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Solved: A 45.2-mg sample of phosphorus reacts with selenium to form 131.6 mg of the

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 98P Chapter 6

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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Problem 98P

A 45.2-mg sample of phosphorus reacts with selenium to form 131.6 mg of the selenide. What is the empirical formula of the phosphorus selenide?

Step-by-Step Solution:
Step 1 of 3

Solution 98P

The empirical formula can be found using the table given below.The number of moles of each elements is obtained and the simplest whole number ratio in which the elements are present in the compound is  found.

 

Elements

Mass of the elements

Atomic weight

Number of moles

P

45.2

31

45.2/31=1.45

Se

86.4

80

86.4/80=1.08

The simplest ratio in which the elements Phosphorous and Selenium is present is 1.3:1(1.45/1.08=1.3).So the whole number ratio should be 4:3 and the empirical formula should be P4Se3

 

Step 2 of 3

Chapter 6, Problem 98P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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Solved: A 45.2-mg sample of phosphorus reacts with selenium to form 131.6 mg of the