Mercury's Help Desk | General Chemistry Notes

Class: CHEM 113.1


Instructor: Jorge Ramos, Ph.D.      

Office Hours: 

As posted in Chemistry Office, Remsen Hall 206

Or by appointment, Email: jr_starwind@netzero.net

NSB-B311, Monday through Friday when classes are in session.

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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
|--Standard Final Exam from ACS
|--Sample Lab Practical

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 at less than 50 milligrams.

     Al(mg)  Cu(mg)
     14      52
     18      67
     20      74
     26      96
     29     107
     32     118
     35     129
     38     140
     41     151
     43     159
     46     170