Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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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

 

Expert Solution
Check Mark
Step 1

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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