Chapter 4 Chemical Formulas

Chapter 4: Chemical Formulas NAMING AND WRITING CHEMICAL FORMULAS - Category 1: Binary ionic compounds where the metal present forms only a single type of positively charged ion (cation) - Metallic ions from groups 1 and 2 of the Periodic Table - These metallic ions have only one type of charge - Composed of a positive ion (cation) that is written first and then a negative ion (anion0 - Ex: 𝐶𝑎𝐶𝑙2 1. Name the cation first and then the anion. 2. The one-atom cation takes its name from the element. Therefore, the calcium ion, 𝐶𝑎2+ , is called calcium and its chemical symbol appears first. 3. The one-atom anion with which the cation combines is named by taking the root of the element’s name and adding -ide. The anion’s name comes second. Therefore, thee chlorine ion, 𝐶𝑙 − , is called chloride. 4. The name of this compound is calcium chloride. - A quick way to determine the formula of a binary ionic compound is to use the criss-cross rule. - - Example 1: To determine the formula for calcium chloride, first write the ionic forms with their associated charges. - Next, move the numerical value of the metal ion’s superscript (without the charge) to the subscript of the nonmetal’s symbol. Then take the numerical value of the nonmetal’s superscript and make it the subscript of the metal as shown above. - NOTE: The numerical value 1 is not shown in the final formula. - Now the chlorine’s 1 as the subscript of the calcium and the calcium’s 2 as the subscript of the chloride. As a result you have 𝐶𝑎𝐶𝑙2 as the final formula for calcium chloride. Example 2: Write the name and formula for the product formed when aluminum reacts with oxygen. First write the name. 1. Name the cation first and then the anion. 2. The one-atom cation takes its name from the name of the element. Therefore, the aluminium ion, 𝐴𝑙 3+ , is called aluminum and its chemical symbol appears first. 3. The one-atom anion w/ which the cation combines is named by taking the root of the element’s name and adding -ide. The anion’s name comes second. Therefore, the oxygen ion, 𝑂2− , is called oxide. 4. The name of this compound is aluminum oxide. To determine the formula for aluminum oxide, first write the ionic forms with their associated charges. Next, move the numerical value of the Al’s superscript (w/o the charge) to the subscript of the O symbol. Do the same w/ the 2 of the O. Now the 2 is the subscript of the aluminum and the 3 as a subscript of the oxygen. You now have 𝐴𝑙2 𝑂3. - - Category 2: Binary ionic compounds where the metal forms more than one type of ionic compounds with a given negatively charged ion (anion) - The metals form more than one ion, each w/ a diff charge - The metallic ions (cation) ionically bind w/ a negatively charged ion (anion) - Example: The compound containing the 𝐹𝑒 2+ ion and the compound containing the 𝐹𝑒 3+ ion both combine w/ chloride ion to form 2 different compounds. You get the formula 𝐹𝑒𝐶𝑙2 for iron(II) chloride. The compound formed using the 𝐹𝑒 3+ ion and the chloride ion is 𝐹𝑒𝐶𝑙3 , which is iron (III) chloride. The names iron (II) chloride and iron (III) chloride are arrived at by using the Roman numerals in parentheses to indicate the charge of the metallic ion used as the cation. Category 1 and 2 Ionic Compounds Formed W/ Polyatomic Ions - A polyatomic ion is a group of elements that act like a single ion when forming a compound. - The bonds within are predominantly covalent. However, the group as a whole has an excessive charge, usually negative, b/c of an excess of electrons. - If the compounds formed w/ the polyatomic ions consist of 3 elements, they are called ternary compounds. - Sometimes an element combines w/ oxygen to form more than just 2 polyatomic ions, the prefix hypois used to name the polyatomic ion w/ the fewest oxygen ions and the prefix per- to name the polyatomic ion w/ the most oxygen ions. Writing Formulas w/ Polyatomic Ions - Simply treat the polyatomic ion as if it were a single anion. - If the cation is from category 1, follow the rules for category 1. If the cation is from category 2, follow the rules for category 2. Example 1: Use the crisscross method to write the formula for calcium sulfate, a category 1 cation and a polyatomic ion. The final formula is 𝐶𝑎𝑆𝑂4. Notice the subscripts “2” are omitted. - Category 3: Binary covalent compounds formed b/w 2 nonmetals Binary covalent compounds are formed b/w 2 nonmetals 1. The first name element in the formula is named first, using the full elemental name. 2. The second element is named as if it were an anion and uses its elemental name. 3. Prefixes are used to denote the number of the second element present. These prefixes are shown in the table below. 4. The prefix mono- is never used for naming the first element. For example, CO is called carbon monoxide, not monocarbon monoxide. - Prefix Number Prefix Number mono- 1 hexa- 6 di- 2 hepta- 7 tri- 3 octa- 8 tetra- 4 nona- 9 penta- 5 deca- 10 To write the formula for binary covalent compounds, use the same steps as when writing the formula of ionic compounds. 1. The symbol of the first element in the formula is written first, followed by the second element. 2. Use the prefix(es) denoted in the name for the number of each element present in the formula. Examples: Sulfur hexafluoride (𝑆𝐹6 ) and Phosphorus trichloride (𝑃𝐶𝑙3 ) OXIDATION STATES AND FORMULA WRITING - An oxidation state is assigned to each member of a formula or polyatomic ion - It’s designated by a small, whole-number superscript preceded by a plus or minus sign - This is not to be confused with the ionic charges used to the right of ionic charge - These charges are directly related to the bonding that occurs in compounds - They're also used to track electron transfers The Rules for Assigning an Oxidation State - Remember: The sum of the oxidation states must be zero for an electrically neutral compound - For an ion, the sum of the oxidation states must equal the charge on the ion 1. The oxidation state of an atom in an element is zero. 2. The oxidation state of a monatomic ion is the same as its charge. 3. The oxidation state of fluorine is -1 in compounds. 4. The oxidation state of oxygen is usually -2 in its compounds. 5. The oxidation state of hydrogen in most compounds is +1. EXAMPLE 1: In 𝑁𝑎2 𝑆𝑂4, what is the oxidation state of sulfur? EXAMPLE 2: What is the oxidation state of chromium in 𝐾2 𝐶𝑟2 𝑂7 ? CHEMICAL FORMULAS: THEIR MEANING AND USE - By using the atomic masses assigned to the elements, we can find the formula mass of a compound. - The formula mass is determined by multiplying the atomic mass units (as a whole number) by the subscript for that element and then adding these values for all the elements in the formula. - If the formula represents the actual makeup of one molecule of the substance, molecular mass can be used as well. - The simplest ratio formula is called the empirical formula, and the actual formula is called the true formula. Example 1: 𝐶𝑎(𝑂𝐻)2 (one calcium amu + two hydrogen amu = formula mass) 1Ca (amu = 40) = 40 2O (amu = 16) = 32 2H (amu = 1) = 2.0 Formula Mass 𝐶𝑎(𝑂𝐻)2 = 74 amu If you have 6.02 × 1023atoms of an element, then the atomic mass units can be expressed in grams, and then the formula mass can be called the molar mass. Example 2: 𝐹𝑒2 𝑂3 2Fe (amu = 56) = 112 3O (amu = 16) = 48.0 - Formula mass 𝐹𝑒2 𝑂3 = 160. Amu - Finding the percentage composition is finding what percent of the total weight of a compound is made up of a particular element. 𝑇𝑜𝑡𝑎𝑙 𝑎𝑚𝑢 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 × 100% =Percentage composition of that element 𝑇𝑜𝑡𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑎𝑚𝑢 To find the percent composition of calcium in calcium hydroxide: 𝐶𝑎 = 32 𝑎𝑚𝑢 × 100% = 54%Calcium 𝐹𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠 = 74 𝑎𝑚𝑢 Example: Find the percent compositions of Cu and 𝐻2 𝑂in the compound 𝐶𝑢𝑆𝑂4 ∙ 5𝐻2 𝑂. First, calculate the formula mass: 1 Cu = 64 amu 1 S = 32 amu 4 O = 64(4 × 16) amu 5𝐻2 𝑂= 90. (5 × 18) amu 250 amu And then find the percentages: 𝐶𝑢 = 64 𝑎𝑚𝑢 Percentage Cu: × 100% = 26% 𝐹𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠 = 250 𝑎𝑚𝑢 36% Percentage 𝐻2 𝑂: 5.0 𝐻2 𝑂 = 90.𝑎𝑚𝑢 𝐹𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠 = 250 𝑎𝑚𝑢 × 100% = When you are given the percentage of each element in a compound, you can find the empirical formula as shown w/ the following example: Given Ba = 58.81%, S = 13.73%, and O = 27.46%. Find the empirical formula. 1. Divide each percent by the amu of the element. a. Ba: 58.8/137 = 0.43, S:13.7/32 = 0.43, O: 27.5/16 =1.72 2. Manipulate numbers to get small whole numbers. Try dividing them all by the smallest first. In this case, divide each by 0.43. a. Ba: 0.43/0.43 = 1, S: 0.43/0.43 = 1, O: 1.72/0.43 = 4 3. The formula is 𝐵𝑎𝑆𝑂4 . WRITING AND BALANCING EQUATIONS - An equation is a simplified way of recording a chemical change. - Chemical symbols and formulas are used to represent the reactants and the products. Example: 𝐻2 + 𝑂2 → 𝐻2 𝑂 - While the left side has 2 atoms of oxygen, the right side only has one. To get the number of atoms of each element represented on the left side to equal the number on the right. - To do this, we can only use coefficients in front of the formula. - THE SUBSCRIPTS IN THE FORMULA MAY NOT BE CHANGED - If we put 2 in front of the 𝐻2 𝑂the number of oxygen atoms represented on the 2 sides of the equation are equal. However, there are now 4 hydrogens on the right side w/ only 2 on the left.This can be corrected by using a coefficient of 2 in front of the 𝐻2 . 2𝐻2 + 𝑂2 → 2𝐻2 𝑂 SHOWING PHASES IN CHEMICAL EQUATIONS - Once an equation is balanced, you can indicate the phases of substances, telling whether each substance is in the liquid phase (𝑙), the gaseous phase (g), or the solid phase (s). - Since many solids will not react to any appreciable extent unless they are dissolved in water, the notation (aq) is used to indicate the substance exists in a water (aqueous) solution. Ex: (aq) + Zn (s) → 𝑍𝑛𝐶𝑙2 (aq) + 𝐻2 (g) WRITING IONIC EQUATIONS - Ionic equations are used to show only the substances that react in the chemical action and b/c they stress the reaction and production of ions Reactants: 2NCl(aq) releases → 2𝐻 + (aq) + 2𝐶𝑙 − (aq) in solution Zn(s) stay as → Zn(s) particles Products: 𝑍𝑛𝐶𝑙2 (aq) releases as 𝐻2 (g) stay as → → 𝑍𝑛2+ (aq) + 2𝐶𝑙 − (aq) in solution 𝐻2 (g) molecules The complete equation: 2𝐻 + (𝑎𝑞) + 2𝐶𝑙 − (𝑎𝑞) + 𝑍𝑛 (𝑠) → 𝑍𝑛2+ (𝑎𝑞) + 2𝐶𝑙 − (𝑎𝑞) + 𝐻2 (𝑔)