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General Chemistry Lab Notes


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Frequently asked Questions




|--Missing a Lab?
|--Reporting results with error bars
|--The empirical formula of a hydrate
|--Balancing RedOx reactions with the ion-electron method
|--Metal Recovery from RedOx reaction:
aluminum metal + copper chloride-> aluminum chloride + copper metal

|--Preparing a standard Sodium Hydroxide solution
|--Lab Files From Last Year

Missing a Lab?

It is best not to miss any labs because a worksheet must be submitted and graded each day.

If you do miss a lab, you have the option to attend another section where the lab that you missed (or will miss) is being carried out. For a list of these sections, check the course listing in the college homepage.

Once you find a lab section, you can ask the instructor if it is OK to stay in the room and do the experiment.

The instructor reserves the right to deny your request. This will likely happen if the room is full.

If the experiment you need to do was done in a previous week, then all of the above mentioned procedure is useless and you will not be able to recover the points for the lab you missed.

Reporting results with error bars

If you are having problems with error bars, download the document about uncertainties in measurement reporting. This document was meant for a Physical Chemistry class but it is all about standard statistics.
Uncertainties in measurement reporting
With sample calculations
[PDF] 2.73 mb

Preparing a standard Sodium Hydroxide solution

The stock solution is usually 2 or 3 mol/L. Let's assume it is 3 mol/L and you want to prepare 400 mL of a Sodium Hydroxide solution with approximate concentration 0.1 mol/L.

Use the standard formula CiVi = CfVf
Ci, Vi = Initial concentration and volume
Cf, Vf = Final concentration and volume
(3 mol/L) Vi = (0.1 mol/L) 400 X 10-3 L 
        (0.1 mol/L) 400 X 10-3 L 
Vi = ----------------------------------
        (3 mol/L)
Vi = 13.3 X 10-3 L
Vi = 13.3 mL
Thus you will need 13.3 mL of 3 mol/L stock solution then add DI water to make 400 mL of solution: (400 - 13.3) = 386.7 mL.

However, the volume does not need to be exact because you need to standardize the solution. How do we calculate the exact concentration of dilute sodium hydroxide from the titration data?

Say you prepared a standard KHP solution by dissolving 4.5 g of KHP in 250 mL of water. Note the KHP does not have to be exactly 4 or 5 g, but the amount must be known exactly.
Now suppose you take 35.00 mL out of the 250 mL KHP solution and titrate with the dilute Sodium Hydroxide solution. The equivalence point is reached after you add 32.50 mL of the dilute Sodium Hydroxide solution into the 35.00 mL of KHP solution.

                             4.5 g
Mol of KHP in 4.5 g = --------------------- = 0.022035 mol 
                            204.22 g/mol) 
                       35
KHP in 35.00 mL = ------------- X 0.022035 mol  = 0.0030849 mol
                      250
Recall that mol of KHP = mol of NaOH
KHP + NAOH --> KNaP + H2O
                         0.0034049 mol
Molarity of NaOH = -------------------------- = 0.09492 mol/L
                        32.50 X 10-3 L 

Thus your standard Sodium Hydroxide solution has a concentration of 0.09492 mol/L, which is very close to the 0.1 mol/L we were aiming to when making the dilution from the 3 mol/L stock solution.

The empirical formula of a hydrate

When 5.061 g of MgSO4•xH2O is heated to evaporate all the water, 2.472 g of MgSO4 is recovered. Find the value of x.

g of MgSO4 = 2.472 g
g of H2O = 5.061 g - 2.472 g = 2.589
Use MgSO4 = 120.4 g/mol
   5.061 g of MgSO4
-------------------- = 0.02053 mol MgSO4
   120.4 g/mol
Use H2 = 18.02 g/mol
   2.589 g of H2O
-------------------- = 0.1437 mol H2O
   18.02 g/mol
Ratio H2O:MgSO4
   0.1437 mol H2O
----------------------- = 7.000
   0.02053 mol MgSO4
Thus there are seven (7) water molecules for every MgSO4 unit.

Empirical formula: MgSO4•7H2O

Balancing RedOx reactions with the ion-electron method

Redox reactions are commonly run in acidic solution, in which case the reaction equations often include H2O(l), OH-(aq) and H+(aq) but you might not know whether the H2O(l), OH-(aq) and H+(aq) are reactants or products.

Ion-Electron method Outline

  • [1] Divide equation into half-reaction
  • [2] Balance atoms other than H and O
  • [3] Balance O by adding H2O
  • [4] Balance H by adding H+
  • [5] Balance net charge by adding e-
  • [6] Make e- gain equal to e- loss and add half reactions
  • [7] Add half reactions and cancel anything that is the same on both sides

    For Basic solution add steps

  • [8] Add to both sides of the equation the same number of OH- as there are H+
  • [9] Combine OH- and H+ to form H2O
  • [10] Cancel any H2O that you can

    For example, the following reaction occurs in acidic solution:
    IO3- + Re → ReO4- + I-

  • [1] Divide equation into half-reaction
    IO3- → I-
    Re → ReO4-
  • [2] Balance atoms other than H and O (no atoms to balance here)
    IO3- → I-
    Re → ReO4-
  • [3] Balance O by adding H2O
    IO3- → I- + 3H2O
    4H2O + Re → ReO4-
  • [4] Balance H by adding H+
    6H+ + IO3- → I- + 3H2O
    4H2O + Re → ReO4- + 8H+
  • [5] Balance net charge by adding e-
    6e- + 6H+ + IO3- → I- + 3H2O
    4H2O + Re → ReO4- + 8H+ + 7e-
  • [6] Make e- gain equal to e- loss and add half reactions
    {6e- + 6H+ + IO3- → I- + 3H2O} X7
    {4H2O + Re → ReO4- + 8H+ + 7e-} X6

    42e- + 42H+ + 7IO3- → 7I- + 21H2O
    24H2O + 6Re → 6ReO4- + 48H+ + 42e-
  • [7] Add half reactions and cancel anything that is the same on both sides
    42e- + 42H+ + 7IO3- + 24H2O + 6Re →
    7I- + 21H2O + 6ReO4- + 48H+ + 42e-

    7IO3- + 3H2O + 6Re → 7I- + 6ReO4- + 6H+

    For Basic solution add steps

  • [8] Add to both sides of the equation the same number of OH- as there are H+
    6OH- + 7IO3- + 3H2O + 6Re → 7I- + 6ReO4- + 6H+ + 6OH-

  • [9] Combine OH- and H+ to form H2O
    6OH- + 7IO3- + 3H2O + 6Re → 7I- + 6ReO4- + 6H2O

  • [10] Cancel any H2O that you can
    6OH- + 7IO3- + 6Re → 7I- + 6ReO4- + 3H2O

    If you need more information about RedOx reactions, download the document below.
    RedOx Reactions
    With sample problems
    [PDF] 2.53 mb

    Metal Recovery from RedOx reaction: aluminum metal + copper chloride-> aluminum chloride + copper metal

    These are the masses of copper recovered when different masses of aluminum foil are reacted with 10 mL of the copper chloride stock solution. The aluminum is the limiting reactant and the mass is kept lower than 60 milligrams.

     -- Mass of Cu obtained in last Exp. 
    mg of Al ... mg of Cu
    14 ............... 52
    18 ............... 67
    20 ............... 70
    22 ............... 80
    25 ............... 87
    26 ............... 94
    29 ............... 100
    30 ............... 105
    32 ............... 118
    35 ............... 129
    36 ............... 125
    38 ............... 140
    39 ............... 135
    40 ............... 140
    41 ............... 145
    43 ............... 159
    44 ............... 155
    46 ............... 165
    48 ............... 170
    53 ............... 186
    55 ............... 195
    

    Lab Files From Last Year


    Exp #3

    Exp #4

    Exp #5

    Exp #6

    Exp #7

    Exp #8

    Exp #9

    Exp #10





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