Living By Chemistry: First Edition Textbook
Living By Chemistry: First Edition Textbook
1st Edition
ISBN: 9781559539418
Author: Angelica Stacy
Publisher: MAC HIGHER
Question
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Chapter U1.22, Problem 4E
Interpretation Introduction

Interpretation:

The formulae for the ionic compounds need to be determined.

Concept introduction:

Some ionic species are composed of more than one atom. These are called polyatomic ions. The entire cluster of atoms shares the charge on a polyatomic ion. Each polyatomic ion has a charge that is used when writing formulae of the compound.

Expert Solution & Answer
Check Mark

Answer to Problem 4E

The following table shows formulae of polyatomic ionic compounds.

Name of compounds Formulae
a. Lithium sulfate Li2SO4
b. Potassiumhydroxide KOH
c. Magnesium nitrate Mg(NO3)2
d. Ammonium sulfate (NH4)2SO4

Explanation of Solution

Given information:

Names of some ionic compounds are given.

Name of compounds
a. Lithium sulfate
b. Potassium hydroxide
c. Magnesium nitrate
d. Ammonium sulfate

To determine the chemical formulas of ionic compounds, rule of zero charge is used. Polyatomic ions do not always combine in 1:1 ratio. When there is more than one of the same polyatomic ion in a formula, the ion is enclosed in parenthesis and a subscript number is written indicating the number of ions in the compound.

  1. Lithium sulfate - The polyatomic ion here is sulfate (SO42-). Lithium has + 1 charge so 2 lithium atoms will combine with 1 sulfate ion. Hence, sum of charges becomes zero. The formula for this compound is Li2SO4.
  2. Potassium hydroxide - The polyatomic ion here is hydroxide (OH-). Potassium has + 1 charge so 1potassium atom will combine with 1 hydroxide ion. Hence, sum of charges becomes zero. The formula for this compound isKOH.
  3. Magnesium nitrate - The polyatomic ion here is nitrate(NO3-). Magnesium has 2 + charge so 1 magnesium atom will combine with 2 nitrate ions. Hence, sum of charges becomes zero. The formula for this compound isMg(NO3)2.
  4. Ammonium sulfate -The polyatomic ions here areammonium (NH4) + and sulfate (SO42-). 2 ammonium ions will combine with 1 sulfate ion. Hence, sum of charges becomes zero. The formula for this compound is(NH4)2SO4.
Conclusion

Ionic compounds with polyatomic ions follow certain rules while writing chemical formulas. Polyatomic ions are a group of atoms bonded together.

Chapter U1 Solutions

Living By Chemistry: First Edition Textbook

Ch. U1.2 - Prob. 1ECh. U1.2 - Prob. 2ECh. U1.2 - Prob. 5ECh. U1.3 - Prob. 1TAICh. U1.3 - Prob. 1ECh. U1.3 - Prob. 2ECh. U1.3 - Prob. 3ECh. U1.3 - Prob. 4ECh. U1.3 - Prob. 5ECh. U1.3 - Prob. 6ECh. U1.4 - Prob. 1TAICh. U1.4 - Prob. 1ECh. U1.4 - Prob. 2ECh. U1.4 - Prob. 3ECh. U1.4 - Prob. 4ECh. U1.4 - Prob. 5ECh. U1.4 - Prob. 6ECh. U1.4 - Prob. 7ECh. U1.4 - Prob. 8ECh. U1.4 - Prob. 9ECh. U1.4 - Prob. 10ECh. U1.4 - Prob. 11ECh. U1.5 - Prob. 1TAICh. U1.5 - Prob. 1ECh. U1.5 - Prob. 2ECh. U1.5 - Prob. 3ECh. U1.5 - Prob. 4ECh. U1.5 - Prob. 5ECh. U1.5 - Prob. 6ECh. U1.5 - Prob. 7ECh. U1.6 - Prob. 1TAICh. U1.6 - Prob. 1ECh. U1.6 - Prob. 2ECh. U1.6 - Prob. 3ECh. U1.6 - Prob. 4ECh. U1.6 - Prob. 5ECh. U1.6 - Prob. 6ECh. U1.7 - Prob. 1TAICh. U1.7 - Prob. 1ECh. U1.7 - Prob. 2ECh. U1.7 - Prob. 3ECh. U1.7 - Prob. 4ECh. U1.7 - Prob. 5ECh. U1.8 - Prob. 1TAICh. U1.8 - Prob. 1ECh. U1.8 - Prob. 2ECh. U1.8 - Prob. 4ECh. U1.8 - Prob. 5ECh. U1.8 - Prob. 6ECh. U1.8 - Prob. 7ECh. U1.9 - Prob. 1TAICh. U1.9 - Prob. 1ECh. U1.9 - Prob. 2ECh. U1.9 - Prob. 5ECh. U1.9 - Prob. 7ECh. U1.10 - Prob. 1TAICh. U1.10 - Prob. 1ECh. U1.10 - Prob. 2ECh. U1.10 - Prob. 3ECh. U1.10 - Prob. 4ECh. U1.10 - Prob. 5ECh. U1.10 - Prob. 6ECh. U1.10 - Prob. 7ECh. U1.10 - Prob. 8ECh. U1.11 - Prob. 1TAICh. U1.11 - Prob. 1ECh. U1.11 - Prob. 2ECh. U1.11 - Prob. 3ECh. U1.11 - Prob. 4ECh. U1.11 - Prob. 5ECh. U1.11 - Prob. 6ECh. U1.11 - Prob. 7ECh. U1.11 - Prob. 9ECh. U1.11 - Prob. 11ECh. U1.11 - Prob. 12ECh. U1.12 - Prob. 1TAICh. U1.12 - Prob. 1ECh. U1.12 - Prob. 2ECh. U1.12 - Prob. 3ECh. U1.12 - Prob. 4ECh. U1.12 - Prob. 5ECh. U1.12 - Prob. 6ECh. U1.12 - Prob. 7ECh. U1.12 - Prob. 8ECh. U1.13 - Prob. 1TAICh. U1.13 - Prob. 1ECh. U1.13 - Prob. 2ECh. U1.13 - Prob. 3ECh. U1.13 - Prob. 4ECh. U1.13 - Prob. 5ECh. U1.13 - Prob. 6ECh. U1.13 - Prob. 7ECh. U1.13 - Prob. 8ECh. U1.13 - Prob. 9ECh. U1.14 - Prob. 1TAICh. U1.14 - Prob. 1ECh. U1.14 - Prob. 2ECh. U1.14 - Prob. 3ECh. U1.14 - Prob. 4ECh. U1.14 - Prob. 5ECh. U1.14 - Prob. 6ECh. U1.14 - Prob. 7ECh. U1.14 - Prob. 8ECh. U1.14 - Prob. 9ECh. U1.14 - Prob. 10ECh. U1.14 - Prob. 11ECh. U1.14 - Prob. 12ECh. U1.14 - Prob. 13ECh. U1.14 - Prob. 14ECh. U1.15 - Prob. 1TAICh. U1.15 - Prob. 1ECh. U1.15 - Prob. 2ECh. U1.15 - Prob. 3ECh. U1.15 - Prob. 4ECh. U1.15 - Prob. 5ECh. U1.15 - Prob. 6ECh. U1.15 - Prob. 7ECh. U1.15 - Prob. 8ECh. U1.15 - Prob. 9ECh. U1.15 - Prob. 10ECh. U1.15 - Prob. 11ECh. U1.15 - Prob. 12ECh. U1.16 - Prob. 1TAICh. U1.16 - Prob. 1ECh. U1.16 - Prob. 2ECh. U1.16 - Prob. 3ECh. U1.16 - Prob. 4ECh. U1.16 - Prob. 5ECh. U1.16 - Prob. 6ECh. U1.17 - Prob. 1TAICh. U1.17 - Prob. 1ECh. U1.17 - Prob. 2ECh. U1.17 - Prob. 3ECh. U1.17 - Prob. 4ECh. U1.17 - Prob. 5ECh. U1.17 - Prob. 6ECh. U1.17 - Prob. 7ECh. U1.17 - Prob. 8ECh. U1.17 - Prob. 9ECh. U1.17 - Prob. 10ECh. U1.17 - Prob. 11ECh. U1.18 - Prob. 1TAICh. U1.18 - Prob. 1ECh. U1.18 - Prob. 2ECh. U1.18 - Prob. 3ECh. U1.18 - Prob. 4ECh. U1.18 - Prob. 5ECh. U1.18 - Prob. 6ECh. U1.18 - Prob. 7ECh. U1.18 - Prob. 8ECh. U1.18 - Prob. 9ECh. U1.18 - Prob. 10ECh. U1.19 - Prob. 1TAICh. U1.19 - Prob. 1ECh. U1.19 - Prob. 2ECh. U1.19 - Prob. 3ECh. U1.19 - Prob. 4ECh. U1.19 - Prob. 5ECh. U1.19 - Prob. 6ECh. U1.19 - Prob. 7ECh. U1.19 - Prob. 8ECh. U1.19 - Prob. 9ECh. U1.19 - Prob. 10ECh. U1.19 - Prob. 11ECh. U1.19 - Prob. 12ECh. U1.19 - Prob. 13ECh. U1.19 - Prob. 14ECh. U1.19 - Prob. 15ECh. U1.19 - Prob. 16ECh. U1.20 - Prob. 1TAICh. U1.20 - Prob. 1ECh. U1.20 - Prob. 2ECh. U1.20 - Prob. 3ECh. U1.20 - Prob. 4ECh. U1.20 - Prob. 5ECh. U1.20 - Prob. 6ECh. U1.20 - Prob. 7ECh. U1.21 - Prob. 1TAICh. U1.21 - Prob. 1ECh. U1.21 - Prob. 2ECh. U1.21 - Prob. 3ECh. U1.21 - Prob. 4ECh. U1.21 - Prob. 5ECh. U1.21 - Prob. 6ECh. U1.21 - Prob. 7ECh. U1.21 - Prob. 8ECh. U1.22 - Prob. 1TAICh. U1.22 - Prob. 1ECh. U1.22 - Prob. 2ECh. U1.22 - Prob. 3ECh. U1.22 - Prob. 4ECh. U1.22 - Prob. 5ECh. U1.22 - Prob. 6ECh. U1.22 - Prob. 7ECh. U1.23 - Prob. 1TAICh. U1.23 - Prob. 1ECh. U1.23 - Prob. 2ECh. U1.23 - Prob. 3ECh. U1.23 - Prob. 4ECh. U1.23 - Prob. 5ECh. U1.24 - Prob. 1TAICh. U1.24 - Prob. 1ECh. U1.24 - Prob. 2ECh. U1.24 - Prob. 3ECh. U1.24 - Prob. 4ECh. U1.24 - Prob. 5ECh. U1.24 - Prob. 6ECh. U1.24 - Prob. 7ECh. U1.24 - Prob. 8ECh. U1.24 - Prob. 9ECh. U1.24 - Prob. 10ECh. U1.24 - Prob. 11ECh. U1.24 - Prob. 12ECh. U1.24 - Prob. 13ECh. U1.25 - Prob. 1TAICh. U1.25 - Prob. 1ECh. U1.25 - Prob. 2ECh. U1.25 - Prob. 3ECh. U1.25 - Prob. 4ECh. U1.25 - Prob. 5ECh. U1.25 - Prob. 6ECh. U1.26 - Prob. 1TAICh. U1.26 - Prob. 1ECh. U1.26 - Prob. 2ECh. U1.26 - Prob. 3ECh. U1.26 - Prob. 4ECh. U1.26 - Prob. 5ECh. U1.26 - Prob. 6ECh. U1.26 - Prob. 7ECh. U1.26 - Prob. 8ECh. U1.26 - Prob. 9ECh. U1.26 - Prob. 10ECh. U1.27 - Prob. 1TAICh. U1.27 - Prob. 1ECh. U1.27 - Prob. 2ECh. U1.27 - Prob. 3ECh. U1.27 - Prob. 4ECh. U1.27 - Prob. 5ECh. U1.27 - Prob. 6ECh. U1.27 - Prob. 7ECh. U1 - Prob. SI1RECh. U1 - Prob. SI2RECh. U1 - Prob. SI3RECh. U1 - Prob. SI4RECh. U1 - Prob. SI5RECh. U1 - Prob. SI6RECh. U1 - Prob. SII1RQCh. U1 - Prob. SII2RQCh. U1 - Prob. SII3RQCh. U1 - Prob. SII4RQCh. U1 - Prob. SIII1RQCh. U1 - Prob. SIII2RQCh. U1 - Prob. SIII3RQCh. U1 - Prob. SIII4RQCh. U1 - Prob. SIV1RQCh. U1 - Prob. SIV2RQCh. U1 - Prob. SIV3RQCh. U1 - Prob. SIV4RQCh. U1 - Prob. SIV5RQCh. U1 - Prob. SIV6RQCh. U1 - Prob. SV1RQCh. U1 - Prob. SV2RQCh. U1 - Prob. SV3RQCh. U1 - Prob. SV4RQCh. U1 - Prob. 1RECh. U1 - Prob. 2RECh. U1 - Prob. 3RECh. U1 - Prob. 4RECh. U1 - Prob. 5RECh. U1 - Prob. 6RECh. U1 - Prob. 7RECh. U1 - Prob. 8RECh. U1 - Prob. 9RECh. U1 - Prob. 10RECh. U1 - Prob. 11RECh. U1 - Prob. 12RE
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