Transcript Slide 1

Molar Mass (M) and Density (d) of Gases
PV = nRT
Density of CO2:
44.0 g/mol
= 1.96 g/L
22.4 L/mol
Which gas would be the most dense?
N2, CO2, He, or O2
x g/mol
d=
22.4 L/mol
How Molar Mass (M) and Density (d) are Related:
PV = nRT
P
n
=
RT
V
MP = nM
=d
RT
V
Hint: always use PV = nRT first and watch your units!
Practice
An experiment shows that a 0.495 g sample
of an unknown gas occupies 127 mL at
98°C and 754 torr pressure. Calculate the
molar mass of the gas.
Solution
(PV = nRT)
The molar m as s is equal to t he num ber of gram s of unknown per mole. W e know t he mass
s ample, s o we need to compute t he number of moles . To do t his , we rearrange t he ideal gas
n
p V
R T
7 54t
 orr 
n 
  1 27 mL  1 L 

 1 00 0mL


 7 60t orr 


1 at m
0 .08 206Lat mK
1
1
 mo l
 3 71K
3
n  4 .14 1 0
N ow we can det ermine the molar mass .
M
m
n
M 
0 .49 5g

3
4 .141
 0
 mo l
M  1 20
g
mo l
mo l
Another Example
The density of a gas containing chlorine and oxygen
has a density of 2.875 g/L at 756 mmHg and 11oC.
What is the most likely formula of the gas?
756 mmHg = 0.995 atm
11oC = 284 K
2.875 g/L
= 67.3g/mol
0.0427mol/L
PV = nRT
(0.995atm)(V) = n (0.08206L.atm/mol.K)(284K)
(0.995atm)
n
= 0.0427mol/L
=
.
.
(0.08206L atm/mol K)(284K)
(V)
Dalton’s Law
•
•
•
•
Gas identity is not important
Mixture of gases obeys ideal gas law
Dependent only on total number of moles
Ptot = P1 + P2 + P3 + …
Dalton’s Law of Partial Pressures
• For a mixture of gases in a container
PTotal = P1 + P2 + P3 + …
Mole Fraction
• Percentage of moles in a mixture
Xi = ni / ntot
• Pi = XiPtot (partial pressure = mole fraction x
total pressure)
Mole Fraction and Partial
Pressure
C1 =
C1 =
P1
P1 + P2 + P3 + …
n1
nTOTAL
=
=
P1
PTOTAL
P1
PTOTAL
Mole Fraction Example
At 25°C, a 1.0 L flask contains 0.030 moles of
nitrogen, 150.0 mg of oxygen, and 4 x 1021
molecules of ammonia.
A. What is the partial pressure of each gas?
B. What is the total pressure in the flask?
C. What is the mole fraction of each?
Partial Pressures
W e'll st art by det ermining t he number of moles of ox y gen and ammonia.
   1 mo l 
  3 2.0 0g 

0mg
00
1



  
n O 2  1 50 .0mg
1 g
3
n O 2  4 .68 8 1 0



23
 0  mo lecu les
 6 .02 21
n N H3  4 1 0  mo lecu les 
21
1 mo l
3
n N H3  7  1 0
mo l
N ow, we us e the ideal gas law to c ompute t he pres sures of eac h gas .
PN 2 
PO 2 

1
 l 0 .08 206L at m K
0 .03 0mo
(2 98K
 )
1
 mo l
1 .0 L
n O 2 R 2 98 K
PN H3 
1 .0 L
n N H3 R 2 98 K
1 .0 L
PO 2  0 .11 5at m
PN H3  0 .2at m
mo l
PN 2  0 .73at m
Total Pressure
To determ ine t he total pres sure, we add t he part ial press ures
Pt ot  PN 2  PO 2  PN H3
Pt ot  1 .0at m
Mole Fractions
W e may det ermine mole f rac tions by us ing m oles or pres sures. I f we us e pres s ures ,
PN 2
XN 2 
Pt ot
PO 2
XO 2 
Pt ot
PN H3
XN H3 
Pt ot
XN 2  0 .73
XO 2  0 .11
XN H3  0 .2
Mole Fractions
U sing moles, we get
n t ot  n O 2  n N 2  n N H3
nN 2
XN 2 
n t ot
nO 2
XN 2 
n t ot
n N H3
XN H3 
n t ot
XN 2  0 .73
XO 2  0 .11
XN H3  0 .2
n t ot  0 .04 1mo l
Practice
A sample of KClO3 is heated and decomposes to
produce O2 gas. The gas is collected by water
displacement at 25°C. The total volume of the
collected gas is 229 mL at a pressure of 754 torr.
How many moles of oxygen formed?
Hint: The gas collected is a mixture so use Dalton’s Law
to calculate the pressure of oxygen then the ideal gas law
to find the number of moles oxygen.
PT = PO2 + PH2O
Solution
To determine t he press ure of2 ,Owe s ubtrac t t he v apor press ure of water atC25f rom the t ot a
press ure. The v apor pres sure of wat er at this t emperat ure is 23. 9 torr.
PO 2   7 54t
 orr  2 3.9t
 orr
PO 2  7 30 .1t orr
N ow we us e t he ideal gas law to determine the number of moles.
nO 2
( 7 30 .1t
 orr)  
PO 2 V
nO 2 
R T
3
n O 2  9 .00 1 0
mo l
  0 .22 9 L

 7 60t orr 
1 at m
1
0 .08 206L at m K
1
 mo l
 2 98 K