temperature in °C
example · °C
V
Sample volume
example · L
n
Amount of gas
calculated · mol
m
Sample mass
calculated · g
Formulas
T
=
t
+
273.15
298.15 = 25 + 273.15
M
=
d
×
0.08206
×
T
P
32.05
=
1.31
×
0.08206
×
298.15
1
P
×
V
=
n
×
0.08206
×
T
1 × 1 = 0.04087 × 0.08206 × 298.15
d
=
m
V
1.31
=
1.31
1
n
=
m
M
0.04087
=
1.31
32.05
Assumptions
•
The gas is ideal: low pressure and high temperature, so its particles take up no room and do not attract each other.
•
R = 0.08206 L·atm/(mol·K), so the pressure is in atm, the volume in liters and the temperature in kelvins (°C + 273.15).
•
Any sample gives the same molar mass. Take 1.00 L: it holds d grams, and PV = nRT counts its moles.
Step-by-step
WE KNOW
d = 1.31 g/L, t = 25 °C, P = 1 atm, V = 1 L
FIND
temperature (T), molar mass (M), amount of gas (n), sample mass (m)
Step 1 · Find temperature (T)
Use T = t + 273.15
The gas laws use kelvins, which start 273.15 degrees below 0 °C.
T
=
25
+
273.15
T
=
298.15
K
Step 2 · Find molar mass (M)
Use M = d × 0.08206 × T ÷ P
Put
n
=
m
M
into
PV
=
nRT:
then
M
=
(
m
V
)
RT
P
,
and
m
V
is
the
density.
M
=
1.31
×
0.08206
×
298.15
1
M
=
0.1075
×
298.15
M
=
32.05
g/mol
Step 3 · Find amount of gas (n)
Use P × V = n × 0.08206 × T
Divide both sides by RT to count the moles in the sample.
n
=
P
×
V
0.08206 ×
T
n
=
1
×
1
0.08206 × 298.15
n
=
1
24.47
n
=
0.04087
mol
Step 4 · Find sample mass (m)
Use d = m ÷ V
Each
liter
of
the
gas
weighs
d
grams.
m
=
d
×
V
m
=
1.31
×
1
m
=
1.31
g
Check
Put the numbers back in. Both sides should match.
298.15 = 25 + 273.15 ✓
32.05
=
1.31
×
0.08206
×
298.15
1
✓
1 × 1 = 0.04087 × 0.08206 × 298.15 ✓
1.31
=
1.31
1
✓
0.04087
=
1.31
32.05
✓
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Real gases: the van der Waals pressure
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