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Get Full Access to Chemistry: The Central Science - 14 Edition - Chapter 3 - Problem 3.48
Get Full Access to Chemistry: The Central Science - 14 Edition - Chapter 3 - Problem 3.48

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# Determine the empirical formulas of the compounds with the following compositions by mass: (a) 55.3% K, 14.6% P, and 30.1% O (b) 24.5% Na, 14.9%

ISBN: 9780134414232 1274

## Solution for problem 3.48 Chapter 3

Chemistry: The Central Science | 14th Edition

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Chemistry: The Central Science | 14th Edition

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Problem 3.48 Determine the empirical formulas of the compounds with the following compositions by mass: (a) 55.3% K, 14.6% P, and 30.1% O (b) 24.5% Na, 14.9% Si, and 60.6% F (c) 62.1% C, 5.21% H, 12.1% N, and the remainder O
Step-by-Step Solution:

Step 1 of 5) Determine the empirical formulas of the compounds with the following compositions by mass: An equation written in this form, with all soluble strong electrolytes shown as ions, is called a complete ionic equation. Notice that K+1aq2 and NO3 -1aq2 appear on both sides of Equation 4.6. Ions that appear in identical forms on both sides of a complete ionic equation, called spectator ions, play no direct role in the reaction. Spectator ions can be canceled, like algebraic quantities, on either side of the reaction arrow, since they are not reacting with anything. Once we cancel the spectator ions, we are left with the net ionic equation, which is one that includes only the ions and molecules directly involved in the reaction: Pb2+1aq2 + 2 I -1aq2 ¡ PbI21s2 [4.7] Because charge is conserved in reactions, the sum of the ionic charges must be the same on both sides of a balanced net ionic equation. In this case, the 2+ charge of the cation and the two 1- charges of the anions add to zero, the charge of the electrically neutral product. If every ion in a complete ionic equation is a spectator, no reaction occurs.

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