Question

2. During a Chem 121 experiment, a student obtained the following data after heating a sample of Calcium Chloride hydrate. Data | Mass --- | --- Weight of crucible and lid (g) | 32.1070 Weight of crucible, lid and hydrate (g) | 36.8283 Weight of crucible, lid and anhydrous salt after | a) 1st heating (g) | 35.7347 b) 2nd heating (g) | 35.6682 c) 3rd heating (g) | 35.6681 Use this information to determine: a) the weight of the hydrated salt (CaCl2•xH2O) b) the weight of the anhydrous salt (CaCl2) Note: the weight after the 2nd and 3rd heating are close enough in value to assume that constant weight has been reached and all the water molecules have been removed c) the weight of water removed d) the number of moles of the anhydrous salt Note: the molar mass of CaCl2 is 110.98 g/mol e) the number of moles of water f) the empirical formula of the hydrate Note: the formula of the hydrate is CaCl2•xH2O and you need to solve for x by finding the ratio of number of moles of H2O to the number of moles of CaCl2

          2. During a Chem 121 experiment, a student obtained the following data after heating a sample of Calcium Chloride hydrate.

Data | Mass
--- | ---
Weight of crucible and lid (g) | 32.1070
Weight of crucible, lid and hydrate (g) | 36.8283
Weight of crucible, lid and anhydrous salt after | 
a) 1st heating (g) | 35.7347
b) 2nd heating (g) | 35.6682
c) 3rd heating (g) | 35.6681

Use this information to determine:
a) the weight of the hydrated salt (CaCl2•xH2O)
b) the weight of the anhydrous salt (CaCl2)
Note: the weight after the 2nd and 3rd heating are close enough in value to assume that constant weight has been reached and all the water molecules have been removed
c) the weight of water removed
d) the number of moles of the anhydrous salt
Note: the molar mass of CaCl2 is 110.98 g/mol
e) the number of moles of water
f) the empirical formula of the hydrate
Note: the formula of the hydrate is CaCl2•xH2O and you need to solve for x by finding the ratio of number of moles of H2O to the number of moles of CaCl2

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2. During a Chem 121 experiment, a student obtained the following data after heating a sample of Calcium Chloride hydrate.

Data | Mass
— | —
Weight of crucible and lid (g) | 32.1070
Weight of crucible, lid and hydrate (g) | 36.8283
Weight of crucible, lid and anhydrous salt after |
a) 1st heating (g) | 35.7347
b) 2nd heating (g) | 35.6682
c) 3rd heating (g) | 35.6681

Use this information to determine:
a) the weight of the hydrated salt (CaCl2•xH2O)
b) the weight of the anhydrous salt (CaCl2)
Note: the weight after the 2nd and 3rd heating are close enough in value to assume that constant weight has been reached and all the water molecules have been removed
c) the weight of water removed
d) the number of moles of the anhydrous salt
Note: the molar mass of CaCl2 is 110.98 g/mol
e) the number of moles of water
f) the empirical formula of the hydrate
Note: the formula of the hydrate is CaCl2•xH2O and you need to solve for x by finding the ratio of number of moles of H2O to the number of moles of CaCl2

Added by Denise L.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Suman K

04/13/2022

Step 1

8283g - 32.1070g Weight of hydrated salt = 4.7213g Show more…

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During Chem 121 experiment, a student obtained the following data after heating a sample of Calcium Chloride hydrate: Data: Weight of crucible and lid: 32.1070g Weight of crucible, lid, and hydrate: 36.8283g Weight of crucible, lid, and anhydrous salt after: a) 1st heating: 35.7347g b) 2nd heating: 35.6682g c) 3rd heating: 35.6681g Use this information to determine: a) The weight of the hydrated salt (CaCl2•xH2O) b) The weight of the anhydrous salt (CaCl2) Note: The weights after the 2nd and 3rd heating are close enough in value to assume that constant weight has been reached and all the water molecules have been removed. c) the weight of water removed d) the number of moles of the anhydrous salt Note: The molar mass of CaCl2 is 110.98 g/mol. e) the number of moles of water f) the empirical formula of the hydrate Note: The formula of the hydrate is CaCl2•xH2O, and you need to solve for x by finding the ratio of the number of moles of H2O to the number of moles of CaCl2.

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Transcript

00:01 Hello guys, so in this question we have been given a data and experiment data of the sample after heating a sample of calcium chloride hydrate.

00:08 So now we have to calculate the weight of hydrated salt from the given data.

00:14 So first one we have to calculate is the weight of hydrated salt, that is calcium chloride.

00:31 So this is the hydrated salt and we have to find.

00:35 So here the weight of crucible lid and hydrate is equal to o.

00:38 So weight of crucible lid and hydrate weight of crucible lid is equal to 36 .8283 minus 32 .1070.

01:01 So when we calculate this we will get the value of weight of hydrated salt.

01:06 So that is equal to 4 .7 -213 grams.

01:10 So this is the final answer to all the first part of the question.

01:13 Now let's move on to the second part of the question where we have to find weight of anhydrous salt.

01:19 We have to find weight of anhydrous salt.

01:26 It is equal to weight of crucible lid and hydrate minus weight of crucible and lid.

01:50 So from this we get 35 .6682 minus 32 .1070.

01:57 And the final answer for this is equal to 3 .5612 grams.

02:03 So this is the final value of the second part of the question.

02:09 Now we have to find the number of moles, number of moles of anhydrousalt.

02:21 So it is equal to mass divided by molar mass and divided by molar mass is equal to 3 .5612 divided by 110 .98.

02:34 So this is the molar mass and the final answer we get is 0 .032088 moles.

02:45 So this is the answer to the third part of the question...

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