Question

1. Calculate the formula weight of the Al [C9H6ON]3 complex that will be formed in this experiment.

2. A student performed a gravimetric analysis of an unknown solid sample by precipitating the aluminum using 8-HQ, forming the Al(HQ)3 complex as you will do in this lab. If the precipitate, after subtracting the crucible and microfilter, weighed 1.2176 g, what is the mass (in g) of aluminum in the precipitate?

3. Now suppose this student told you that she had sissolved the unknown in 250.00 mL, but only used 10.00 mL of that 250.00 mL solution to do the precipitation with 8-HQ. How many grams of Al were in the ORIGINAL 250.00 mL solution (i.e., in the original unknown sample)?

4. Supposed another student told you he calculated the number of grams of Al in the original 250.00 mL solution to be 1.788 g, and that he prepared that solution by dissolving 4.1200 g of a solid unknown in water. What is the wt % Al³⁺ in the original solid sample?

Answer 1

Formula weight:

Atomic weights:

Al = 26.98 g/mol

C = 12 g/mol

O = 16 g/mol

N = 14 g/mol

For this complex:

FW Al [C9H6ON]3 = 26.98 + (27*12 + 18*1 + 3*16 + 3*14) = 458.98 g/mol

For part B, if the precipitate is 1.2176 g, then calculate first the moles of the complex:

moles complex = 1.2176 / 458.98 = 2.652x10^-3 mol

For the mass of Al = 2.652x10^-3 * 26.98 = 0.0715 g of Al in the precipitade.

For art C, simply do the rule of 3:

if in 10 mL ------> 0.0715 g then in 250 mL:

0.0715 g * (250/10) = 1.7875 g of Al in the original sample.

For part D,the %Al would be:

%Al = 1.788 g / 4.1200 g * 100 = 43.398%

Hope this helps.