Living by Chemistry
2nd Edition
ISBN: 9781464142314
Author: Angelica M. Stacy
Publisher: W. H. Freeman
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Question
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
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.
- 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.
- 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.
- 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.
- 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
Ch. U1.1 - Prob. 1TAICh. U1.1 - Prob. 1ECh. U1.1 - Prob. 2ECh. U1.1 - Prob. 3ECh. U1.1 - Prob. 4ECh. U1.1 - Prob. 5ECh. U1.1 - Prob. 6ECh. U1.1 - Prob. 7ECh. U1.1 - Prob. 8ECh. U1.2 - Prob. 1TAI
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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. C1.1RECh. U1 - Prob. C1.2RECh. U1 - Prob. C1.3RECh. U1 - Prob. C1.4RECh. U1 - Prob. C1.5RECh. U1 - Prob. C1.6RECh. U1 - Prob. C2.1RECh. U1 - Prob. C2.2RECh. U1 - Prob. C2.3RECh. U1 - Prob. C2.4RECh. U1 - Prob. C3.1RECh. U1 - Prob. C3.2RECh. U1 - Prob. C3.3RECh. U1 - Prob. C3.4RECh. U1 - Prob. C4.1RECh. U1 - Prob. C4.2RECh. U1 - Prob. C4.3RECh. U1 - Prob. C4.4RECh. U1 - Prob. C5.1RECh. U1 - Prob. C5.2RECh. U1 - Prob. C5.3RECh. U1 - Prob. C5.4RECh. 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. 12RECh. U1 - Prob. 1STPCh. U1 - Prob. 2STPCh. U1 - Prob. 3STPCh. U1 - Prob. 4STPCh. U1 - Prob. 5STPCh. U1 - Prob. 6STPCh. U1 - Prob. 7STPCh. U1 - Prob. 8STPCh. U1 - Prob. 9STPCh. U1 - Prob. 10STPCh. U1 - Prob. 11STPCh. U1 - Prob. 12STPCh. U1 - Prob. 13STPCh. U1 - Prob. 14STPCh. U1 - Prob. 15STPCh. U1 - Prob. 16STPCh. U1 - Prob. 17STPCh. U1 - Prob. 18STPCh. U1 - Prob. 19STPCh. U1 - Prob. 20STPCh. U1 - Prob. 21STPCh. U1 - Prob. 22STP
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