Unformatted text preview: 11/16/2011
11/16/2011 CHE 133
MAKEMAKEUP LABORATORY EXERCISE (12/5 or 7)
Eligible Students are ONLY those who have
excused absences from Gasometric Determination
of NaHCO3
in a
Mixture either a TEST (105 point)
or
or
PRELIMINARY (55 point) Exercise
MUST SIGN UP ON SHEET OUTSIDE OF
DR.
DR. AKHTAR’S OFFICE (213 CHEM)
BY November 24 Last Update: 11/16/2011 12:02 PM (details will follow) 1 ? QUESTIONS ? 2 Objective:
Determine the percent of NaHCO3
percent
in a mixture by Gasometry How can the composition of a mixture be
determined by measuring the volume of gas
evolved when that mixture undergoes a chemical
reaction? Concepts:
Ideal Gas Law
Vapor Pressure Henry’s Law
Stoichiometry Techniques:
Capture Gaseous Product
Corrections to volume What principles must be considered in using the
gas volume as a measure of the amount of gas
liberated? Apparatus:
Gas Syringe Thermometer Barometer 3 The Exercise is Conceptually Simple.
Simple. 4. Use Ideal Gas Law to get number of moles of CO2, nCO2 The unknowns consist of a uniform Pv=nRT mixture of NaHCO3 and NaCl n CO2 = Pv CO2
CO2 / RT 5. From Stoichiometry of reaction, get moles of NaHCO3
nNaHCO3 = nCO2 1. Weigh Sample, wSample.
2. Do Chemistry – Reaction with HCl 6. From number of moles, get weight
From
get
wNaHCO3 = nNaHCO3 * 84.0 g / mol NaHCO3 (s) + H+ (aq) Na+ (aq) + H2O (l) + CO2 (g)
+ Cl (aq)
+ Cl (aq)
NaCl(s) 4 7. Compute Percent Composition of Sample Na+ (aq) + Cl(aq) PctNaHCO3 = 100 * wNaHCO3 / wSample (NaCl will dissolve in, but not react with, HCl) 3. Capture Liberated Gas and Measure its volume, vCO2
5 6 1 11/16/2011
11/16/2011 The Basic Experimental Arrangement But  it Involves Some Important Concepts Limiting Reagents
Gas Mixtures – Partial Pressure
Gas Solubility – Henry’s Law Syringe Does CO2 behave ideally?
Accuracy & Reproducibility NaHCO3 / NaCl HCl
7 How Much HCl Insures that Unknown is the Limiting
Reagent?
NaHCO3 (s) + H+ (aq) Na+ 1 mol NaHCO3 (aq) + H2O (l) + CO2 (g) 8 How Much CO2 is produced?
You weigh ~ 0.2 g of unknown.
What is maximum volume of CO2 we can expect?
( with P = 1.0 atm and T = 25oC = 298oK ) 1 mol HCl Stoichiometery is 1 : 1. Assume unknown is pure NaHCO3
200 mg NaHCO3/(84 mg/mmol ) = 2.4 mmol max of unknown need 2.4 mmol of HCl max to consume it On the
NaHCO3 (s) + H+ (aq) Na+ (aq) + H2O (l) + CO2 (g)
analytical
balance
balance
1 mol NaHCO3 1 mol CO2
0.200 g = 200 mg = 2.4 mmol HCl is 1.0 M = 1.0 mmol /mL
Therefore, need at most 2.4 mmol/ 1.0 mmol/L = 2.4 mL v = n R T/ P = 2.4 X 0.0821 X 298/1.0 = 59 mL
(Syringe capacity = 60 mL but unknowns < 100% NaHCO3) Are using 10 mL of 1.0 M HCl ( 10 X 1.0 = 10 mmol )
10
– a significant excess Unknown will be the limiting reagent 9 How
How much H2O(l) is produced by the reaction? At STP: 1 mole occupies 22.4 L
1 mmole occupies 22.4 mL 10 What Percent of Gas Volume is the H2O we Produce
in the Reaction? NaHCO3 (s) + H+ (aq) Na+ (aq) + H2O (l) + CO2 (g)
Reaction produces water.
Is volume of water produced enough to large enough
to affect volume of gas measured (max 60 mL)? 100 X 0.042 mL
 = 0.07 %
60 mL 1 mol NaHCO3 1 mol H2O
Still, considering pure NaHCO3,
(200 mg NaHCO3 = 2.4 mmol) Volume of water produced by the reaction is
< 0.1% of volume of gas collected We produce at most 2.4 mmol of liquid H2O
2.4
Is this volume significant compared to the
~ 60 mL of CO2 produced?
11 12 2 11/16/2011
11/16/2011 Gas Mixtures – Partial Pressure What’s in the System?
Initial (mmol) Final H2O (l) Species 556 mmol 556 + x mmol HCl (aq) 10 mmol 10 – x mmol x mmol Reaction is conducted in a closed system
• at constant external pressure
 atmospheric (P ~ 1 atm)
• at constant temperature
 room temperature (T ~ 25oC) 0 mmol NaHCO3 (s)
NaCl (s) y mmol 0 mmol Air (g) w mmol w mmol NaCl (aq) 0 mmol
mmol x + y mmol
mmol CO2 (g) 0 mmol x  z mmol CO2 (aq) 0 mmol z mmol Initially:
System contains air & water (HCl)
water
Pressure in system is potentially due to:
• water (from HCl)
PH2O
• the air in the system
Pair
And after reaction,
• the liberated CO2 Patm PH2O + Pair = Patm PH2O, Pair and PCO2 PH2O + Pair + PCO2 = Patm 13 14 What is the Magnitude of PH2O?
Table showing vapor pressure of water as a
vapor
Vapor
function
function of temperaturePressure of Water lab.
is posted in P P H2O
PIH2O ,T, vI, nair PH2O is a
function of
temperature
temperature Vapor Pressure (mm Hg) 60 PFH2O,T, vF, nair,
H2O P nCO2
nCO2s H2O(l) H2O(l) 50 Initially, we have
P = PIair + PIH2O Finally, we have
P = PFair + PFH2O + PFCO2
P,T & nair don’t change, so
PIH2O = PFH2O = PH2O and 40
30
20
10 P – PH2O = PIair 0
0 5 10 15 20 25 30 35 40 Temperature (oC) Over the range 20oC – 30oC, PH2O increases from:
17.5 to 31.8 mm Hg
(0.023 to 0.042 atm)
2.3% to 4.2% for P ~1 atm P = n RT/v P – PH2O = PFair + PFCO2 PIair = PFair + PFCO2
PFCO2 = PIair  PFair 15 nCO2 RT/vF = nair RT/vI  nair RT/vF 16 What are the various volumes in the exercise? nCO2 RT/vF = nair RT/vI  nair RT/vF
nCO2 /vF = nair /vI  nair /vF
nCO2 vSYS: The volume of
just the large test
tube and the right
angle elbow = nair vF (1/vI  1/vF) But, nair = (P  PH2O) vI / RT
P
nCO2 = [(P  PH2O) vI / RT] vF (1/vI  1/vF)
P
nCO2 = (P  PH2O) (vF  vI) / RT
P You will
measure
this! Equation 4 vI:
The gas phase*
volume of the entire
closed system before
reaction What is partial pressure of CO2 at end of the reaction?
PCO2 = nCO2 RT/vF PCO2 = [(P  PH2O) (vF  vI) /RT ] RT/vF
P
PCO2 = (P  PH2O) (vF – vI)/vF vSYS initial vSYS – vHCl + vsyringe Equation 6
17 *We must exclude the volume of the liquid HCl but can ignore the solid NaHCO3
18 3 11/16/2011
11/16/2011 Gas Solubility – Henry’s Law vF: The gas phase*
volume of the entire
closed system after
reaction nCO2s vSYS – vHCl + vfinal
syringe *We must exclude the
volume of the HCl again. Equation 4 final
initial
vF – vI = vSYS – vHCl + vsyringe – vSYS + vHCl  vsyringe
final
initial
vF – vI = vSYS – vHCl + vsyringe – vSYS + vHCl  vsyringe
1 initial
vsyringe
final
vsyringe 19 PCO2 =
vI = ( P  PH2O ) ( 1 – vI / vF) SCO2 = kH * PCO2 ( kH = 3.2 X 102 mol / Latm )
Suppose:
• vSYS  volume of large test tube & right angle elbow is
SYS
95.0 mL
• Initial syringe reading is 5.0 mL and the final reading is
53.0 mL
• We use 10.0 mL of HCl NOTEthat: :
Note THAT mol of liberated CO2 dissolves in water.
Using Henry’s Law, can calculate concentration
of CO2 dissolved in water. • P = 1.000 atm, = We can estimate the deviation from ideality by
examining the van der Waals constants. vF = 95.0 mL – 10.0 mL + 53.0 mL = 138.0 mL
( 1.000 – 0.026) ( 1 – 90.0 / 138.0) PCO2 = 20 Can CO2 be treated as an Ideal Gas? 90.0 mL PCO2 = 0.339 atm ( P + a / v2 ) ( v – b ) = RT
v We use 10.0 mL of HCl CO2 nCO2s = 10.0 mL * 0.011 mmol/mL = 0.11 mmol 0.6% 0.18% Small corrections compared to others 21 Review
variable 1 mol For CO2 at Room Temperature, the corrections
to P and v are:
a / v2 b / v SCO2 = 3.2 X 102 mol/Latm * 0.339 atm = 0.011 M Issue 0.026
0.026 Hg
20 mmatm Departure of gases from Ideality? Equation 6 95.0 mL – 10.0 mL + 5.0 mL PH2O = • T = 22oC 22 Reproducibility
(Factors affecting Precision) Effect Wsample analytical balance (0.0004 /0.3000) ~0.1% volume of H2O Produced vCO2 0.1% Non Ideality of CO2 PCO2 0.6% vapor Pressure of Water PCO2 2.3 – 4.2% Solubility of CO2 nCO2 3.0% PCO2 barometer (1 / 760) ~0.1% PH2O table of values (1 / 760) ~0.1% T thermometer (1/ 300)* ~0.3% (0.5 / 50) ~1% vCO2 If we seek accuracy to within 1% syringe Precision should be about 1%
Accuracy should be less than 3%
23 * Assuming no ambient temperature changes – see Prob 2
24 4 11/16/2011
11/16/2011 Calculations
Weight of Sample
Volume of gaseous CO2
v
Pressure,
P = 752 mm Hg =
Temperature, T = 23oC =
[email protected] 23oC (from Table) 21 mm Hg 0.2147 g
45.7 mL
0.989 atm
296 K
0.028 atm mmol CO2 (gas) = (P  PH2O)v / RT
mmol CO2 (liquid) (Henry’s Law) 1.81 mmol
0.10 mmol Tot CO2 1.91 mmol To calculate the Henry’s Law correction, we need the
Volume of the System. Suppose it is 100.0 mL
That makes the initial volume,
VI = 100.0 10.0 + 5.0
= 95.0 mL
and the final volume,
VF = 100.0 10.0 + 50.7
= 140.7 mL Vsys = 100.0 mL
Syringe:
Initial = 5.0 mL
Final = 50.7 mL PCO2 = ( P  PH2O )( 1 – vI / vF)
= ( 0.989 – 0.028)( 1 – 95.0 / 140.7) = 0.312 atm
SCO2 = 3.2 X 102 * 0.312 atm = 0.010 M 25 Calculations Calculated Weight of Sample
v
Volume of gaseous CO2
Pressure,
P = 752 mm Hg =
Temperature, T = 23oC =
[email protected] 23oC (from Table) 21 mm Hg 0.2147 g
45.7 mL
0.989 atm
296 K
0.028 atm mmol CO2 (gas) = (P  PH2O)v / RT
mmol CO2 (liquid) (Henry’s Law) 1.81 mmol
0.10 mmol Tot CO2
1.91 mmol
mmol NaHCO3 (from stoichiometry) 1.91 mmol
0.160 g
Weight of NaHCO3 1.91 X 84.0
% NaHCO3 = 100 X 0.160 / 0.2147 = 74.5 %27 nCO2s = 10.0 mL * 0.010 mmol/mL = 0.10 mmol 26 Procedure  Notes
Everyone should weigh ~ 200 mg = 0.2 g
– ACCURATELY – for their initial run
Depending on your sample, you may need to adjust
the weight in subsequent runs to insure that you
get between 30 and 50 mL of CO2, but not more
than 50 mL.
Test that system is airtight before using
Set syringe at 5.0 mL initially – read to 1 decimal
– remember to subtract from final volume
Do test run  then 4 which you report. 28 Last Exercise Determination of NaHCO3
in a Mixture
Part 2  Gravimetric
Gravimetric
Final Exercise – 105 points
Do SUSB054  Prelab Assignment 2 Final Quiz will be given at the beginning of
the checkout laboratory meeting
29 5 ...
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 ROBERTSCHNEIDER
 Chemistry, ml, mmol

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