So I am a little confused on I may need help finding the mole ratio and possibly the empirical formula? The following data were collected from an experiment to synthesize lead iodide. A 0.150 g sample of lead, was dissolved in 10.0 mL of 3 M nitric acid. An excess of a sodium iodide solution was added to the dissolved lead ions. A yellow precipitate lead iodide was was formed in this reaction. See the table below for the data collected and the calculations in determining the empirical formula of lead iodide from this experiment. A Mass of empty beaker 26.975 g B Mass of lead + beaker 27.125 g C Mass of lead iodide formed + beaker 27.492 g D Mass of lead (B-A) 0.150 g E Mass of lead iodide (C-A) 0.517g F Mass of iodine (E-D) 0.367g G Moles of Pb element 0.000724g H Moles of I element 0.00289g J Mole ratio of I to Pb K Empirical formula for PbxIy
So I am a little confused on I may need help finding the mole ratio and possibly the empirical formula?
The following data were collected from an experiment to synthesize lead iodide.
A 0.150 g sample of lead, was dissolved in 10.0 mL of 3 M nitric acid. An excess of a sodium iodide solution was added to the dissolved lead ions. A yellow precipitate lead iodide was was formed in this reaction. See the table below for the data collected and the calculations in determining the empirical formula of lead iodide from this experiment.
|
A |
Mass of empty beaker |
26.975 g |
|
B |
Mass of lead + beaker |
27.125 g |
|
C |
Mass of lead iodide formed + beaker |
27.492 g |
|
D |
Mass of lead (B-A) |
0.150 g |
|
E |
Mass of lead iodide (C-A) |
0.517g |
|
F |
Mass of iodine (E-D) |
0.367g |
|
G |
Moles of Pb element |
0.000724g |
|
H |
Moles of I element |
0.00289g |
|
J |
Mole ratio of I to Pb |
|
|
K |
Empirical formula for PbxIy |
The number of moles of iodine is = 0.00289 mole
The number of moles of lead is = 0.000724 mole
The mole ratio of I to Pb =?
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