{"id":1151,"date":"2018-07-06T02:53:34","date_gmt":"2018-07-06T02:53:34","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/?post_type=chapter&#038;p=1151"},"modified":"2018-09-07T16:08:54","modified_gmt":"2018-09-07T16:08:54","slug":"formulas-of-ionic-compounds","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/formulas-of-ionic-compounds\/","title":{"raw":"5.2 Formulas of Ionic Compounds","rendered":"5.2 Formulas of Ionic Compounds"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Determine formulas for simple ionic compounds<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn every ionic compound, the total number of positive charges of the cations equals the total number of negative charges of the anions. Thus, ionic compounds are electrically neutral overall, even though they contain positive and negative ions. We can use this observation to help us write the formula of an ionic compound. The formula of an ionic compound must have a ratio of ions such that the numbers of positive and negative charges are equal.\r\n<div class=\"textbox examples\">\r\n<h3>Example 1:\u00a0<strong>Predicting the Formula of an Ionic Compound<\/strong><\/h3>\r\nThe gemstone sapphire (Figure 1) is mostly a compound of aluminum and oxygen that contains aluminum cations, Al<sup>3+<\/sup>, and oxygen anions, O<sup>2\u2212<\/sup>. What is the formula of this compound?\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211106\/CNX_Chem_02_06_Sapphire1.jpg\" alt=\"This is a photograph of a ring with a sapphire set in it.\" width=\"325\" height=\"248\" \/> Figure 1. Although pure aluminum oxide is colorless, trace amounts of iron and titanium give blue sapphire its characteristic color. (credit: modification of work by Stanislav Doronenko)[\/caption]\r\n\r\n[reveal-answer q=\"960063\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"960063\"]\r\n\r\nBecause the ionic compound must be electrically neutral, it must have the same number of positive and negative charges. Two aluminum ions, each with a charge of 3+, would give us six positive charges, and three oxide ions, each with a charge of 2\u2212, would give us six negative charges. The formula would be Al<sub>2<\/sub>O<sub>3<\/sub>.\r\n\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nPredict the formula of the ionic compound formed between the sodium cation, Na<sup>+<\/sup>, and the sulfide anion, S<sup>2\u2212<\/sup>.\r\n\r\n[reveal-answer q=\"974651\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"974651\"]Na<sub>2<\/sub>S[\/hidden-answer]\r\n\r\n<\/div>\r\nMany ionic compounds contain polyatomic ions (Table 1) as the cation, the anion, or both. As with simple ionic compounds, these compounds must also be electrically neutral, so their formulas can be predicted by treating the polyatomic ions as discrete units. We use parentheses in a formula to indicate a group of atoms that behave as a unit. For example, the formula for calcium phosphate, one of the minerals in our bones, is Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>. This formula indicates that there are three calcium ions (Ca<sup>2+<\/sup>) for every two phosphate [latex]\\left({\\text{PO}}_{4}{}^{3-}\\right)[\/latex] groups. The [latex]{\\text{PO}}_{4}{}^{3-}[\/latex] groups are discrete units, each consisting of one phosphorus atom and four oxygen atoms, and having an overall charge of 3-. The compound is electrically neutral, and its formula shows a total count of three Ca, two P, and eight O atoms.\r\n<table id=\"fs-idp278664880\" class=\"span-all\" summary=\"This table has three columns labeled \u201ccharge,\u201d \u201cname,\u201d and \u201cformula.\u201d Ammonium has a charge of positive 1 and the formula N H subscript 4 superscript plus sign. Acetate has a charge of negative 1 and the formula C subscript 2 H subscript 3 O subscript 2 superscript negative sign. Cyanide has a charge of negative 1, and the formula C N superscript negative sign. Hydroxide has a charge of negative 1 and the formula O H superscript negative sign. Nitrate has a charge of negative 1 and the formula N O subscript 3 superscript negative sign. Nitrite has a charge of negative 1 and the formula N O subscript 2 superscript negative sign. Perchlorate has a charge of negative 1 and the formula C l O subscript 4 superscript negative sign. Chlorate has a charge of negative 1 and the formula C l O subscript 3 superscript negative sign. Chlorite has a charge of negative 1 and the formula C l O subscript 2 superscript negative sign. Hypochlorite has a charge of negative 1 and the formula C l O superscript negative sign. Permanganate has a charge of negative 1 and the formula M n O subscript 4 superscript negative sign. Hydrogen carbonate, or bicarbonate has a charge of negative 1 and the formula H C O subscript 3 superscript negative sign. Carbonate has a charge of negative 2 and the formula C O subscript 3 superscript 2 negative sign. Peroxide has a charge of negative 2 and the formula O subscript 2 superscript 2 negative sign. Hydrogen sulfate, or bisulfate, has a charge of negative 1 and the formula H S O subscript 4 superscript negative sign. Sulfate has a charge of negative 2 and the formula S O subscript 4 superscript 2 negative sign. Sulfite has a charge of negative 2 and the formula S O subscript 3 superscript 2 negative sign. Dihydrogen phosphate has a charge of negative 1 and the formula H subscript 2 P O subscript 4 superscript negative sign. Hydrogen phosphate has a charge of negative 2 and the formula H P O subscript 4 superscript 2 negative sign. Phosphate has a charge of negative 3 and the formula P O subscript 4 superscript 3 negative sign.\">\r\n<thead>\r\n<tr>\r\n<th colspan=\"7\">Table 1. Common Polyatomic Ions<\/th>\r\n<\/tr>\r\n<tr>\r\n<th width=\"10%\">Charge<\/th>\r\n<th width=\"20%\">Name<\/th>\r\n<th width=\"10%\">Formula<\/th>\r\n<th width=\"10%\"><\/th>\r\n<th width=\"10%\">Charge<\/th>\r\n<th width=\"30%\">Name<\/th>\r\n<th width=\"10%\">Formula<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>1+<\/td>\r\n<td>ammonium<\/td>\r\n<td>[latex]{\\text{NH}}_{4}{}^{+}[\/latex]<\/td>\r\n<td><\/td>\r\n<td>1\u2212<\/td>\r\n<td>permanganate<\/td>\r\n<td>[latex]{\\text{MnO}}_{4}{}^{-}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1\u2212<\/td>\r\n<td>acetate<\/td>\r\n<td>[latex]{\\text{C}}_{2}{\\text{H}}_{3}{\\text{O}}_{2}{}^{-}[\/latex]<\/td>\r\n<td><\/td>\r\n<td>1\u2212<\/td>\r\n<td>hydrogen carbonate, or bicarbonate<\/td>\r\n<td>[latex]{\\text{HCO}}_{3}{}^{-}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1\u2212<\/td>\r\n<td>cyanide<\/td>\r\n<td>[latex]\\text{CN}^-[\/latex]<\/td>\r\n<td><\/td>\r\n<td>2\u2212<\/td>\r\n<td>carbonate<\/td>\r\n<td>[latex]{\\text{CO}}_{3}{}^{2-}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1\u2212<\/td>\r\n<td>hydroxide<\/td>\r\n<td>[latex]\\text{OH}^-[\/latex]<\/td>\r\n<td><\/td>\r\n<td>2\u2212<\/td>\r\n<td>peroxide<\/td>\r\n<td>[latex]{\\text{O}}_{2}{}^{2-}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1\u2212<\/td>\r\n<td>nitrate<\/td>\r\n<td>[latex]{\\text{NO}}_{3}{}^{-}[\/latex]<\/td>\r\n<td><\/td>\r\n<td>1\u2212<\/td>\r\n<td>hydrogen sulfate, or bisulfate<\/td>\r\n<td>[latex]{\\text{HSO}}_{4}{}^{-}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1\u2212<\/td>\r\n<td>nitrite<\/td>\r\n<td>[latex]{\\text{NO}}_{2}{}^{-}[\/latex]<\/td>\r\n<td><\/td>\r\n<td>2\u2212<\/td>\r\n<td>sulfate<\/td>\r\n<td>[latex]{\\text{SO}}_{4}{}^{2-}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1\u2212<\/td>\r\n<td>perchlorate<\/td>\r\n<td>[latex]{\\text{ClO}}_{4}{}^{-}[\/latex]<\/td>\r\n<td><\/td>\r\n<td>2\u2212<\/td>\r\n<td>sulfite<\/td>\r\n<td>[latex]{\\text{SO}}_{3}{}^{2-}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1\u2212<\/td>\r\n<td>chlorate<\/td>\r\n<td>[latex]{\\text{ClO}}_{3}{}^{-}[\/latex]<\/td>\r\n<td><\/td>\r\n<td>1\u2212<\/td>\r\n<td>dihydrogen phosphate<\/td>\r\n<td>[latex]{\\text{H}}_{2}{\\text{PO}}_{4}{}^{-}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1\u2212<\/td>\r\n<td>chlorite<\/td>\r\n<td>[latex]{\\text{ClO}}_{2}{}^{-}[\/latex]<\/td>\r\n<td><\/td>\r\n<td>2\u2212<\/td>\r\n<td>hydrogen phosphate<\/td>\r\n<td>[latex]{\\text{HPO}}_{4}{}^{2-}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1\u2212<\/td>\r\n<td>hypochlorite<\/td>\r\n<td>[latex]\\text{ClO}^-[\/latex]<\/td>\r\n<td><\/td>\r\n<td>3\u2212<\/td>\r\n<td>phosphate<\/td>\r\n<td>[latex]{\\text{PO}}_{4}{}^{3-}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n&nbsp;\r\n<div class=\"textbox examples\">\r\n<h3>Example 2:\u00a0<strong>Predicting the Formula of a Compound with a Polyatomic Anion<\/strong><\/h3>\r\nBaking powder contains calcium dihydrogen phosphate, an ionic compound composed of the ions Ca<sup>2+<\/sup> and [latex]{\\text{H}}_{2}{\\text{PO}}_{4}{}^{-}[\/latex]. What is the formula of this compound?\r\n\r\n[reveal-answer q=\"807500\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"807500\"]\r\n\r\nThe positive and negative charges must balance, and this ionic compound must be electrically neutral. Thus, we must have two negative charges to balance the 2+ charge of the calcium ion. This requires a ratio of one Ca<sup>2+<\/sup> ion to two [latex]{\\text{H}}_{2}{\\text{PO}}_{4}{}^{-}[\/latex] ions. We designate this by enclosing the formula for the dihydrogen phosphate ion in parentheses and adding a subscript 2. The formula is Ca(H<sub>2<\/sub>PO<sub>4<\/sub>)<sub>2<\/sub>.\r\n\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nPredict the formula of the ionic compound formed between the lithium ion and the peroxide ion, [latex]{\\text{O}}_{2}{}^{2-}[\/latex] (Hint: Use the periodic table to predict the sign and the charge on the lithium ion.)\r\n\r\n[reveal-answer q=\"571395\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"571395\"]Li<sub>2<\/sub>O<sub>2<\/sub>[\/hidden-answer]\r\n\r\n<\/div>\r\nBecause an ionic compound is not made up of single, discrete molecules, it may not be properly symbolized using a <em>molecular<\/em> formula. Instead, ionic compounds must be symbolized by a formula indicating the <em>relative numbers<\/em> of its constituent cations. For compounds containing only monatomic ions (such as NaCl) and for many compounds containing polyatomic ions (such as CaSO<sub>4<\/sub>), these formulas are just the empirical formulas introduced earlier in this chapter. However, the formulas for some ionic compounds containing polyatomic ions are not empirical formulas. For example, the ionic compound sodium oxalate is comprised of Na<sup>+<\/sup> and [latex]{\\text{C}}_{2}{\\text{O}}_{4}{}^{2-}[\/latex] ions combined in a 2:1 ratio, and its formula is written as Na<sub>2<\/sub>C<sub>2<\/sub>O<sub>4<\/sub>. The subscripts in this formula are not the smallest-possible whole numbers, as each can be divided by 2 to yield the empirical formula, NaCO<sub>2<\/sub>. This is not the accepted formula for sodium oxalate, however, as it does not accurately represent the compound\u2019s polyatomic anion, [latex]{\\text{C}}_{2}{\\text{O}}_{4}{}^{2-}[\/latex].\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\r\nThe total number of positive charges of the cations equals the total number of negative charges of the anions. Thus, ionic compounds are electrically neutral overall, even though they contain positive and negative ions.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<ol>\r\n \t<li>For each of the following pairs of ions, write the symbol for the formula of the compound they will form:\r\n<ol style=\"list-style-type: lower-alpha\">\r\n \t<li>Ca<sup>2+<\/sup>, S<sup>2- \u00a0<\/sup><\/li>\r\n \t<li>[latex]{\\text{NH}}_{4}{}^{+}[\/latex], [latex]{\\text{SO}}_{4}{}^{2-}[\/latex]<\/li>\r\n \t<li>Al<sup>3+<\/sup>, Br<sup>- \u00a0<\/sup><\/li>\r\n \t<li>Na<sup>+<\/sup>, [latex]{\\text{HPO}}_{4}{}^{2-}[\/latex]<\/li>\r\n \t<li>Mg<sup>2+<\/sup>, [latex]{\\text{PO}}_{4}{}^{3-}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>For each of the following pairs of ions, write the symbol for the formula of the compound they will form:\r\n<ol style=\"list-style-type: lower-alpha\">\r\n \t<li>K<sup>+<\/sup>, O<sup>2-<\/sup><\/li>\r\n \t<li>[latex]{\\text{NH}}_{4}{}^{+}[\/latex], [latex]{\\text{PO}}_{4}{}^{3-}[\/latex]<\/li>\r\n \t<li>Al<sup>3+<\/sup>, O<sup>2-<\/sup><\/li>\r\n \t<li>Na<sup>+<\/sup>, [latex]{\\text{CO}}_{3}{}^{2-}[\/latex]<\/li>\r\n \t<li>Ba<sup>2+<\/sup>, [latex]{\\text{PO}}_{4}{}^{3-}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"265027\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"265027\"]\r\n\r\n1.\u00a0(a) CaS; (b) (NH<sub>4<\/sub>)<sub>2<\/sub>CO<sub>3<\/sub>; (c) AlBr<sub>3<\/sub>; (d) Na<sub>2<\/sub>HPO<sub>4<\/sub>; (e) Mg<sub>3<\/sub> (PO<sub>4<\/sub>)<sub>2<\/sub>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<strong>ionic bond:\u00a0<\/strong>electrostatic forces of attraction between the oppositely charged ions of an ionic compound\r\n\r\n<strong>ionic compound:\u00a0<\/strong>compound composed of cations and anions combined in ratios, yielding an electrically neutral substance\r\n\r\n<strong>monatomic ion:\u00a0<\/strong>ion composed of a single atom\r\n\r\n<strong>oxyanion:\u00a0<\/strong>polyatomic anion composed of a central atom bonded to oxygen atoms\r\n\r\n<strong>polyatomic ion:\u00a0<\/strong>ion composed of more than one atom","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Determine formulas for simple ionic compounds<\/li>\n<\/ul>\n<\/div>\n<p>In every ionic compound, the total number of positive charges of the cations equals the total number of negative charges of the anions. Thus, ionic compounds are electrically neutral overall, even though they contain positive and negative ions. We can use this observation to help us write the formula of an ionic compound. The formula of an ionic compound must have a ratio of ions such that the numbers of positive and negative charges are equal.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1:\u00a0<strong>Predicting the Formula of an Ionic Compound<\/strong><\/h3>\n<p>The gemstone sapphire (Figure 1) is mostly a compound of aluminum and oxygen that contains aluminum cations, Al<sup>3+<\/sup>, and oxygen anions, O<sup>2\u2212<\/sup>. What is the formula of this compound?<\/p>\n<div style=\"width: 335px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211106\/CNX_Chem_02_06_Sapphire1.jpg\" alt=\"This is a photograph of a ring with a sapphire set in it.\" width=\"325\" height=\"248\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. Although pure aluminum oxide is colorless, trace amounts of iron and titanium give blue sapphire its characteristic color. (credit: modification of work by Stanislav Doronenko)<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q960063\">Show Answer<\/span><\/p>\n<div id=\"q960063\" class=\"hidden-answer\" style=\"display: none\">\n<p>Because the ionic compound must be electrically neutral, it must have the same number of positive and negative charges. Two aluminum ions, each with a charge of 3+, would give us six positive charges, and three oxide ions, each with a charge of 2\u2212, would give us six negative charges. The formula would be Al<sub>2<\/sub>O<sub>3<\/sub>.<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>Predict the formula of the ionic compound formed between the sodium cation, Na<sup>+<\/sup>, and the sulfide anion, S<sup>2\u2212<\/sup>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q974651\">Show Answer<\/span><\/p>\n<div id=\"q974651\" class=\"hidden-answer\" style=\"display: none\">Na<sub>2<\/sub>S<\/div>\n<\/div>\n<\/div>\n<p>Many ionic compounds contain polyatomic ions (Table 1) as the cation, the anion, or both. As with simple ionic compounds, these compounds must also be electrically neutral, so their formulas can be predicted by treating the polyatomic ions as discrete units. We use parentheses in a formula to indicate a group of atoms that behave as a unit. For example, the formula for calcium phosphate, one of the minerals in our bones, is Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>. This formula indicates that there are three calcium ions (Ca<sup>2+<\/sup>) for every two phosphate [latex]\\left({\\text{PO}}_{4}{}^{3-}\\right)[\/latex] groups. The [latex]{\\text{PO}}_{4}{}^{3-}[\/latex] groups are discrete units, each consisting of one phosphorus atom and four oxygen atoms, and having an overall charge of 3-. The compound is electrically neutral, and its formula shows a total count of three Ca, two P, and eight O atoms.<\/p>\n<table id=\"fs-idp278664880\" class=\"span-all\" summary=\"This table has three columns labeled \u201ccharge,\u201d \u201cname,\u201d and \u201cformula.\u201d Ammonium has a charge of positive 1 and the formula N H subscript 4 superscript plus sign. Acetate has a charge of negative 1 and the formula C subscript 2 H subscript 3 O subscript 2 superscript negative sign. Cyanide has a charge of negative 1, and the formula C N superscript negative sign. Hydroxide has a charge of negative 1 and the formula O H superscript negative sign. Nitrate has a charge of negative 1 and the formula N O subscript 3 superscript negative sign. Nitrite has a charge of negative 1 and the formula N O subscript 2 superscript negative sign. Perchlorate has a charge of negative 1 and the formula C l O subscript 4 superscript negative sign. Chlorate has a charge of negative 1 and the formula C l O subscript 3 superscript negative sign. Chlorite has a charge of negative 1 and the formula C l O subscript 2 superscript negative sign. Hypochlorite has a charge of negative 1 and the formula C l O superscript negative sign. Permanganate has a charge of negative 1 and the formula M n O subscript 4 superscript negative sign. Hydrogen carbonate, or bicarbonate has a charge of negative 1 and the formula H C O subscript 3 superscript negative sign. Carbonate has a charge of negative 2 and the formula C O subscript 3 superscript 2 negative sign. Peroxide has a charge of negative 2 and the formula O subscript 2 superscript 2 negative sign. Hydrogen sulfate, or bisulfate, has a charge of negative 1 and the formula H S O subscript 4 superscript negative sign. Sulfate has a charge of negative 2 and the formula S O subscript 4 superscript 2 negative sign. Sulfite has a charge of negative 2 and the formula S O subscript 3 superscript 2 negative sign. Dihydrogen phosphate has a charge of negative 1 and the formula H subscript 2 P O subscript 4 superscript negative sign. Hydrogen phosphate has a charge of negative 2 and the formula H P O subscript 4 superscript 2 negative sign. Phosphate has a charge of negative 3 and the formula P O subscript 4 superscript 3 negative sign.\">\n<thead>\n<tr>\n<th colspan=\"7\">Table 1. Common Polyatomic Ions<\/th>\n<\/tr>\n<tr>\n<th style=\"width: 10%;\">Charge<\/th>\n<th style=\"width: 20%;\">Name<\/th>\n<th style=\"width: 10%;\">Formula<\/th>\n<th style=\"width: 10%;\"><\/th>\n<th style=\"width: 10%;\">Charge<\/th>\n<th style=\"width: 30%;\">Name<\/th>\n<th style=\"width: 10%;\">Formula<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>1+<\/td>\n<td>ammonium<\/td>\n<td>[latex]{\\text{NH}}_{4}{}^{+}[\/latex]<\/td>\n<td><\/td>\n<td>1\u2212<\/td>\n<td>permanganate<\/td>\n<td>[latex]{\\text{MnO}}_{4}{}^{-}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>1\u2212<\/td>\n<td>acetate<\/td>\n<td>[latex]{\\text{C}}_{2}{\\text{H}}_{3}{\\text{O}}_{2}{}^{-}[\/latex]<\/td>\n<td><\/td>\n<td>1\u2212<\/td>\n<td>hydrogen carbonate, or bicarbonate<\/td>\n<td>[latex]{\\text{HCO}}_{3}{}^{-}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>1\u2212<\/td>\n<td>cyanide<\/td>\n<td>[latex]\\text{CN}^-[\/latex]<\/td>\n<td><\/td>\n<td>2\u2212<\/td>\n<td>carbonate<\/td>\n<td>[latex]{\\text{CO}}_{3}{}^{2-}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>1\u2212<\/td>\n<td>hydroxide<\/td>\n<td>[latex]\\text{OH}^-[\/latex]<\/td>\n<td><\/td>\n<td>2\u2212<\/td>\n<td>peroxide<\/td>\n<td>[latex]{\\text{O}}_{2}{}^{2-}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>1\u2212<\/td>\n<td>nitrate<\/td>\n<td>[latex]{\\text{NO}}_{3}{}^{-}[\/latex]<\/td>\n<td><\/td>\n<td>1\u2212<\/td>\n<td>hydrogen sulfate, or bisulfate<\/td>\n<td>[latex]{\\text{HSO}}_{4}{}^{-}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>1\u2212<\/td>\n<td>nitrite<\/td>\n<td>[latex]{\\text{NO}}_{2}{}^{-}[\/latex]<\/td>\n<td><\/td>\n<td>2\u2212<\/td>\n<td>sulfate<\/td>\n<td>[latex]{\\text{SO}}_{4}{}^{2-}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>1\u2212<\/td>\n<td>perchlorate<\/td>\n<td>[latex]{\\text{ClO}}_{4}{}^{-}[\/latex]<\/td>\n<td><\/td>\n<td>2\u2212<\/td>\n<td>sulfite<\/td>\n<td>[latex]{\\text{SO}}_{3}{}^{2-}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>1\u2212<\/td>\n<td>chlorate<\/td>\n<td>[latex]{\\text{ClO}}_{3}{}^{-}[\/latex]<\/td>\n<td><\/td>\n<td>1\u2212<\/td>\n<td>dihydrogen phosphate<\/td>\n<td>[latex]{\\text{H}}_{2}{\\text{PO}}_{4}{}^{-}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>1\u2212<\/td>\n<td>chlorite<\/td>\n<td>[latex]{\\text{ClO}}_{2}{}^{-}[\/latex]<\/td>\n<td><\/td>\n<td>2\u2212<\/td>\n<td>hydrogen phosphate<\/td>\n<td>[latex]{\\text{HPO}}_{4}{}^{2-}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>1\u2212<\/td>\n<td>hypochlorite<\/td>\n<td>[latex]\\text{ClO}^-[\/latex]<\/td>\n<td><\/td>\n<td>3\u2212<\/td>\n<td>phosphate<\/td>\n<td>[latex]{\\text{PO}}_{4}{}^{3-}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2:\u00a0<strong>Predicting the Formula of a Compound with a Polyatomic Anion<\/strong><\/h3>\n<p>Baking powder contains calcium dihydrogen phosphate, an ionic compound composed of the ions Ca<sup>2+<\/sup> and [latex]{\\text{H}}_{2}{\\text{PO}}_{4}{}^{-}[\/latex]. What is the formula of this compound?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q807500\">Show Answer<\/span><\/p>\n<div id=\"q807500\" class=\"hidden-answer\" style=\"display: none\">\n<p>The positive and negative charges must balance, and this ionic compound must be electrically neutral. Thus, we must have two negative charges to balance the 2+ charge of the calcium ion. This requires a ratio of one Ca<sup>2+<\/sup> ion to two [latex]{\\text{H}}_{2}{\\text{PO}}_{4}{}^{-}[\/latex] ions. We designate this by enclosing the formula for the dihydrogen phosphate ion in parentheses and adding a subscript 2. The formula is Ca(H<sub>2<\/sub>PO<sub>4<\/sub>)<sub>2<\/sub>.<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>Predict the formula of the ionic compound formed between the lithium ion and the peroxide ion, [latex]{\\text{O}}_{2}{}^{2-}[\/latex] (Hint: Use the periodic table to predict the sign and the charge on the lithium ion.)<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q571395\">Show Answer<\/span><\/p>\n<div id=\"q571395\" class=\"hidden-answer\" style=\"display: none\">Li<sub>2<\/sub>O<sub>2<\/sub><\/div>\n<\/div>\n<\/div>\n<p>Because an ionic compound is not made up of single, discrete molecules, it may not be properly symbolized using a <em>molecular<\/em> formula. Instead, ionic compounds must be symbolized by a formula indicating the <em>relative numbers<\/em> of its constituent cations. For compounds containing only monatomic ions (such as NaCl) and for many compounds containing polyatomic ions (such as CaSO<sub>4<\/sub>), these formulas are just the empirical formulas introduced earlier in this chapter. However, the formulas for some ionic compounds containing polyatomic ions are not empirical formulas. For example, the ionic compound sodium oxalate is comprised of Na<sup>+<\/sup> and [latex]{\\text{C}}_{2}{\\text{O}}_{4}{}^{2-}[\/latex] ions combined in a 2:1 ratio, and its formula is written as Na<sub>2<\/sub>C<sub>2<\/sub>O<sub>4<\/sub>. The subscripts in this formula are not the smallest-possible whole numbers, as each can be divided by 2 to yield the empirical formula, NaCO<sub>2<\/sub>. This is not the accepted formula for sodium oxalate, however, as it does not accurately represent the compound\u2019s polyatomic anion, [latex]{\\text{C}}_{2}{\\text{O}}_{4}{}^{2-}[\/latex].<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<p>The total number of positive charges of the cations equals the total number of negative charges of the anions. Thus, ionic compounds are electrically neutral overall, even though they contain positive and negative ions.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<ol>\n<li>For each of the following pairs of ions, write the symbol for the formula of the compound they will form:\n<ol style=\"list-style-type: lower-alpha\">\n<li>Ca<sup>2+<\/sup>, S<sup>2- \u00a0<\/sup><\/li>\n<li>[latex]{\\text{NH}}_{4}{}^{+}[\/latex], [latex]{\\text{SO}}_{4}{}^{2-}[\/latex]<\/li>\n<li>Al<sup>3+<\/sup>, Br<sup>&#8211; \u00a0<\/sup><\/li>\n<li>Na<sup>+<\/sup>, [latex]{\\text{HPO}}_{4}{}^{2-}[\/latex]<\/li>\n<li>Mg<sup>2+<\/sup>, [latex]{\\text{PO}}_{4}{}^{3-}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>For each of the following pairs of ions, write the symbol for the formula of the compound they will form:\n<ol style=\"list-style-type: lower-alpha\">\n<li>K<sup>+<\/sup>, O<sup>2-<\/sup><\/li>\n<li>[latex]{\\text{NH}}_{4}{}^{+}[\/latex], [latex]{\\text{PO}}_{4}{}^{3-}[\/latex]<\/li>\n<li>Al<sup>3+<\/sup>, O<sup>2-<\/sup><\/li>\n<li>Na<sup>+<\/sup>, [latex]{\\text{CO}}_{3}{}^{2-}[\/latex]<\/li>\n<li>Ba<sup>2+<\/sup>, [latex]{\\text{PO}}_{4}{}^{3-}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q265027\">Show Selected Answers<\/span><\/p>\n<div id=\"q265027\" class=\"hidden-answer\" style=\"display: none\">\n<p>1.\u00a0(a) CaS; (b) (NH<sub>4<\/sub>)<sub>2<\/sub>CO<sub>3<\/sub>; (c) AlBr<sub>3<\/sub>; (d) Na<sub>2<\/sub>HPO<sub>4<\/sub>; (e) Mg<sub>3<\/sub> (PO<sub>4<\/sub>)<sub>2<\/sub><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Glossary<\/h2>\n<p><strong>ionic bond:\u00a0<\/strong>electrostatic forces of attraction between the oppositely charged ions of an ionic compound<\/p>\n<p><strong>ionic compound:\u00a0<\/strong>compound composed of cations and anions combined in ratios, yielding an electrically neutral substance<\/p>\n<p><strong>monatomic ion:\u00a0<\/strong>ion composed of a single atom<\/p>\n<p><strong>oxyanion:\u00a0<\/strong>polyatomic anion composed of a central atom bonded to oxygen atoms<\/p>\n<p><strong>polyatomic ion:\u00a0<\/strong>ion composed of more than one atom<\/p>\n","protected":false},"author":6181,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1151","chapter","type-chapter","status-publish","hentry"],"part":1128,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1151","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/users\/6181"}],"version-history":[{"count":6,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1151\/revisions"}],"predecessor-version":[{"id":1623,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1151\/revisions\/1623"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/parts\/1128"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1151\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/media?parent=1151"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=1151"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/contributor?post=1151"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/license?post=1151"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}