Chem 1411- Lab Practical 2
Heat gained by calorimeter
J hot - J cold
Theoretical yield
(474.38 g/mol)(weight of Al foil used) / molar mass of Al = 14.16g
Heat gained by cold water
(mass)(4.18 J/gC)(temp of cold water) = J
Heat lost by water
(mass)(4.18 J/gC)(temp of hot water) = J
Pair
(moles of air in grad cylinder)(R)(K) / (corrected vol in mL x10^-3)
specific heat
1. (mass)(4.18 J/gC)(Temp of water)= J 2. (Calorimeter Constant)(Temp of water) = J 3. Add 1. and 2. together 4. (mass of metal sample)(Temp of metal) = gC 5. step 3 divided by step 4 = specific heat J/gC
Why must the oven temp be kept below 100 degrees Celcius when drying your product?
1. The compound could decompose at any higher temp. 2. The H2O attached to the alum would partially evaporate.
Volume of gas collected
50 mL- buret reading of vol of H2 collected and then add that to the volume of buret below 50 mL mark to get the total vol of H2 collected
You added rather than subtracted the vapor pressure of water to your final pressure reading.
Adding pressure to the total instead of subtracting would lead to higher vapor pressure, which would result in higher Mg mass.
Experimental Errors for Lab 9
Adding too much KMnO4 would cause it to become dark purple and would cause the molarity to be low
Calculate the energy of the 285 nm photon that is associated with skin cancer and the energy of the 290 nm photons associated with essential vitamin D3 production.
E= hv E=hc/v E285= (6.626x10^-34)(2.998x10^8)/(285x10^-9) = 6.97x10^-19J E290= (6.626x10^-34)(2.998x10^8)/(290x10^-9) = 6.85x10^-19J
2 regions of electrons
EG: linear 180 Hyb: sp MG: Polarity: depends
6 regions of electrons
EG: octahedral 90, 180 Hyb: sp3d2 MG: 1 lone pair= square pyramidal Polar 2 lone pairs= square planar depends
4 regions of electrons
EG: tetrahedral Hyb: sp3 MG: 1 lone pair= trigonal pyramid 2 lone pairs= bent <109.5 Polar
5 regions of electrons
EG: trigonal bipyramidal 90,120,180 Hyb: sp3d MG: 1 lone pair= see saw Polar 2 lone pairs= t-shaped Polar 3 lone pairs= liner 180 depends
3 regions of electrons
EG: trigonal planar 120 Hyb: sp2 MG: 1 lone pair= bent <120 Polar
What was the purpose of washing your product with either ice-cold water or alcohol?
Excess sulfuric acid has to be removed by the ice cold water or alcohol. Diluting the acid with water would increase the temp of the solution causing the solubility of alum
If you thought you reached the end point, but the color faded within 15 seconds and you did not add more KMnO4?
High molarity of KMnO4 since the end point is not reached and less volume of KMnO4 solution was used
A mercury discharge tube produces numerous strong spectral bands, however the spectrum of the mercury-filled fluorescent lights in the lab shows only three strong bands of color. Why?
Lamps contain mercury that is coated with a fluorescent substance. High energy UV rays are produced when an electric current passes through. The UV rays react with coating, which produces the visible light. The electrons emitted from the coating is what makes the red, blue, green light.
If you had not rinsed your buret with the stock KMnO4 solution before beginning the titration?
Low molarity since the solution is slightly diluted
If you had added the KMnO4 too rapidly and your end point was an opaque purpleish-brown rather than a clear solution with a hint of purple?
More volume of KMnO4 would be used causing the molarity of KMnO4 to be low
Experimental Errors for Lab 11
Moving filter paper off watch glass caused some little pieces to fall off. In the oven some H20 could've evaporated off the alum causing a smaller isolated weight.
Why does the Neon discharge lamp produce more distinct spectral lines tan the hydrogen lamp?
Neon is transferring more electrons than Hydrogen, producing more spectral lines of color, which means more energy is being transferred.
If you had been unable to complete the titration in a single period, but had stored the KMnO4 solution for several days before returning and picking up where you left off on the experiment without additional standardization?
The KMnO4 would oxidize if you stored it. This causes MnO4 to form in the solution. The MnO2 causes the apparent concentration of the solution to change as it is used. This would require more KMnO4 to titrate the oxalic acid and it would lead to a lower calculated molarity
Your instructor had inverted two numbers when reading the pressure and reported the barometric pressure was 765 torr when in fact the correct room pressure was 756 torr.
The mass of Mg metal present indicated higher amount of mass acutally present since . the barometric pressure reported was more than the actual pressure.
Why do KCL and CaCl2 produce different flame colors even though K+ and Ca2+ are isoelectronic?
The number of protons are not the same for K+ and Ca2+. Each has different energy levels causing different flame colors.
How would your data have been affected if you had failed to dry your calorimeter between experiments?
The total heat absorbed would be greater, while the specific heat would be lower for than the actual value.
You did not add the "void volume" of Hydrogen to the volume of Hydrogen in the graduated part of the buret.
The void volume is supposed to be added to the final volume of Hydrogen. So by not adding it would cause a lower final volume and a lower Mg mass.
Your partner tightly held the gas-filled portion of the buret in his hand while you were trying to determine the difference in the water levels.
The water level read may be incorrect since the buret may not be exactly vertical. This would cause the volume of gas to be lower as well as the mass of Mg metal calculated
Why is it important not to reduce the volume too much?
The water would evaporate and this would denature the crystals since they were partially made of water
You did not correct for the difference in water levels.
There will be more moles of H2 and this will give you a higher calculated mass of Mg metal.
What would have been the effect on your calculation of the specific heat of the metal had you forgotten to add the correction for your calculation?
There would be a higher specific heat of metal since the heat of reaction lost by metal would be higher
If you had added twice as much sulfuric acid and water as was called for in the procedure?
There would be no impact. KMnO4 only reacts with the oxalic acid. The molarity would remain the same
frequency
V=C/Y (2.998x10^8m/s)/(700x10^-9) = 4.28x10^14s^-1
Why is water so much higher than might have been predicted based upon its molecular weight?
Water has a higher change in Hvap than hexane because it takes more energy to break hydrogen bonds for H2O than the bonds in hexane
Why is it necessary to reduce the volume of your reaction mixture after the addition of sulfuric acid?
We reduced the volume of the mixture to turn the water into vapor form. This caused the concentration of the solution to increase without the water breaking the alum apart, which allows crystals to form.
Percent Yield
actual/ theoretical
Percent error
calculated change in Hvap of water - the literature of change of Hvap of water / the literature of change of Hvap of water
Calorimeter constant
heat gained in calorimeter / Tcold
correction of volume
volume in mL - 0.20 mL
Actual Yield
weight of filer paper and alum - weight of filter paper