Experiment 1
Sunday, November 27, 2005
3:59 PM
Need to Know
• Know the reaction equations, how the reactions occur (ex. spontaneously, by heating, etc.) and
the physical descriptions and color of both the reactants and products
o Ex. CuO (s) (black solid) + H2SO4 (aq) + H2O --> Cu(SO)4 (aq) (blue solution) + H2O
• Know how to calculate percent yield (i.e. CuSO4 was obtained from the reaction of CuO with
sulfuric acid. If 2.5 g of CuSO4 was obtained from 5.0 g of CuO, what is the percent yield?)
Purpose
• To synthesize different copper compounds
Theory
• Many organic and inorganic compounds are synthesized by the chemical industry even though
they can be found in nature, because a limited natural supply or expensive extraction process
may make synthesis more economical
• Things we have to take into consideration when synthesizing:
o Availability of equipment
o Percentage yield
o Value of by-products
• This experiment illustrates the synthesis of several copper compounds from metallic copper:
o Cu -> Cu(NO3)2 -> Cu(OH)2 -> CuO -> CuSO4--5H2O -> Cu
• We expect to get the same mass of copper at the end than what we started with…in order to do
so, we must prevent loss by:
o Avoiding spattering while boiling
o Not leaving product on the sides of beakers
o Not spilling the product
o Purifying precipitates by washing efficiently then drying completely before weighing
Procedure
• Part 1: Synthesis of Copper(II) Nitrate and Copper(II) Hydroxide
o Cu + 4HNO3 -> Cu(NO3)2 + 2NO2 + 2H2O
Reaction notes:
•
It is SPONTANEOUS
We carry this out in the fume hood and swirl the reaction mixture to
remove any gases trapped in the solution
o Cu(NO3)2 + 2NaOH -> Cu(OH)2 + 2NaNO3
• Reaction notes:
It is SPONTANEOUS
The solution should be basic (alkaline) after the addition
Cu(OH)2, the product, is a gelatinous precipitate
Cu(OH)2, the product, is BLUE
• Part 2: Synthesis of Copper(II) Oxide
o Cu(OH)2 –Δ-> CuO + H2O
• Reaction notes:
It is BY HEATING
We want to convert Cu(OH)2 to CuO because it is a LESS GELATINOUS
precipitate than Cu(OH)2 is, and thus easier to isolate
The BLUE Cu(OH)2 becomes BLACK CuO
If heating doesn't do the trick we add MORE NaOH
o
Filtration
• Filter with a suction filter flask and a Buchner funnel Wash the CuO with water both to get it out of the beaker and because it is wet
•
with a solution which contains NaNO3 and NaOH, and we want to get rid of it
• Part 3: Synthesis of Copper(II) Sulfate
o
CuO + H2SO4 –Δ-> CuSO4 + H2O
• Reaction notes:
It is BY HEATING
As the CuSO 4orms, it dissolves into Cu + S4 , and the Cu ion gets
hydrated to become Cu(H O2 4+
The BLACK CuO will dissolve into a BLUE solution
• Part 4: Synthesis of Copper
o CuSO4 + Zn -> ZnSO4 + Cu
• Reaction notes:
It is SPONTANEOUS
Zinc is more chemically active than copper and displaces copper(II)
ions from solutions, meaning that it is better at combining with SO4
than Cu is
The solid visible consists of unreacted zinc metal and copper metal
(the product)
The BLUE solution will turn WHITE/CLEAR
o Zn + 2HCl -> ZnCl2 + H2
• Reaction notes:
It is SPONTANEOUS
The purpose of this is to remove excess zinc metal from the previous
reaction
We know this reaction is over when we don't see anymore bubbles,
because that is the formation of H2 happening
Questions to Understand
• Why would we allow the last traces of water to evaporate slowly rather than rapidly expelling the
water by intense heating?
o "At the end of the experiment, the final traces of water were allowed to evaporate
instead of intensely heating the copper because we wanted to prevent any heat-
catalyzed oxidation reactions from occurring which would have converted the metallic
copper to oxides (for example, CuO), thus introducing an impurity into our sample."
Experiment 2
Wednesday, November 30, 2005
9:51 AM
Need to Know
• Be able to calculate molecular weights as done for your report
• Know what substances (and their phases) are present at each stage of the experiment
• Understand what errors may have occurred and how specific errors will affect the calculated
molecular weight
Purpose
• To determine the molecular weight of volatile liquids using the Dumas Method
Theory
• A volatile liquid is one that evaporates easily and allows us to use the Dumas Method to
determine its molecular weight
o The Dumas Method assumes that the vapor obeys the Ideal Gas Law: PV=nRT
o This means that liquids with WEAKER intermolecular forces will be more accurate than
those with stronger ones (why? I don't know…) • The idea here is that we will place a volatile liquid into an Erlenmeyer flask where the boiling
point of the liquid is above room temperature (so it won't boil spontaneously) but below water
(so that we can put it in a boiling water bath and make it boil)
o We cover the flask with foil but prick a hole to allow gas to escape
o When we heat the liquid, it will evaporate and gas will escape from the flask until there
is only so much inside that the pressure inside the flask EQUALS the atmospheric
pressure of the lab outside the flask
o Once we reach this point, we can use the PV=nRT equation because:
• We know P: it is the atmospheric pressure of the lab
• We know V: it is the volume of the flask because there is going to be just
enough gas left in there as needed to fill up the flask
• We know R: it is a constant
• We know T: the temperature inside the flask will be equal to the temperature of
the water bath outside it
o
Thus we can calculate the molar amount of gas
• Then we weigh the beaker to find how much the gas weighs
• Now we know the molar amount and the weight, and we can calculate
molecular weight!
Procedure
• Set up the Erlenmeyer flask with the volatile liquid inside
Boil the water bath and then put the flask in at a 45o angle (if not, we will not be able to tell
•
when the liquid has evaporated!)
o The water should be SLOWLY boiling (or else we'll lose it)
o If any water gets into the flask, it's game over…we must re-start the experiment (think
about why)
• As soon as all the liquid has disappeared, continue heating for 1 more minute and then remove
the flask
• Let it sit for 15-20 minutes so that all the vapor and condense back into liquid form
• Weigh the flask to find the weight of the liquid and we're good to go with the PV=nRT equation!
Questions to Understand
• What substances and phases of the substances are present in the flask after the methanol is
added and the flask covered with foil?
o Liquid methanol
o Gas methanol (there is always a LITTLE evaporation)
o Air (oxygen gas)
• What substances and phases of the substances are present in the flask just before you remove it
from the hot water bath?
o Gas methanol
• NO liquid because it has all disappeared!
• NO oxygen because the evaporation of the methanol has pushed it all out!
• What substances and phases of the substances are present in the flask when it is weighed after
cooling?
o Liquid methanol (because it has condensed by now!)
o Gas methanol (again, there is always a little evaporation)
o Air (because after all, the thing is open to the environment)
• What factors may contribute to the lack of accuracy in your calculated molecular weights?
o I don't know, what do you think?
Experiment 3
Sunday, December 04, 2005
10:14 PM Experiment 3 – Acid-Base Titrations: Identification of an Unknown Solid Acid
Need to Know
• Be able to perform simple stoichiometric calculations (like the pre-lab questions)
• Know how to do the calculations in the standardization of a NaOH solution with oxalic acid
• Know how to calculate the molecular weight of an unknown acid
• Know what the different terms used in the manual concerning titrations and standardized
solutions mean (primary and secondary standards, mono-, di-, and tri-protic acids, weak vs.
strong acids and bases)
• Know the weak and strong acids and bases listed in Table 1
Theory
• The fundamental process occurring in an acid-base reaction is the transfer of a proton (H+) from
the acid to the base
• Bronsted and Lowry’s definitions:
• An acid is a substance that can donate protons
• A base is a substance that can accept a proton
Neutralization is a proton transfer from acid to base
•
• Acid strength is related to the ability of a substance to give up protons (i.e. a strong acid is one
which gives up protons very easily)
• A conjugate base is what the acid becomes after it has given up a proton
• The strength of an acid is inversely related to the strength of its conjugate base (think
about why this must be true)
• Know Table 1, for goodness’ sake (pg. 33)
• Strong acid + strong base neutralization reaction always goes to completion
• Weak acid + weak base neutralization reaction DOES NOT go to completion
• Strong acid + weak base (or vice versa) neutralization reaction always goes to completion
• Why? Because what little of the weaker substance dissociates is quickly neutralized by
the stronger acid/base (to form a salt and water), and thus to maintain the equilibrium,
some more of the weaker substance must dissociate until ultimately everything is gone
• However, the overall equations for the different types of neutralization reactions are different!
• With a strong acid + strong base, they both completely dissociate and so the overall
+ -
reaction is simply H3O + OH -> 2H O 2
• The Na and Cl dissociate at the beginning and don’t even participate in the
reaction, so they are SPECTATOR IONS
• However, with stron
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