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3.11a Mass of H in Ca(H2PO4)2

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Chemistry

University of Kentucky

4.7 (745 ratings) | 75K Students Enrolled

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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 (745 ratings)

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FV

Aug 26, 2019

Very clear in the explanations provided, it is a complete course to really introduce to basic concepts of chemistry such as stechiometry, thermodynamic, atom structures and reactions.

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!

From the lesson

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

Taught By

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

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Transcript

In this problem we're going to find the mass in grams of hydrogen found in a sample of calcium dihydrogen phosphate. And the first thing we want to look at is kind of come up with our plan of how we're going to approach this problem. We see that we have our grams of our compound. And I want to go from grams of my compounds to moles of compound because I know there's a relationship between the moles of the compounds and the moles of hydrogen in that compound. And what we're, that's going to allow us to do is use that ratio to find moles of hydrogen and then we can find grams of hydrogen. So this is the plan we're going to take. The first thing we're going to have to do is in order to get from grams to moles is we're going to have to know the molar mass. So we look at calcium, which has a molar mass of 40.01, and there's one of those in the compound. We have hydrogen, and there are four of those, each with a mass of 1.08. For our phosphorus, there are two phosphorus, each with a mass of 30.97. And our oxygen and there are 8 of those times 16. And so what we can actually do is add all those up and we get a molar mass of 233.98 grams per mole. So now we know the relationship for this compound, for grams and moles. So we're able to get from grams of the compound to moles of the compound. So, now let's look and see how we're going to set this up now that we have our molar mass. And we said we're going to start with our grams of our compound, which is given in the problem, and so that is 8.92 grams of Ca(H2PO4)2. And we're going to use our molar mass, our 233.98 grams per mole of the compound. Then we're going to say, well for every one mole of Ca(H2PO4)2 we have, we have 4 mols of hydrogen. And where this comes from is these subscripts. Here we have two mols of hydrogen for one of these ions, and we have two of those ions per unit of the compound. So I know I have four mols of hydrogen for every one mol of the entire compound. Now we can use our molar mass of hydrogen, which is 1.008 grams per mole of hydrogen. And I'm going to go back and check through my units, just to make sure everything is canceling out correctly. Here I have grams of the compound. Cancel that with grams. Mols and mols cancel. Mols of hydrogen, mols of hydrogen cancel, and I've got grams of hydrogen remaining. Make sure when you're cancelling out mols that you cancel out mols of hydrogen with mols of hydrogen or mols of the compound with mols of the compound. They have to be like units and like substance. Now, I can do the calculation, and what I end up with is that I have 0.1537 grams of hydrogen. And because my initial number here only has three significant figures, I need to round that to 0.154 grams of hydrogen found in my 8.92 grams of calcium dihydrogen phosphate.

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