Chapter 7 Writing Formulas & Naming Compounds Unit 4 Chapter 7 Writing Formulas & Naming Compounds

Chemical Formulas and Chemical Compounds Chapter 7

Chapter 7 – Section 1: Chemical Names and Formulas Chemical Formulas A chemical formula indicates the number of each kind of atom in a chemical compound. when there is no subscript next to an atom, the subscript is understood to be 1. Examples: octane — C8H18 aluminum sulfate — Al2(SO4)3 there are 18 hydrogen atoms in the molecule. there are 8 carbon atoms in the molecule. Parentheses surround the polyatomic ion to identify it as a group. There are 3 SO4- groups. there are 2 aluminum atoms in the formula unit.

Chemical Formulas Sample Problem Chapter 7 – Section 1: Chemical Names and Formulas Chemical Formulas Sample Problem Count the number of atoms in the following chemical formulas: Solution: Ca(OH)2 KClO3 NH4OH Fe2(CrO4)3 1 Calcium, 2 Oxygens, and 2 Hydrogens 1 Potassium, 1 Chlorine, and 3 Oxygens 1 Nitrogen, 5 Hydrogens, and 1 Oxygen 2 Irons, 3 Chromiums, and 12 Oxygens

Chapter 7 – Section 1: Chemical Names and Formulas Cations Atoms with 1, 2, or 3 valence electrons tend to lose them to form positive ions, which are called a cations.

Chapter 7 – Section 1: Chemical Names and Formulas Anions Atoms with 5, 6, or 7 valence electrons tend to gain more in order to have an octet (8 electrons ) in their outer shell. Gaining extra electrons forms negative ions, called anions.

Chapter 7 – Section 1: Chemical Names and Formulas Monoatomic Ions Monoatomic Ions are ions formed from a single atom. Some main-group elements tend to form covalent bonds instead of ions (ex. C and Si.) +1 +2 +3 -3 -2 -1

Nomenclature Flowchart Chapter 7 – Section 1: Chemical Names and Formulas Nomenclature Flowchart Compounds Compounds Ionic Ionic Molecular Acids Prefix System Polyatomic Ions Stock System Binary Acids Oxyacids Binary Binary Simple (Main Group Elements) Simple (Main Group Elements) Stock System (d-Block Elements)

Writing Binary Ionic Compounds Chapter 7 – Section 1: Chemical Names and Formulas Writing Binary Ionic Compounds Binary Compounds are composed of two elements. Rules for writing binary ionic compounds: Write the symbols for the ions, and their charges. Note: The cation is always written first. Cross over the charges (use the absolute value of each ion’s charge as the subscript for the other ion.) Simplify the numbers and remove the 1’s. Example: aluminum oxide The correct formula for aluminum oxide is Al3+ O2– 2 3 Al2O3

Naming Binary Ionic Compounds Chapter 7 – Section 1: Chemical Names and Formulas Naming Binary Ionic Compounds The name of the cation is given first, followed by the name of the anion. Monatomic cations are identified simply by the element’s name. For monatomic anions, the ending of the element’s name is dropped, and the ending -ide is added. Examples: cation anion Al2O3 aluminum oxide KF potassium fluoride

Binary Ionic Compounds Sample Problem Chapter 7 – Section 1: Chemical Names and Formulas Binary Ionic Compounds Sample Problem Write chemical formulas for : Magnesium Iodide Calcium Oxide Write the correct names for: Li2S ZnCl2 Solution: Mg2+ I – MgI2 1 2 Ca2+ O 2– CaO 2 2 Hint: Always divide subscripts by their largest common factor . Lithium Sulfide Lithium Zinc Chloride Zinc

Nomenclature Flowchart Chapter 7 – Section 1: Chemical Names and Formulas Nomenclature Flowchart Compounds Compounds Ionic Ionic Molecular Acids Prefix System Polyatomic Ions Stock System Binary Acids Oxyacids Binary Binary Simple (Main Group Elements) Stock System (d-Block Elements) Stock System (d-Block Elements)

Chapter 7 – Section 1: Chemical Names and Formulas The Stock System Most d-block elements (transition metals) can form 2 or more ions with different charges. To name ions of these elements, scientists use the Stock system, designed by Alfred Stock in 1919. The system uses Roman numerals to indicate an ion’s charge. Example: Fe2+ Fe3+ iron(II) iron(III) Visual Concept

Stock System Naming Sample Problem A Chapter 7 – Section 1: Chemical Names and Formulas Stock System Naming Sample Problem A Write the formula and give the name for the compound formed by the ions Cr3+ and F–. Solution: Write the ions side by side, cation first. Cross over the charges to give subscripts. Chromium forms more than one ion, so its name must include the charge as a Roman numeral. Cr3+ F – 1 3 CrF3 Chromium (III) Fluoride

Stock System Naming Sample Problem B Chapter 7 – Section 1: Chemical Names and Formulas Stock System Naming Sample Problem B Write chemical formulas for : Tin (IV) Iodide Iron (III) Oxide Write the correct names for: VF3 CuO Solution: Sn4+ I – SnI4 1 4 Fe3+ O 2– Fe2O3 2 3 3+ - V F3 Vanadium (III) Fluoride - 2 + 2 Cu O Copper (II) Oxide Hint: “Uncross” subscripts to get the charges of the ions. Be sure to verify the charge of the anion.

Nomenclature Flowchart Chapter 7 – Section 1: Chemical Names and Formulas Nomenclature Flowchart Compounds Compounds Ionic Ionic Molecular Acids Prefix System Polyatomic Ions Polyatomic Ions Stock System Binary Acids Oxyacids Binary Simple (Main Group Elements) Stock System (d-Block Elements)

Chapter 7 – Section 1: Chemical Names and Formulas Polyatomic Ions Common Polyatomic Ions A polyatomic ion is a charged group of covalently bonded atoms. Common endings are -ate or -ite, but there are exceptions. For more than 1 polyatomic ion, use parentheses with the subscript on the outside. Example: Al2(SO4)3 There are 3 sulfate ions in this compound Visual Concept

Polyatomic Ions Sample Problem Chapter 7 – Section 1: Chemical Names and Formulas Polyatomic Ions Sample Problem Write chemical formulas for : Calcium Hydroxide Tin (IV) Sulfate Write the correct names for: (NH4)3 PO4 Cu(NO3)2 Solution: Ca2+ OH – Ca(OH)2 1 2 Sn4+ SO4 2– Sn(SO4)2 2 4 Hint: Remember to divide subscripts by their largest common factor . Ammonium Phosphate - 2+ Cu (NO3)2 Copper(II) Nitrate Hint: “Uncross” subscripts to get the charges of the ions.

Nomenclature Flowchart Chapter 7 – Section 1: Chemical Names and Formulas Nomenclature Flowchart Compounds Compounds Ionic Molecular Molecular Acids Prefix System Prefix System Polyatomic Ions Stock System Binary Acids Oxyacids Binary Simple (Main Group Elements) Stock System (d-Block Elements)

Chapter 7 – Section 1: Chemical Names and Formulas The Prefix System Molecular compounds are composed of covalently-bonded molecules. The old prefix system is still used for molecular compounds. Name the prefix, then the element. Anions end in -ide. The prefix mono- usually isn’t used for cations. Examples: P4O10 CO tetraphosphorus decoxide carbon monoxide

The Prefix System Sample Problem Chapter 7 – Section 1: Chemical Names and Formulas The Prefix System Sample Problem Write chemical formulas for : dinitrogen trioxide carbon tetrabromide Write the correct names for: As2S3 PCl5 Solution: N2O3 CBr4 diarsenic trisulfide phosphorus pentachloride

Nomenclature Flowchart Chapter 7 – Section 1: Chemical Names and Formulas Nomenclature Flowchart Compounds Compounds Ionic Molecular Acids Acids Prefix System Stock System Binary Acids Binary Acids Polyatomic Ions Oxyacids Oxyacids Binary Simple (Main Group Elements) Stock System (d-Block Elements)

Chapter 7 – Section 1: Chemical Names and Formulas Acids An acid is a certain type of molecular compound. All acids start with H (e.g. HCl, H2SO4). Acids can be divided into two categories: Binary acids are acids that consist of H and a non-metal. (e.g. HCl.) Oxyacids are acids that contain H and a polyatomic ion that includes O (e.g. H2SO4.)

Chapter 7 – Section 1: Chemical Names and Formulas Binary Acids General rules for naming a binary acid: Begin with the prefix hydro-. Name the anion, but change the ending to –ic. Add acid to the name. Examples: HCl, hydrochloric acid. HBr, hydrobromic acid. H2S, hydrosulfuric acid.

Chapter 7 – Section 1: Chemical Names and Formulas Oxyacids General rules for naming an oxyacid : Name the polyatomic ion. Replace -ate with -ic or -ite with -ous Add acid to the name. Examples: H2SO4, sulfuric acid. H2SO3, sulfurous acid. HNO3, nitric acid. HNO2, nitrous acid.

Naming Acids Sample Problem Chapter 7 – Section 1: Chemical Names and Formulas Naming Acids Sample Problem Write the correct name for each of the following: HF HNO2 H2S H2SO4 H3PO4 Type of Acid: Name: binary acid hydro fluorine ic acid oxyacid nitrite ous acid binary acid hydro sulfur ic acid oxyacid sulfate uric acid oxyacid phosphate oric acid

Nomenclature Flowchart Chapter 7 – Section 1: Chemical Names and Formulas Nomenclature Flowchart Compounds Compounds Ionic Molecular Molecular Acids Prefix System Stock System Stock System Binary Acids Polyatomic Ions Oxyacids Binary Simple (Main Group Elements) Stock System (d-Block Elements)

Chapter 7 – Section 2: Oxidation Numbers In order to indicate the general distribution of electrons among covalently bonded atoms, oxidation numbers are assigned to the atoms. Unlike ionic charges, oxidation #’s do not represent actual electrons gained or lost. Many elements can have different oxidation #’s depending on what they’re combined with. The 9 Oxidation #’s of Nitrogen

Rules for Assigning Oxidation Numbers Chapter 7 – Section 2: Oxidation Numbers Rules for Assigning Oxidation Numbers The sum of the oxidation numbers for a neutral compound equals zero. The sum of the oxidation numbers for an ion equals the charge of the ion. Atoms in a pure element are zero. The most electronegative element in a compound is assigned a negative number equal to the charge it would have as an anion. Hydrogen is always either +1 or -1.

Assigning Oxidation Numbers Sample Problem Chapter 7 – Section 2: Oxidation Numbers Assigning Oxidation Numbers Sample Problem Assign oxidation numbers to each atom in the following compounds or ions: UF6 H2SO4 c. ClO3- Solution: 6+ 1- -1 x 6 = -6 F = -1, U = +6 + 6 1+ 6+ 2- -2 x 4 = -8 O = -2, H = +1, S = +6 +1 x 2 = +2 + 6 5+ 2- O = -2, Cl = +5 -2 x 3 = -6 + 5 -1

Using Oxidation Numbers in Naming Chapter 7 – Section 2: Oxidation Numbers Using Oxidation Numbers in Naming The Stock System is actually based on oxidation numbers. It can be used as an alternative to the prefix system for naming molecular compounds. Prefix system Stock system PCl3 phosphorus trichloride phosphorus(III) chloride PCl5 phosphorus pentachloride phosphorus(V) chloride N2O dinitrogen monoxide nitrogen(I) oxide NO nitrogen monoxide nitrogen(II) oxide Mo2O3 dimolybdenum trioxide molybdenum(III) oxide

Using Oxidation Numbers in Naming Sample Problem Chapter 7 – Section 2: Oxidation Numbers Using Oxidation Numbers in Naming Sample Problem Write the correct prefix system name and the correct Stock system name for each of the following: As2S3 SO3 Prefix System: Stock System: 3+ 2- diarsenic trisulfide arsenic (III) sulfide -2 x 3 = -6 +3 x 2 = +6 + 6+ 2- sulfur trioxide sulfur (VI) oxide -2 x 3 = -6 +6 +

Chapter 7 – Section 3: Using Chemical Formulas Formula Masses The formula mass of any compound is the sum of the masses of all the atoms in its formula. example: Formula mass of water, H2O: H2 = 1.0 amu x 2 = 2.0 amu. O = + 16.0 amu. 18.0 amu A compound’s molar mass is numerically equal to its formula mass. Only the units are different. (Ex: Molar mass of H2O = 18.0 g.)

Molar Masses Sample Problem Chapter 7 – Section 3: Using Chemical Formulas Molar Masses Sample Problem Determine the molar mass of each of the following compounds: Al2S3 Ba(OH)2 Solution: Al2 = 27.0 x 2 = 54.0 g S3 = 32.1 x 3 = + 96.3 g 150.3 g Ba = 137.3 g O2 = 16.0 x 2 = 32.0 g H2 = 1.0 x 2 = + 2.0 g 171.3 g

Molar Mass as a Conversion Factor Chapter 7 – Section 3: Using Chemical Formulas Molar Mass as a Conversion Factor The molar mass of a compound can be used as a conversion factor to convert between moles and grams for a given substance. Example: What is the mass of 2.5 moles of H2O? molar mass of H2O = 18.0 g/mol conversion factor given 18.0 g H2O 2.5 mol H2O x = 45 g H2O 1 mol H2O

Molar Mass as a Conversion Factor Sample Problem Chapter 7 – Section 3: Using Chemical Formulas Molar Mass as a Conversion Factor Sample Problem Calculate the moles in 1170 g of copper (II) nitrate. Solution: 2+ - Cu (NO3) Cu(NO3)2 1. Determine the correct formula: 2 2. Calculate the molar mass: Cu = 63.5 g N2 = 14.0 x 2 = 28.0 g 3. Convert from g to mol: O6 = 16.0 x 6 = + 96.0 g Conversion factor 187.5 g Given 1 mol Cu(NO3)2 1170 g Cu(NO3)2 x = 6.24 mol Cu(NO3)2 187.5 g Cu(NO3)2

Chapter 7 – Section 3: Using Chemical Formulas Percent Composition The percentage by mass of each element in a compound is known as the percent composition of the compound. mass of element in compound % of element = x 100 molar mass of compound Visual Concept

Percent Composition Sample Problem Chapter 7 – Section 3: Using Chemical Formulas Percent Composition Sample Problem Find the percentage composition of copper(I) sulfide, Cu2S. Find the molar mass of Cu2S: Find the percentage by mass of each element: Cu2 = 63.5 x 2 = 127.0 g Solution: S = + 32.1 g 159.1 g 127.0 g % Cu = x 100 = 79.8% Cu 159.1 g 32.1 g % S = x 100 = 20.2% S 159.1 g