A hydrated substance was found to have the following percent composition: Na = 16.1 /23.0 0.7/0.34% = 2 C = 4.2/12.01 0.349/0.349 = 1 O = 16.8/16 1.05/0.349 = 3 H2O = 62.9/18 3.494/0.349 = 10 A sample of 5.82g of this substance was heated in a crucible. Calculate the mass of the anhydrous salt that would remain in the crucible.
A hydrated substance was found to have the following percent composition: Na = 16.1 /23.0 0.7/0.34% = 2 C = 4.2/12.01 0.349/0.349 = 1 O = 16.8/16 1.05/0.349 = 3 H2O = 62.9/18 3.494/0.349 = 10 A sample of 5.82g of this substance was heated in a crucible. Calculate the mass of the anhydrous salt that would remain in the crucible.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Data Table
|
Mass of empty crucible and lid |
35.194g |
|
Mass of crucible, lid, and hydrate |
42.559g |
|
Mass of hydrate |
7.365g |
|
Mass of crucible; lid, and sample after first heating |
41.414g |
|
Mass of crucible; lid, and sample after second heating |
41.409g |
|
Mass of anhydrous salt after all heatings |
6.211g |
|
Mass of water given off |
1.154g |
|
Mass of one mole anhydrous salt |
106 |
- A hydrated substance was found to have the following percent composition:
Na = 16.1 /23.0 0.7/0.34% = 2
C = 4.2/12.01 0.349/0.349 = 1
O = 16.8/16 1.05/0.349 = 3
H2O = 62.9/18 3.494/0.349 = 10
- A sample of 5.82g of this substance was heated in a crucible. Calculate the mass of the anhydrous salt that would remain in the crucible.
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