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Chem 205 Laboratory Reports.
Chem 205 Laboratory Reports.
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Lab reports for Chem 205 (Concordia) Introductory Chemistry I.
Chemistry 205
Experiment 1: Densities of Organic Liquids
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Laboratory Report
Data and Observations
Ethyl
acetate
Hexane
Acetone
Unknow n
Mass of empty vial and cap (g)
12.583
11.719
11.856
12.431
Mass of vial, cap and liquid (g)
13.494
12.456
12.771
13.263
Mass of liquid (g)
0.911
0.737
0.915
0.832
Initial pipette reading (mL)
0
0
0
0
Final pipette reading (mL)
-1.0
-1.0
-1.0
-1.0
Volume of liquid delivered
1.0
1.0
1.0
1.0
Does it dissolve in water?
no
no
yes
yes
If not, does it sink or float?
floats
floats
ø
ø
Calculation and Analysis
1.
Densities of organic solvants
Ethyl
acetate
Hexane
Acetone
Unknow n
Calculated density (g/mL)
0.911
0.737
0.915
0.832
Accepted value (g/mL)
0.902
0.659
0.791
see below
9.98
11.7
15.7
see below
Relative error ( % )
1. The average relative errors of the densities is 12.36%. It is therefor impossible to
identify the unknown liquid. The possibilities are:
- tert-butyl methyl ether (0.741 g/mL, 12.2 % of relative error),
ethanol ( 0.785 g/mL, 5.98 % of relative error),
tert-butanol (0.786 g/mL, 5.85 % of relative error),
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acetone (0.791 g/mL, 5.18 % of relative error),
toluene ( 0.867 g/mL, 4.03 % of relative error),
ethyl acetate (0.902 g/mL, 7.76 % of relative error).
But, because the calculated densities gotten in this experiment procedure are over the
accepted values, it cannot be toluene nor ethyl acetate. And since the melting point of tertbutanol is 25˚C
, the only possibilities left are acetone, ethanol and tert-butyl methyl ether. All of which
being soluble in water
, there is no more information to tell them apart.
3.The relative error of the unknown liquid density is 5.18 % if it is acetone, 5.98 % if it is
ethanol or 0.741 % if it is tert-butyl methyl ether.
4.
Densities of ethyl acetate
Run
Density Experiment
Absolute Deviation
1
0.911 g/mL
0.001 g/mL
2
0.886 g/mL
0.026 g/mL
3
0.886 g/mL
0.026 g/mL
4
0.906 g/mL
0.006 g/mL
5
1.01 g/mL
0.098 g/mL
6
0.875 g/mL
0.037 g/mL
Averages
0.912 g/mL
0.032 g/mL
5. The average absolute deviation is 0.032 g/mL. This is a measure of how precise the
measurements are. Meanwhile, the average error of my measurement is 9.98% This says
how accurate I was compared to the accepted value. In order to improve the accuracy of
the accepted value, the absolute deviation of a set of values is more useful than one
single measure compared to the actual accepted value. That is, considering the value we
have here, it is possible to consider 0.912 ± 0.032 g/mL as an accepted value, since it
was calculated using six independent runs of a same procedure. To obtain the best
possible value, different procedures have to be repeated over and over, each time
minimizing errors until a set of values with a minimal average deviation is obtained.
6. Since the average value of set no. 1 is lower than the accepted value, this means that
either the balance was registering masses too low, or the pipette was delivering a
smaller volume than indicated. That is proved by the fact that if exactly 1 mL of water is
weighted by the hypothetical balance, it would give a mass of lets say 0.97 g, therefor a
density lower than the accepted value. And if the hypothetical pipette delivers instead of 1
mL, 0.97 mL, a perfectly zeroed balance reads 0.97 g, and the calculated density would
be, again, smaller than the accepted value.
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7. The syringes are usually less precise than pipettes when it comes to measurement.
This is why we use pipettes
Chemistry 205
Experiment 2: Densities of Organic Liquids
Prelaboratory Questions
1. a) Make the different acids (hydrochloric and acetic) react with a limiting amount of steel
wool. Observe any changes in color, smell, etc.
b) Observe the change in gas volume created by the reaction of acetic acid with steel
wool, where acetic acid is the limiting reactant. (Place the wool in a test tube, add a few
drops of acid, reverse the test tube in a beaker half-filled with water, note changes in water
level)
c) Make 4 mL of sodium carbonate and 1 mL of calcium chloride react together, observe
changes in weight. Add sulfuric acid, observe further changes. (weight all the solutions
and all the containers together before and after each reaction)
2. -Calcium chloride: CaCl₂ , solid.
-Sodium carbonate: CaCO₃ , solid.
-Sodium bicarbonate: Ca(CO₃)₂ , solid.
-Carbon dioxide: CO₂ , gas.
-The compound which, when dissolved in water, yields hydrochloric acid: HCl, gas.
-The compound which, when dissolved in water, yields sulfuric acid: H₂SO₄ , liquid.
4.Observation: “The copper sheet [...] changes from the shiny pinkish color of pure copper
to the familiar green color of old copper rooves.”
Speculation: “The copper sheet [...] slowly reacts with water and carb on
dioxide in the air [...].”
Chemistry 205
Experiment 2: Making Scientific Observations
Laboratory Report
Objective:
Differentiate ob servations, speculations and conclusions when ob serving reactions
implying acids, b asses and metal.
Observations and Data:
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1. Action of hydrochloric acid on steel wool:
Before reaction: The acid is a clear liquid, with a light vinegary smell. The steel wool is dull
gray.
During reaction: There are little gas b ub b les forming on the steel wool.
After reaction: The acid still looks the same, it doesn’t smell as much. The steel seams a
little b righter, b ut it can b e b ecause it is wet.
Action of acetic acid on steel wool:
Before reaction: The acid is a colorless, strongly smelling liquid. The steel wool is dull
gray.
During reaction: No ob servations.
After reaction: Nothing changes.
1. Data for quantitative examination of action of acetic acid on steel wool:
Length of the inside tube: 72 mm
Inside diameter of the tube: 8 mm
Volume of the test tube: 14476 mm²
Initial reading of the level of water in the reaction tube: 0 mm
Final reading of the level of water in the reaction tube: 10 mm
Volume of level change: 2010 mm²
Percentage change in the gas volume in the reaction tube: 14 %
1. Data for reaction involving sodium carbonate:
Mass of empty glassware: 79.491 g
Mass of beaker, Erlenmeyer and test tubes before reactions: 85.780g
Observations upon adding the calcium chlorine solution: Formation of a white precipitate.
The mixture looks more viscous.
Mass of beaker, Erlenmeyer and test tubes after adding the calcium chlorine solution:
85.778 g
Mass change: -0.002 g
Observations upon adding the sulfuric acid: There is a violent reaction taking place,
forming a lot of gas and a white powder. The mixture is very cloudy.
Mass of beaker, Erlenmeyer and test tubes after adding the sulfuric acid: 85.694 g
Mass change: -0.084 g
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1. Procedure for disposal of the steel wool residues:
a) Neutralize the acid b y slowly pouring sodium b icarb onate solution in the test tub e until
it stops b ub b ling.
b) Place a sheet of b rown paper in the sink, away from the drain
c) Pour the solution in the sink, making sure any steel residue sicks on the paper.
d) If the principal steel wool b all doesn’t fall b y itself, place the test tub e upside-down on
the paper and gently tap it to dislodge the b all.
e) Rinse the tub e with some water, repeat step c. Let water run for a few seconds.
f) Dispose of the steel wool in the designated container, throw the paper in regular
garb age.
Conclusions and Interpretation of Results
1. Based on notes about strong and weak acids, it is possible to say HCl is a stronger
acid than acetic acid. A strong acid is a substance that, when dissolved in water,
separates into H⁺ and a anion. With the observations made, an easy speculate would be
that strong acids reacts more with metals than weak acids. furthermore, this hypothesis
becomes possible: the gas produced by the reaction with HCl is H₂; after a long period of
time, Cl will create a deposit on the Fe as the concentration of hydrogen ions lower.
2. The gas volume in the test tube diminished, proving that a gas was used during the
reaction, probably oxygen, to form an iron oxide. If HCl had been used, maybe the reaction
would have been faster.
3. Unlike a cylinder, the test tube has a rounded bottom. This means that the volume of a
test is smaller than the volume of the cylinder with corresponding dimensions. But, this
error is only at the bottom. At the top, the tube opens up a little and sometimes, even has a
tip, creating an irregular volume for the opening. When both those things are taken in
consideration, our calculation has a net negative error. The real value is a larger number
than the calculated one.
4. Based on the “Water Solubility of Ionic Compounds” table, the precipitate created upon
addition of calcium chloride in the solution of sodium carbonate is calcium carbonate,
because salts of carbonate are insoluble and calcium does not figure in the soluble
compounds. At that stage, a small change of mass was noted. This means, somewhere
in the reaction, a gas was produced. But, no bubbling was observed. When sulfuric acid is
added, a possible reaction can be that one of the H⁺ from the acid creates a covalent
bond with the carbonate ions, and has as product calcium bicarbonate. But, this
hypothesis does not explain why there is a gas produces and a loss of mass. An
explanation to that would be that in contact with some ion the solution, the hydrogen ions
bonds together and gaseous hydrogen is formed. In the solubility table, we can see that
calcium chloride is insoluble. This leads to an other possibility: after the addition of the
sulfuric acid; hydrogen “steals” oxygen from carbonate, creating water and carbon dioxide.
In this case, the second precipitate would be calcium sulfate.
5. “Matter can neither be created nor destroyed” (from week 4’s PowerPoint notes)
The experiments conducted in the Erlenmeyer flask do not provide evidence of this law.
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There was a net diminution of mass from the reactants to the products. This ever goes
against the law. But, we have evidence that a gas was produced. We can conclude that
the difference of masses come from the fact the gas was not weighted alongside the
other products after the reactions. But, this reaction is not a good evidence of the law of
conservation of matter since it if not happening in a closed system.
Chemistry 205
Experiment 3: Separation of a mixture
Prelaboratory Questions
1. Procedure:
a. Remove the iron filings
1.Place the mixture in a weighing dish. Note the mass.
2.Using a magnet, get all the iron on one side of the dish. Do not put the magnet in the
mixture, work through the dish.
3.Pour out the mixture in a flask, holding the iron back with the magnet. Note the mass.
a. Recuperate naphthalene
1.Fill in a test tube with ice, fit the adapter on it. Place the test tube in the flask.
2.Put the flask, with it’s side-arm open, in a 75-80˚C water bath. Fit a pipette bulb on the
side-arm of the flask
3.Change melted ice for frozen one in the test tube regularly.
4.When all the naphthalene is securely crystalized, remove the test tube, scrape off the
crystals and weight them.
a. Separate the solids
1.Weight the flask, add water to the solution.
2.Connect everything like in the picture in the manual.
3.Pour out the liquid though the filter. Start step d with urea solution.
4. Place the filtrate in a beaker, add water to particles left in the flask, pour it all in the
beaker. Let the sand sink and the plastic float. Separate.
a. Crystalize urea
1.Place a stopper on the flask, using the water aspirator, make sure air flows.
2. Slowly heat up solution without letting it bubble.
2. Urea is the only substance of the mixture that is dissolved in water, because it it the only
one one that is left when we get to the point of evaporating the solvent.
3. Since urea is a polar molecule (it dissolved in water) and naphthalene is not (since it
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was not dissolved), any non-polar solvent would dissolve naphthalene, like hexane or
dichloromethane but urea would stay in the water. Then, by decantation and evaporation,
both solutes can be be extracted and isolated.
4. a.Moth balls tend to sublimate when left in an open area for long.
b.Naphthalene can be detected using smell.
c.That is a result of its volatility.
5. Iron is the only element in the mixture: naphthalene and urea are compounds, sand
and polypropylene are themselves mixtures. This said, there might be trace of a pure
element in the sand.
1. Sand is likely to contain: gemstones, metals, rocks, granite, shell fragments, etc.
In terms of chemical elements, we will find: sulfur, oxygen, carbon, hydrogen, iron, copper,
nickel, silver, etc.
Chemistry 205
Experiment 3: Separation of a mixture
Laboratory Report
Objective
Separate components of a mixture using their physical properties.
Procedure
1. Weight approximately 1 g of the mixture in a weighting dish.
1.1 Re-zero the the balance before each measurement
1.2 Note the mass of the dish
1.3 Pour some of the mixture in the dish.
1.4 Weight and add or remove until the balance read “mass of the dish + ~1 g”
1.5 Calculate the exact of mixture used.
1. Separate the iron from the mixture
2.1 In the weighting dish, push all the mixture to one side
2.2 Place the side of the u-shaped magnet on the bottom of the dish, near the edge of the
heap of mixture.
2.3 Slowly drag the magnet to the opposite side of the dish.
2.4 Repeat the two previous steps until most of the iron is separated from the mixture.
Use a scoopula to scrape the mixture back in the heap as needed.
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2.5 Scoop the iron into a weighted flask.
2.6 Repeat this process until all of the iron has been dragged out of the mix.
1. Separate the naphthalene from the mixture
3.1 Pour the rest of the mixture in an erlenmeyer flask.
3.2 Prepare the apparatus found at page 3 - 21 of the lab manual
3.3 When the water temperature reaches 75˚C, turn down the heat and put on the side
bulb.
3.4 Make naphthalene condensate
3.4.1 Use the pipette to draw out melted ice. Add more ice as needed.
3.4.2 Wipe the water of the rubber stopper
3.6 Keep repeating step 3.4 until all the naphthalene has solidly crystalized on the cold
test tube.
3.7 Take out the assemblage but keep the water bath.
3.8 Scrape off the naphthalene in a weighted flask
1. Separate urea
4.1 Prepare the appropriated filtration apparatus as found on page 3 - 22w
4.2 Pour out the solution that is now found in the flask from previous step on the filter.
Water comes from condensation
4.3 Place the flask with the filtrated urea solution in the water bath. Connect to the water
aspirator. Turn the aspirator on.
4.4 Wait for the water to evaporate, do steps 5 and 6.
4.5 In this case, water did not evaporate. Discard remaining solution in the sink at the end
of the experiment.
1. Separate the polypropylene
5.1 Add water to the sand and polypropylene mix. Pour it all in a small beaker
5.2 Let it rest for a while.
5.3 Using the filtrations apparatus, pour only about half of the water, the polypropylene
should all fall on the filter paper.
5.4 Scrape any remaining polypropylene from the side of the beaker.
5.5 Let the filter paper dry. Transfer the polymer in a weighted flask.
1. Dry sand
6.1 Carefully pour out any supernatant water.
6.2 Place the beaker on the hot plate until sand is dry.
6.3 Transfer the sand in a weighted flask.
Data and Observations
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Mass of mixture used
1.123 g
Mass of iron recovered
0.730 g
Percent in mixture
65.0%
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Mass of naphthalene recovered
0.001 g
Percent in mixture
0.09%
Mass of urea recovered
0g
Percent in mixture
0%
Mass of polypropylene recovered
Percent in mixture
0.053 g
4.7%
Mass of sand recovered
0.196 g
Percent in mixture
17.4%
Total mass of material recovered
0.980 g
Total percent recovery
87.3%
The principal difficulties encountered during the procedure were:
-When trying to separate the iron from the rest, particles are dragged the metal. This part
must then be repeated a number of times.
-During that, naphthalene keeps sublimating and is lost.
-For the set of naphthalene sublimation, water kept condensation on the rubber stopper
and it slipped into the erlenmeyer flask, which get the mixture wet and later, made it
impossible to transfer out of the flask into a beaker.
-When trying to isolate urea, time became a problem as water evaporated too slowly and
urea was never recuperated.
-For the steps toward the separation of the polypropylene from sand, many problems
arose. Fist, once on the filter paper, it was almost impossible to get the particles of it
without scraping little pieces of paper along.
-Then, for decantation, some sand appeared to be floating, with the polypropylene instead
of sinking as expected. This might be explained by the fact that, with there very small
masses, sand particles can be supported by the force of the water membrane like the one
that allows water spiders to stay afloat.
Additional Questions
1. Yes, the order in which the separation was performed was important. some steps
could have been interchanged, but not all. For example, naphthalene could have, and
should have been sublimated first. But, urea couldn’t have been separated first, because
the polypropylene would have melted at water’s boiling temperature.
2. If iron was replaced with magnesium, another method of separation should have been
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found because magnesium is a non-magnetic metal. On the other hand, nickel reacts just
like iron in contact of magnets, which means the method would have worked anyways
with nickel instead of iron.
3. Gold has a highest density then sand, so when the pan in swirled, sand will raise
higher on the sides because of centrifugal forces than gives a body an acceleration
inversely proportional to its mass.
Chemistry 205
Experiment 4a: Preparation of Lead(II) Iodide
Prelaboratory Questions
1.-Transfer 7.0 mL of potassium iodide solution and 2.0 mL of lead(II) nitrate solution in a
beaker using a graduated transfer pipette and a volumetric pipette respectively. Note
observations before, during and after the reaction.
-Prepare and weight a filter paper. Place the paper in a funnel and the funnel in an
erlenmeyer flask.
-Pour the solution and the suspended solid on the filter paper, making sure as much
solid as possible is poured out. Rinse any remaining precipitate with the filtrate. Filter
again. Discard the filtrate.
-Rinse the filter paper with 2 mL of pure water. Note the appearance of the solid. Discard
the filtrate.
-Repeat this last step using acetone.
-Let the filter paper and the solid dry in a beaker under the fume hood for about half an
hour.
-When everything is dry, weight the paper and lead(II) iodine. Note the mass of the solid.
Try transferring the solid in a weighted watch glass. Calculate the mass of solid. Discard
the product
2. Pb²⁺ (aq) + 2 NO₃⁻ (aq) + 2 K⁺ (aq) + 2 I⁻ (aq)
Pb²⁺ (aq) + 2 I⁻ (aq)
PbI₂ (s)
PbI₂ (s) + 2 K⁺ (aq) + 2 NO₃⁻(aq)
3.The limiting reactant lead(II) nitrate, because only 2 mL is used agains 7 mL of the other
reactant while both have the same molarity and the molar proportion is 1:2, not 2:7. This
means 1/7 of the potassium iodide will not react.The theoretical yield of the experiment is
therefor of 88.8% ( (7+2-1) mol / (7+2) mol )
Chemistry 205
Experiment 4b: Synthesis of manganese(II) chloride
Prelaboratory Questions
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1. -Weight about 0.05 to 0.1 g of manganese in a carefully weighted beaker. Note
mass.
-Add about 1.5 mL of hydrochloric acid to the metal using a Pasteur pipette and working
under a fume-hood. Wait for 5 minutes. Note any change in temperature.
-Dry the content of the beaker by heating it on a hot plate. Note its appearance.
-Let the beaker cool in the desiccator. Weight it.
-Reheat the beaker for 3-5 minutes. Repeat previous step. Do this until two consecutive
values have less then 0.003 g between them. Note the mass.
-Dissolve the manganese(II) chloride in water, dispose in provided container.
2. The molar mass of pure solid iron is: 55.85 g/mol.
This means 0.05 g represents 8.95255 x 10-4 moles.
The molar mass of FeCl2 is: 126.74 g/mol.
This means 0.145 g represents 1.14407 x10-3 moles.
The molar mass of FeCl3 is: 162.20 g/mol.
This means 0.145 g represents 8.93958 x 10-4 moles.
Since this reaction has a ratio of 1 mole of product per mol of iron, the chloride of iron
obtained was FeCl3, ferric chloride.
3. If the yield in the experiment had been less then 100 percent, n, the number of Cl per
molecule would had have to be higher, because the molar mass would have had to be
greater to compensate the fact the number of moles was smaller for the same mass of
products.
Chemistry 205
Experiment 4: Preparation of Some Metal Halides
Laboratory Report
Objective
Create lead(II) iodine and manganese(II) chloride using precipitation and redox reactions.
Introduction
A metal halide is a compound made of a metal and an element from the halogens, group
VII A. Some are used in light bulbs to complete the radiation spectrum of mercury. The
most common ways of creating metal halides are by precipitation of and insoluble halide
or by reduction of the solid metal.
A precipitation is a reaction is an aqueous solution were the cation from one reagents
couples up with the anion from the other to form an insoluble product. They other ions are
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called spectator ions and are found in the supernatant solution after the reaction.
A redox occurs when once specie gains electrons that are lost by the other one in the
reaction. It is in fact a couple of reaction: an oxidation, where electrons are lost and a
reduction, where electrons are gained. The names comes from the fact oxygen is a
common oxidizing agent. Redox reactions are often seen in solutions, but can also be the
result of the interaction between a solid and a gas, amongst other.
Procedure
A. Preparation of lead(II) iodine
1. Mix the reagent solutions.
1.1Transfer 7.0 mL of potassium iodide solution from the stock bottle into a clean beaker
using the dedicated 10 mL graduated transfer pipette.
1.2 Add 2.0 mL of lead(II) nitrate solution in a beaker using a volumetric pipette.
1.3 Swirl gently to mix.
1.4 Note observations before, during and after the reaction.
2. Filter the product.
2.1 Prepare and weight a filter paper. Place the paper in a funnel and the funnel
in an
erlenmeyer flask.
2.2 Keep the solid in suspension using the rubber policeman.
2.3 Pour the solution and the suspended solid on the filter paper.
2.4 Wipe the rubber policeman on the filter.
2.5 Rinse any remaining precipitate with the filtrate.
2.6 Repeat filtration process on the same filter until the beaker is clean.
1. Purify the precipitate
3.1 Rinse the filter paper with distilled water.
3.2 Note the appearance of the solid and the filtrate.
3.3 Rinse the filter paper with acetone.
3.4 Discard the filtrate in the container.
3.5 Let the filter dry.
(Start part B of the experiment)
1. Isolate the product
4.1 Unfold the filter paper and place it on top of a beaker.
4.2 While waiting for product B to cool, hold the paper above the warm hot plate.
4.3 When the paper is dry and the powder starts to get loose, note mass
4.4 Carefully transfer PbI2 on the weighted glass watch. Note mass.
1.Weight about 0.05 to 0.1 g of manganese in a carefully weighted beaker. Note mass.
1.1 Note mass of empty beaker
1.2 Add some manganese
1.3 Adjust volume until the mass falls in the required bracket.
1. Add hydrochloric acid
2.1 Work under a fume hood.
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2.2 Using the provided bottle, add 30 drops of HCl
2.3 Let everything react for 5 minutes. Note observations.
3. Isolate the product
3.1 Dry the content of the beaker by heating it on a hot plate.
3.2 Note its appearance.
3.3 Let the beaker cool in the desiccooler. Weight it.
3.5 Reheat the beaker for 5 minutes.
3.6 Repeat steps 3.3 and 3.4 two times, finish by cooling
3.7 Note the mass.
Discard of products:
-Put the filter and the PbI2 in the designated container.
-Dissolve the manganese(II) chloride in water, dispose in provided container.
Observations
A. Addition of lead(II) nitrate to potassium iodide in solution.
-Reagent solutions are both clear.
-When they are mixed, a bright yellow precipitate is formed.
-The supernatant solution is cloudy.
-The wet filtrated solid has the texture and look of paint pigments.
-There is a silvery substance floating on the the solution/covering the precipitate.
-After rinsing with water, the filtrate has a slight yellow color.
A. Addition of HCl(aq) to manganese powder.
-Sizzles, violent reaction.
-Gas is produced.
-Reaction is endothermic
-Water condensates on the sides of the beaker.
-Solution is dark-green.
-When dry, a pinkish solid is left.
Data and Calculation
A. Data for the preparation of lead(II) iodide
Volume of lead(II) nitrate solution used
Moles of lead(II) ions
0.00050 mol
Volume of potassium iodide solution used
7.0 mL
Moles of iodide ions
0.00175 mol
Theoretical yield of lead iodide (moles)
0.00050 mol
Mass of dry filter paper
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Mass of filter paper plus dry lead(II) iodide
1.117 g
Mass of lead(II) iodide prepared
0.275 g
Molar mass of lead(II) iodide
360.22 g/mol
Moles of lead(II) iodide prepared: “total yield”
0.000763 mol
Percent total yield of lead(II) iodide
152%
Mass of empty watch glass
18.634 g
Mass of watch glass plus lead(II) iodide
18.844 g
Mass of lead(II) iodide isolated
0.21 g
Moles of lead(II) iodide isolated: “isolated yield”
Percent isolated yield of lead(II) iodide
Moles of lead(II) iodide lost during isolation
Percent total yield lost during isolation process
0.000583 mol
76.4%
0.000180 mol
23.6%
A. Data for the synthesis of manganese(II) chloride
Mass of empty beaker
47.173 g
Mass of beaker plus manganese
47.230 g
Mass of manganese used
0.058 g
Moles of manganese used
1.05 x 10 -3
Theoretical yield of manganese(II) chloride (moles)
1.05 x 10-3
Mass of empty beaker (same os above)
47.173 g
Mass of beaker plus manganese(II) chloride
47.298 g
Mass of manganese(II) chloride prepared
Molar mass of manganese(II) chloride
Moles of manganese(II) chloride prepared
Percent yield of manganese(II) chloride
0.126 g
125.84 g/mol
1.00 x 10-3
95.2%
Conclusion
This experiment had a yield of 152% for the lead(II) iodide (75.4% of which was isolated)
and of 95.2% for the manganese(II) chloride. If fact there is nothing in this experiment that
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proves the products are rally what they are claimed to be, but, it is assumed the provided
equations are the most likely to have happened and the reagents do not react with the
atmosphere.
For the preparation of lead(II) iodide, not only the total yield was recorded, but also the
isolated yield. This value is very important because it is the working value. If the product
was prepared for resale or for further reaction and that 23.6% was lost in the filtration, the
procedure wouldn’t be very good and changes should be consider to obtain a higher
yield. In this experimental process, the product wasn’t of further use, so the isolated yield
is not that important. To improve the isolated yield, the filter paper could be changed for
one with less stickiness or methods other than filtration could be used altogether. A
substance could be found were the spectators ions have an higher Kps then in water and
then, water could be evaporated to leave only the precipitate. Or, solutions of pure Pb2+
and I- could have been prepared by precipitating the other ions by reactions that leave
those reagents as spectator ions.
If a precipitation reaction was conducted using lead(II) nitrate and potassium sulfate, the
product would be lead(II) sulfate, an insoluble compound. That reaction can be
summarized by this equation: Pb(NO3)2(aq) + K2SO4
PbSO4(s) + 2
NO3SO4(aq).
If the same proportions of lead ions and anion were to be used, the yield would be of
0.152 g. (PbSO4(s) has a molar mass of 303.26 g/mol)
After the precipitation reaction, the concentration of each ion was: 0.11 mol/L I⁻ (
); 0.14 mol/L NO3⁻ (
); 0.18 mol/L K⁺ (
); 0 mol/L Pb2+.
The redox reaction can be written as the addition of an oxidation and a reduction. The
electron lost by Mn(s) during the oxidation (shown as loose electrons on the right hand
side of the first equation) are gained by the oxidizing agent H+ during the reduction.
Mn0(s)
+
Mn2+(aq) + 2 e1-(aq)
H20(g) + 2 Cl1-(aq)
2 H1+(aq) + 2 Cl1-(aq) + 2 e1-(aq)
Mn0(s) + 2 HCl0(aq)
MnCl20(aq) + H20(g)
In this reaction, manganese is the limiting reagent. This means the real product obtained
is a solution of hydrochloric acid and manganese(II) chloride in water. The acid and the
water are then evaporated to isolate the halide. If not all of it is evaporated, the percent
yield of the reaction can be shown as being more then 100% like in part A of this
experiment where the product wasn’t truly dry, leading to an aberrant yield of 152%.
Back to the manganese compound. If, on the other hand, the halide is overheated and
start degrading, creating, as suggested MnO, the error produced will be negative. The
molar mass of MnCl2 being greater than the one of MnO, the product will get lighter and
Cl2 might be produced.
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To minimize those sources of error, it could be temptation to let water evaporate slowly, or
to let the solid cool slowly on the open air. But MnCl2 tends to be hydrated. This is, to form
bounds with water molecules as a solid without dissolving. The hydrated form probably
does not have the pink color of the anhydrous form, but, most importantly, it has a greater
molar mass.
With the method used, it is more likely the product go decomposed then any other of the
possible errors cited above.
If a spill of HCl happened, possible steps to clean it up would be:
-Neutralization with NaOH .
-Dabbing with a cloth.
References
Simpson, Robert S.. Lighting control-technology and applications. Boston,
Massachusetts: Focal Press, 2003.
Encyclopaedia Britannica online “sodium hydroxide” on Nov. 7 2010.
http://www.britannica.com/
Kotz, John C., et al. Chemistry and Chemical Reactivity. 6th edition. Belmont, California:
Thomson Brooks/Cole.
Chemistry 205
Experiment 5: Solubility
Prelaboratory Questions
1. Procedure (N.B. note observations throughout)
Put a small amount of each substance in clean vials.
- Pour 1 mL of water to the substances, try to make them dissolve, add water if needed.
Clean the stirrer between each use to avoid contamination.
Test conductivity.
Clean the conductivity tester between each use and test in distilled water.
Add 1 drop of HCl to each vial.
Neutralize and discard of the solutions.
Repeat from the start with NaOH instead of HCl
Chemistry 205
Experiment 5: Solubility
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Laboratory Report
Observations and conclusions
Magnesium
Observations
Conclusions
Appearance Fine gray powder
Substance type: Element
With water
Dissolution equation:
Powder doesn’t
dissolve at all.
Suspends for about
2 seconds.
Mg(s) + H2O → Mg(s) (no reaction)
Conductivity None
With acid
Gas-forming,
exothermic solution.
Dissolves.
Reaction:
Mg(s) + 2H+(aq) +2Cl-(aq) → MgCl2(aq) + H2(g)
(Redox)
With base
No reaction.
Reaction:
Mg(s) + Na+(aq) + OH-(aq) → Mg(s) + Na+(aq) +
OH-(aq) (no reaction)
Theoretical Solubility: None in water, reacts with acid.
Calcium Oxide
Observations
Conclusions
Appearance White, chalky,
soft powder.
Substance type: Ionic compound (Metal oxide)
With water
Dissolve
partially, turbid
white saturated
solution.
Dissolution equation:
CaO(s) + H2O(l)→ Ca(OH)2(aq) (Acid/Base)
Conductivity Good
With acid
Dissolve to a
clear solution.
Forms gas.
Reaction:
Ca(OH)2(aq) + 4 H+ (aq) + 4 Cl-(aq) → CaCl2(aq) + 2
H2O(l) + H2(g) + 2 Cl-(aq) (Acid/Base)
With base
No changes.
Reaction:
Ca(OH)2(aq) + Na+(aq) + OH-(aq) → Ca2+(aq) + 3 OH-(aq)
+ Na+(aq) (no reaction)
Theoretical Solubility: Soluble in water, form a weak acid.
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Sodium chloride
Observations
Conclusions
Appearance White small
cubic crystals
Substance type: Ionic compound.
With water
Dissolution equation:
NaCl(s) → Na+ (aq) +Cl-(aq) (no reaction)
Dissolves.
Conductivity Weak
With acid
Nothing
changes.
Reaction:
Na+(aq) + H+(aq) + 2 Cl-(aq) → Na+(aq) + H+(aq) + 2 Cl-(aq)
(no reaction)
With base
Nothing
changes.
Reaction:
2 Na+(aq) + Cl-(aq) + OH-(aq) → 2 Na+(aq) + Cl-(aq) +
OH-(aq)(no reaction)
Theoretical Solubility: Soluble in water.
Tin
Observations
Conclusions
Appearance Irregular, bronze
colored, metallic
granules.
Substance type: Element (metal)
With water
Does not
dissolve
Dissolution equation:
Sn(s) → Sn(s) (no reaction)
Conductivity None
With acid
Granules turn
silvery, liquid is
cloudy.
Reaction:
Sn(s) + H+(aq) +Cl-(aq) → Sn(s) + H+(aq) +Cl-(aq) (low
reactivity) (Redox)
With base
Granules turns
silvery, phases
of clear liquids.
Reaction:
Sn(s)+ Na+(aq) + OH-(aq) → Sn(s)+ Na+(aq) + OH-(aq)
(low reactivity) (Redox)
Theoretical Solubility: None in water.
Ammonium chloride
Observations
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Appearance White tiny grains.
With water
Clear solution.
Conductivity Good
Substance type: Ionic compound (cation and
halogen)
Dissolution equation:
NH4Cl(s)→ NH4 +(aq) + Cl-(aq) (no reaction)
With acid
Less viscous,
clear solution.
Reaction:
NH4 +(aq) + 2 Cl-(aq) + H+(aq) → NH4 +(aq) + 2 Cl-(aq) +
H+(aq) (no reaction)
With base
Meniscus
increases.
Reaction:
NH4 +(aq) + 2 Cl-(aq) + Na+(aq) + OH-(aq) → NH3 (aq) +
H2O(l) + 2 Cl-(aq) + Na+(aq) (Acid/Base)
Theoretical Solubility: Soluble in water.
Sucrose
Observations
Conclusions
Appearance White crystal-ish
granules.
Substance type: Covalent compound (carbon
hydrate)
Dissolution equation:
C6H12O6(s)→ C6H12O6(aq) (no reaction)
With water
Clear solution.
Conductivity None
With acid
Clear solution.
Reaction:
C6H12O6(aq) + H+(aq) +Cl-(aq) → C6H12O6(aq) + H+(aq)
+Cl-(aq) (no reaction)
With base
Clear solution.
Reaction:
C6H12O6(aq) + Na+(aq) + OH-(aq) → C6H12O6(aq) +
Na+(aq) + OH-(aq) (no reaction)
Theoretical Solubility: Soluble in water.
Sodium bicarbonate
Observations
Conclusions
Appearance White powder
Substance type: Ionic compound
With water
Dissolution equation:
NaHCO3(s)→ Na+(aq) + HCO3-(aq) (no reaction)
Clear solution
Conductivity Good
With acid
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Reaction:
Na+(aq) + HCO3-(aq) + H+(aq) +Cl-(aq) → Na+(aq) +
HCO3-(aq) + H+(aq) +Cl-(aq) (no reaction)
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With base
Clear solution.
Reaction:
2 Na+(aq) + HCO3-(aq) + OH-(aq) → 2 Na+(aq) +
CO32-(aq) + H2O(l) (Acid/Base)
Theoretical Solubility: Soluble in water.
Nickel (II) chloride
Observations
Conclusions
Appearance Pale green
powder.
Substance type: Ionic compound
With water
Dissolution equation:
NiCl2(s) → Ni2+(aq) + 2 Cl-(aq) (no reaction)
Light applegreen solution.
Conductivity Good
With acid
Nothing changes
Reaction:
Ni+(aq) + 2 Cl-(aq) + H+(aq) +Cl-(aq) → Ni+(aq) + H+(aq) + 3
Cl-(aq) (no reaction)
With base
Turquoise
precipitate.
Reaction:
Ni+(aq) + 2 Cl-(aq) + Na+(aq) + OH-(aq) → NaCl(aq) +
ClOH(aq) + NiOH(s) (Precipitation)
Theoretical Solubility: Soluble in water.
Potassium permanganate
Observations
Conclusions
Appearance Dark purple
Substance type: Ionic compound
metallic needles.
Dissolution equation:
With water
Purple solution.
KMnO4(s) →K+(aq) + MnO4-(aq) (no reaction)
Conductivity Good
With acid
Solution
becomes dark
orange.
Reaction: 2 K+(aq) + 2 MnO4-(aq) + 16 H+(aq) + 16
Cl-(aq) → 2 Mn2+(aq) + 8 H2O(l) + 5 Cl2(aq) + 6 Cl-(aq) +
2 K+(aq) (Redox)
With base
Purple solution
becomes darker.
Reaction:
K+(aq) + MnO4-(aq) + Na+(aq) + OH-(aq) → K+(aq) +
MnO4-(aq) + Na+(aq) + OH-(aq) (no reaction)
Theoretical Solubility: Soluble in water.
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Additional Questions
1. This experiment proves the solubility trends exposed in this course’s manual:
Chemistry and Chemical Reactivity (Kelps??). All the ionic compounds acted
according to the rules learned. Covalent compounds, were soluble according to
their polarity.
2. Solutes that tend to be insoluble in pure water but soluble is acid or base are
metals.That is because there is a redox reaction between a transition metal and
an acid that turn the metal into its ion. Substances that are only soluble in bases
are rare.
Chemistry 205
Experiment 6a: Analysis of an Unknown Oxide of Copper
Prelaboratory Questions
1. -Place 0.15 to 0.20 g of copper oxide in a weighted beaker. Note the exact mass
of the substance.
-Add 2.5 mL of HCl. Swirl mixture and slowly heat until all the solid had dissolved.
-Add 2.5 mL of water and 0.1 g of aluminum. Note observations.
-Isolate the precipitated copper by decantation, rinse with acetone, dry on a low
intensity hot plate.
2. (Molar masses from the Periodic Table of the Element provided at the Chem
205 mid term exam.)
CuO:
Molar mass is 63.55 + 15.999 = 79.55 g/mol
Percent mass of copper: 63.55 / 79.55 * 100 = 79.88 %
Cu2O
Molar mass is ( 2 * 63.55 ) + 15.999 = 143.10
Percent mass of copper: ( 2 * 63.55 ) / 143.10 * 100 = 88.82 %
3. The color of CuO is: black
The color of Cu2O is: reddish-brown
4.The color of Cu2+ (aq) is : blue
The color of Cu+ (aq) is : red
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Chemistry 205
Experiment 6b: Gravimetric Determination of
Carbonate or Bicarbonate
Prelaboratory Questions
1.-Place about 0.2 g of the substance in a weighted beaker. Note the exact mass.
Add HCl drop by drop until the reaction stops. Add 2-3 drops extra.
-Boil the solution until truly dry. Let cool down. Note mass of remaining solid.
2.The boiling points are:
NaCl: 1413 ˚C
CO2: -78.5 ˚C
HCl: 109 ˚C
H2O: 100˚C
This information shows that NaCl will be the last compound to boil in the present
solution, therefor making evaporation an easy way to isolate the desired product. If
the product we wanted to isolate had a boiling point close to the ones of other
present substances, other means of separations would have had to be found.
3.
38 % HCl = 38 g per 100 g
100 g /1.2 g/mL = 83.3 mL
38 g / 83.3 mL = 0.465 g/mL
0.465 g/mL x 1000 mL/L = 456 g/L
465 g/L / 36.46 g/mol = 125 mol/L
The molarity of concentrated HCl is 125 M.
4.
a)Na2CO3(s) + 2 HCl(aq) → 2 NaCl(aq) + H2O(l) + CO2(g)
Molar mass: 105.987 g/mol
Moles in 0.2 g: 0.2 g / 105.987 g/mol = 1.88702 x 10-3 mol
Concentration of the acid: 6 M = 0.006 mol/mL
Volume of acid needed for reaction: (1.88702 x 10-3 mol / 0.006 mol/mL) x 2= 0.629
mL
0.63 mL of HCl are needed.
b)NaHCO3(s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)
Molar mass: 84.005 g/mol
Moles in 0.2 g: 0.2 g / 84.005 g/mol = 2.3808 x 10 -3 mol
Concentration of the acid: 6 M = 0.006 mol/mL
Volume of acid needed for reaction: 2.3808 x 10-3 mol / 0.006 mol/mL = 0.3968 mL
0.40 mL of HCl are needed.
5. In baking powder, NaHCO3 is used conjointly with monocalcium phosphate to
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create an acid-base reaction that yields Co2. The balanced equation of this
reaction is:
2 NaHCO3(s) + Ca(PO4)2(s) → 2 Na+(aq) + 2 CO2(g) + 2 H2O(l) + Ca+(aq) + ½H2(g) + 2 PO43-(aq)
In some recipes, baking soda is used alone because the batter is acidic enough
to make rise without Ca(PO4)2. In that case, the equation is:
NaHCO3(aq) + H+(aq) → Na+(aq) + CO2(g) + H2O(l)
Chemistry 205
Experiment 6A: Analysis of an Unknown Oxide of Copper
Laboratory Report
Data and Observations
1. Upon addition of HCl, unknown #1: Turns dark green, with excess solute. The
solution look lighter as the solute dissolves. Extra HCl is needed. There is
condensation on the beaker. When water is added, the color is more diluted and
look bluer. Some of the copper oxide never dissolves.
2. Upon addition of the aluminum to the solution: reacts immediately, forms a gas.
tiny droplets of copper are formed. Heat is produced, solution is clear, with gray
suspensions. Aluminum dissolves. Reaction slows down with time. The copper
are balls are soft and spongelike.
3.
Mass of beaker (g)
47.194
Mass of beaker plus copper oxide (g)
47.366
Mass of copper oxide (g)
0.172
Mass of beaker plus copper (g)
47.340
Mass of copper obtained (g)
0.146
Moles of copper obtained (mol)
Mass of oxygen in the copper oxide (g)
0.026
Calculation and Analysis
4. Based on the prelab, the unknown substance would be CuO, copper(II) oxide
because the powder form is black.
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5. The oxide copper is 84.9% copper and 15.1% oxygen. It is impossible to
determine which oxide it is based on this result because the choices are 79.88%
copper or 88.82% and the result about the average for CuO or Cu2O.
6. The research done for the pre-laboratory does not support the result of the
laboratory because the percent error is too great. (6.28% for CuO or 4.41% for
Cu2O) it is evenm closer to the wrong result then the right one.
Additional questions
1. Moles of copper oxide: 0.146/63.55 = 2.2974 x 10-3 mol
Moles of HCl needed: CuO + 2 HCl -> Cu 2+ + H2O + 2 Cl- (aq) : 2 *2.2974 x 10-3 =
4.54 x 10-3
mL of HCl 6M needed: 4.54 x 10-3 /6 *1000 = 0.765 mL
Percent of solution used that was in excess: (2.5-0.765)/2.5 *100 = 69.3 %
Because more solution was added without noting volume, more then 69.3% of the
HCl
8. The observed empirical formula is Cu1.4O1
(84.9/63.55) / (15.1/15.99) = 1.4146
Experiment 6B: Gravimetric Determination of
Carbonate or Bicarbonate
Laboratory Report
Data and Observations
1.
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Mass of empty beaker (g)
49.965
Mass of beaker and unknown (g)
50.208
Mass of unknown (g)
0.243
Molar mass of sodium carbonate (g/mol)
105.987
Mole of unknown assuming it is sodium carbonate (mol)
2.29 x 10-3
Molar mass of sodium hydrogen carbonate (g/mol)
84.005
Mole of unknown assuming it is sodium hydrogen carbonate
(mol)
2.89 x 10-3
Mass of beaker and sodium chloride (g)
50.131
Mass of sodium chloride (g)
0.160
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Molar mass of sodium chloride (g/mol)
58.44
Moles of sodium chloride (mol)
2.738 x 10-3
1. The predicted yield of NaCl is 0.268 g for sodium carbonate.
1. The predicted yield of NaCl is 0.169 g for sodium hydrogen carbonate.
4. The most probable identity of the unknown is sodium hydrogen carbonate
because the real yield of NaCl is closer (±0.009 g) to the one predicted for
NaHCO3(s) then the one predicted for Na2CO3(s) (± 0.108 g).
Additional Question
5. If sodium hydrogen carbonate was hydrated, its molar mass would have been
greater (102 g/mol) and therefor, closer to the one of sodium carbonate, but the
yield of NaCl per mole would still have differ a great deal because
Na2CO3(s) + 2 HCl(aq) → 2 NaCl(aq) + H2O(l) + CO2(g)
NaHCO3 H2O(s) + HCl(aq) → NaCl(aq) + 2 H2O(l) + CO2(g)
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Ottawa's
best.
They
tube, so
why
don't
you(tube)?
Things I
want to
do with
celebs
(and less
celeb)
p...
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New hot app: Facebook Albums To PDF
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