Draw the Lewis structure for each organic compound from

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Problem 98E Chapter 9

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

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Problem 98E

Draw the Lewis structure for each organic compound from its condensed structural formula.

a. C3H8

b. CH3OCH3

c. CH3COCH3

d. CH3COOH

e. CH3CHO

Step-by-Step Solution:

Solution 98 E Step 1 of 5 The compounds in this exercise are all organic molecules. We have to assume that exists a chain of atoms of Carbon that can be bonded each other by single, double or triple bonds. Moreover, we have to remember that every atom of Carbon has the possibility to form four bond maximum because it has only 4 electrons of valence.; Oxygen instead, has six electons of valence that are organized as two lone pairs and 2 single electrons that can be used to form two single bonds or a double bond. Step 2 of 5 C38s a molecule belonging to the family of alkanes (general formula C H n n+2). In Lewis structure we have a chain formed by three atoms of Carbon (each of them is bonded to the other by a single covalent bond). The terminal atoms of Carbon are bonded to three atoms of Carbon each, while the central atom of Carbon forms a bond with two atoms of H. We can see the molecule in the figure below Since we have 3 carbons and 8 Hydrogens we have Total electrons= 3(Valence number of C) + 8(valence number of H)= 3(4) + 8(1)= 20 We use all 20 electrons to form single covalent bonds Step 2 of 5 CH OCH is an ether. Ethers have a general formula that is R-O-R. Thus the Oxygen atom is a 3 3 bridge between two Carbon chains (in this case we have only one atom of carbon per chain) Oxygen carries two lone pairs and it can form two single covalent bonds. Since both atoms use one of their valence electrons to form a bond with O, their other three bonds are formed with three atoms of H for each other. thus the Lewis structure is Since we have 2 carbons, one O and 6 Hydrogens we have Total electrons= 2(Valence number of C) + 6(valence number of H) + (valence electrons of O)= 2(4) + 6(1) + 6= 20. In the Lewis structure above, every connecting line counts as a couple of bonding electrons. Thus in this molecule we can confirm we are using 20 electrons of valence. Step 3 of 5 CH COCH is a ketone. Ketones have have a general formula that is RCOR. The CO means a 3 3 Carbonyl group, in which a C forms a double bond with an atom of Oxygen, while the atom of Oxygen keeps two lone pairs. The R groups in this case are two other methyl groups (R=CH ) 3 that are linked to the Carbon of Carbonyl group thanks to a single bond. Moreover, we have to remember that in methyl group Carbon forms three other single bonds with H atoms. For what we’ve said. The structural formula for this compound is Since we have 3 carbons, one O and 6 Hydrogens we have Total electrons= 3(Valence number of C) + 6(valence number of H) + (valence electrons of O)= 3(4) + 6(1) + 6= 24. In the Lewis structure above, every connecting line counts as a couple of bonding electrons. Thus in this molecule we can confirm we are using 24 electrons of valence.

Step 4 of 5

Chapter 9, Problem 98E is Solved
Step 5 of 5

Textbook: Chemistry: A Molecular Approach
Edition: 3rd
Author: Nivaldo J. Tro
ISBN: 9780321809247

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