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3.10b Mass of CaCl2 containing chlorine

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Química

Universidade de Kentucky

4.7 (640 classificações) | 65K alunos inscritos

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This course is designed to cover subjects in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. This course is a precursor to the Advanced Chemistry Coursera course. Areas that are covered include atomic structure, periodic trends, compounds, reactions and stoichiometry, bonding, and thermochemistry.

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4.7 (640 classificações)

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TM

Jun 21, 2017

This is a great course! Very well explained! It's managing to keep my attention on 100% and that's not an easy task for anybody. Thank you for sharing your knowledge with us!

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Dec 30, 2016

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Na lição

Week 3

Now that we know the structure of an atom, we can explore how atoms combine to form either molecular or ionic compounds. Then we will learn the rules of nomenclature that ensure that a compound is named according to IUPAC rules. We will end this unit by looking at quantitative relationships for compounds including the molar mass of and mass percent of an element in a compound.

3.01 Elements vs. Compounds 14:40

3.02 Types of Chemical Bonds 7:25

3.03 Chemical Formulas 7:37

3.04 Formula of ionic compounds 12:46

3.04a Formula of the compound formed from aluminum and phosphorus 1:19

3.04b Formula of the compound formed from ammonium and phosphate 1:27

3.05 Nomenclature 6:26

3.06 Nomenclature of Ionic Compounds 10:42

3.06a IUPAC Name for K2SO4 0:57

3.06b IUPAC Name for NiCO3 1:54

3.06c Formula for manganese(II) flouride 0:52

3.06d IUPAC Name for AgNO3 1:04

3.06e Formula for chromium(III) oxide 1:00

3.07 Nomenclature of Molecular Compounds 8:41

3.07a IUPAC Name for PCl3 1:21

3.07b IUPAC Name for N2O1:01

3.07c Formula for boron triflouride 0:40

3.08 Nomenclature of Acids 5:54

3.09 Molar Mass of Compounds 11:41

3.09a Molecular Mass of Al2(Cr2O7)31:47

3.09b Mass of a sample of vitamin C 2:39

3.09c Moles of sucrose within a sugar cube 2:03

3.10 Mass percent composition 8:55

3.10a Mass percent of chromium 1:53

3.10b Mass of CaCl2 containing chlorine 2:29

3.11 Relationships from Chemical Formulas 11:39

3.11a Mass of H in Ca(H2PO4)2 3:30

3.11b Mass of NH3 2:13

3.12 Chemical formulas from experimental data 9:46

3.12a Empirical formula of a compound containing sulfur and oxygen 2:13

Ministrado por

Dr. Allison Soult
Lecturer
Dr. Kim Woodrum
Sr. Lecturer

So in this problem we're trying to find the mass of calcium chloride that

would contain 25 grams of chlorine.

The key thing we have to remember here is that the percentages are going to

be true for a sample, regardless of the sample size.

So if I calculate the percentage of chlorine in any calcium chloride sample,

it's going to be the same for all of them.

And so I can actually do that,

find the percentage of chlorine in calcium chloride by using our molar mass of

the compound relative to the molar mass of the chlorine in that compound.

So I'm going to first look at the molar mass of the whole compound,

I have 1 calcium which has a mass of 40.01.

And then I have chlorine, which has a mass of 35.45, and there are 2 of them.

And when I add these values together, I end up with 110.91 grams per mol.

Now I want to find the percent chlorine in my sample.

And so I take the 2 times 35.45, or

70.90 divided by the total mass, 10, 110.91 and

what that gives me is a percent of chlorine equal to 63.93%.

So I know that's going to be true in all samples of

Calcium Chloride including one that contains 25 grams of chlorine.

So now I can say 63.93% chlorine equals the part over the whole.

The part in this case is going to be the 25 grams,

because our problem tells us that this is 25 grams of just the chlorine.

So I know that's going to be in my numerator, and

then I have x grams of CaCl2.

And so I can divide by 100, and so I end up

with 0.6393 equals 25 over x.

And I can rearrange and solve this and find that x

is going to be equal to 39.29 grams of calcium chloride.

And we can actually easily go back and check our work, because we can take 25

divided by 39.29, and we should get an answer of approximately 63.93%.

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