Thursday, August 27

Tomorrow is Game Day (i.e. we are doing a lab). Today we went over the pre-lab for the synthesis of copper and iodine lab. I would like it if you tried the mock calculations for tomorrow. I messed up during 2nd hour by doing one of the calculations wrong, so I am going to make the correction here to get people on the right path. All calculations pertain to the scenario's given in the pre-lab.

First scenario:

First, 7.634 g. minus 7.608 g. is the correct way to find the grams of copper consumed in the reaction. (0.026 g. of Cu which will become 0.00041 mol Cu).

Second, (my screw up) 7.732 g. minus 7.608 g. (the mass of the copper at the end of the lab) is the correct way to find the mass of the copper iodide compound. (0.124 g. of Cu(x)I(y) compound)

Third, the mass of the iodine consumed in the lab is 0.124 g. minus 0.026 g. This gives a mass of 0.098 g. of iodine which will become 0.00077 mol I.

Another way to find the mass of iodine would be to take the mass of the compound (7.732 g.) minus the original mass (7.634 g.). This too give 0.098 g. of iodine. This is the most direct way and I should have thought of it in class. Any mass added to the original copper must come from the addition of iodine. OOPS!

The first scenario's empirical formula is Cu(0.00041/0.00041=1) I(0.00077/0.00041=1.88)
-->CuI(2)

Second Scenario (calculated the same way)

Cu --> 7.608 g. - 7.583 g. = 0.025 g. Cu --> 0.00039 mol Cu
I --> 7.661 g. - 7.608 g. = 0.053 g. I --> 0.00042 mol I

Cu (0.00039/0.00039=1) I (0.00042/0.00039=1.08) -->CuI

Remember, copper will only have two possible oxidation numbers, (+1) or (+2), so tomorrow's lab is intended to prove one of them. Therefore, your data may need to be massaged to meet the possible ratio's that correspond to the probable oxidation states of copper. Examples: 1.67 would not be multiplied by 3, but rather in this scenario, rounded up to "2". The value 1.25 would not be multiplied by 4, but rather rounded down to "1".