Unit 4 Chapter 7 Writing Formulas & Naming Compounds

CHAPTER 7 Chemical Formulas and Chemical Compounds

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 — C 8 H 18 aluminum sulfate — Al 2 (SO 4 ) 3 Chapter 7 – Section 1: Chemical Names and Formulas there are 8 carbon atoms in the molecule. there are 18 hydrogen atoms in the molecule. there are 2 aluminum atoms in the formula unit. Parentheses surround the polyatomic ion to identify it as a group. There are 3 SO 4 - groups.

Chemical Formulas Sample Problem Count the number of atoms in the following chemical formulas: Solution: a.Ca(OH) 2 b.KClO 3 c.NH 4 OH d.Fe 2 (CrO 4 ) 3 Chapter 7 – Section 1: Chemical Names and Formulas 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

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.) Chapter 7 – Section 1: Chemical Names and Formulas

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

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

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: Chapter 7 – Section 1: Chemical Names and Formulas Al 2 O 3 KF cationanion aluminum potassium oxide fluoride

Binary Ionic Compounds Sample Problem Write chemical formulas for : a.Magnesium Iodide b.Calcium Oxide Write the correct names for: a.Li 2 S b.ZnCl 2 Chapter 7 – Section 1: Chemical Names and Formulas Solution: Hint: Always divide subscripts by their largest common factor. Lithium Zinc Mg 2+ 2 I –I – 1 Ca 2+ 2 O 2– 2 MgI 2 CaO Lithium Sulfide Zinc Chloride

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

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 The system uses Roman numerals to indicate an ion’s charge. Example:Fe 2+ Fe 3+ Chapter 7 – Section 1: Chemical Names and Formulas iron(II) iron(III) Visual Concept

Stock System Naming Sample Problem A Write the formula and give the name for the compound formed by the ions Cr 3+ 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. Chapter 7 – Section 1: Chemical Names and Formulas Cr 3+ 3 F –F – 1 CrF 3 Chromium (III) Fluoride

2 I –I – O 2– Stock System Naming Sample Problem B Write chemical formulas for : a.Tin (IV) Iodide b.Iron (III) Oxide Write the correct names for: a.VF 3 b.CuO Chapter 7 – Section 1: Chemical Names and Formulas Solution: Hint: “Uncross” subscripts to get the charges of the ions. Vanadium (III) Copper (II) Sn Fe SnI 4 Fe 2 O 3 Fluoride Oxide Be sure to verify the charge of the anion. V 3+ F3F3 - Cu + O - 2

Nomenclature Flowchart Compounds Ionic Molecular Binary Polyatomic Ions Prefix System Acids Stock System Simple (Main Group Elements) Stock System (d-Block Elements) Binary Acids Oxyacids Chapter 7 – Section 1: Chemical Names and Formulas Compounds Ionic 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: Al 2 (SO 4 ) 3 Chapter 7 – Section 1: Chemical Names and Formulas There are 3 sulfate ions in this compound Common Polyatomic Ions Visual Concept

(NO 3 ) 2 Write chemical formulas for : a.Calcium Hydroxide b.Tin (IV) Sulfate Write the correct names for: a.(NH 4 ) 3 PO 4 b.Cu(NO 3 ) 2 Sn 4+ SO 4 2– OH – Polyatomic Ions Sample Problem Chapter 7 – Section 1: Chemical Names and Formulas Solution: Hint: Remember to divide subscripts by their largest common factor. Ammonium Copper(II) Ca Ca(OH) 2 Sn(SO 4 ) 2 Phosphate Nitrate Hint: “Uncross” subscripts to get the charges of the ions. Cu 2+ -

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

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: P 4 O 10 CO Chapter 7 – Section 1: Chemical Names and Formulas tetraphosphorusdecoxide carbonmonoxide

The Prefix System Sample Problem Write chemical formulas for : a.dinitrogen trioxide b.carbon tetrabromide Write the correct names for: a.As 2 S 3 b.PCl 5 Chapter 7 – Section 1: Chemical Names and Formulas Solution: diarsenic phosphorus N2O3N2O3 CBr 4 trisulfide pentachloride

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

Acids An acid is a certain type of molecular compound. All acids start with H (e.g. HCl, H 2 SO 4 ). Acids can be divided into two categories: 1.Binary acids are acids that consist of H and a non-metal. (e.g. HCl.) 2.Oxyacids are acids that contain H and a polyatomic ion that includes O (e.g. H 2 SO 4.) Chapter 7 – Section 1: Chemical Names and Formulas

Binary Acids General rules for naming a binary acid: 1.Begin with the prefix hydro-. 2.Name the anion, but change the ending to –ic. 3.Add acid to the name. Examples: HCl, hydrochloric acid. HBr, hydrobromic acid. H 2 S, hydrosulfuric acid. Chapter 7 – Section 1: Chemical Names and Formulas

Oxyacids General rules for naming an oxyacid : 1.Name the polyatomic ion. 2.Replace -ate with -ic or -ite with -ous 3.Add acid to the name. Examples: H 2 SO 4, sulfuric acid. H 2 SO 3, sulfurous acid. HNO 3, nitric acid. HNO 2, nitrous acid. Chapter 7 – Section 1: Chemical Names and Formulas

sulfate Naming Acids Sample Problem Write the correct name for each of the following: a.HF b.HNO 2 c.H 2 S d.H 2 SO 4 e.H 3 PO 4 Type of Acid: binary acid oxyacid Name: hydro nitrite Chapter 7 – Section 1: Chemical Names and Formulas fluorine ic acid binary acidhydrosulfuric acid ous acid oxyacid uric acid oxyacidphosphate oric acid

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

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. Chapter 7 – Section 2: Oxidation Numbers The 9 Oxidation #’s of Nitrogen

Rules for Assigning Oxidation Numbers 1.The sum of the oxidation numbers for a neutral compound equals zero. 2.The sum of the oxidation numbers for an ion equals the charge of the ion. 3.Atoms in a pure element are zero. 4.The most electronegative element in a compound is assigned a negative number equal to the charge it would have as an anion. 5.Hydrogen is always either +1 or -1. Chapter 7 – Section 2: Oxidation Numbers

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

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. Chapter 7 – Section 2: Oxidation Numbers Prefix systemStock system PCl 3 phosphorus trichloridephosphorus(III) chloride PCl 5 phosphorus pentachloridephosphorus(V) chloride N2ON2Odinitrogen monoxidenitrogen(I) oxide NOnitrogen monoxidenitrogen(II) oxide Mo 2 O 3 dimolybdenum trioxidemolybdenum(III) oxide

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

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, H 2 O: H 2 = 1.0 amu x 2 = 2.0 amu. O = amu amu A compound’s molar mass is numerically equal to its formula mass. Only the units are different. (Ex: Molar mass of H 2 O = 18.0 g.) Chapter 7 – Section 3: Using Chemical Formulas

Molar Masses Sample Problem Determine the molar mass of each of the following compounds: a.Al 2 S 3 a.Ba(OH) 2 Solution: Al 2 = 27.0 x 2 = 54.0 g Chapter 7 – Section 3: Using Chemical Formulas S 3 = 32.1 x 3 = g g Ba = g O 2 = 16.0 x 2 = 32.0 g H 2 = 1.0 x 2 = g g

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 H 2 O? molar mass of H 2 O = 18.0 g/mol Chapter 7 – Section 3: Using Chemical Formulas given conversion factor 2.5 mol H 2 O g H 2 O mol H 2 O 45 g H 2 O = x

2 Molar Mass as a Conversion Factor Sample Problem Calculate the moles in 1170 g of copper (II) nitrate. Solution: Given Conversion factor 1170 g Cu(NO 3 ) 2 mol Cu(NO 3 ) 2 g Cu(NO 3 ) mol Cu(NO 3 ) 2 = x Chapter 7 – Section 3: Using Chemical Formulas Cu = 63.5 g O 6 = 16.0 x 6 = g g (NO 3 ) Cu Determine the correct formula: 2. Calculate the molar mass: - Cu(NO 3 ) 2 N 2 = 14.0 x 2 = 28.0 g 3. Convert from g to mol:

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

Percent Composition Sample Problem Find the percentage composition of copper(I) sulfide, Cu 2 S. 1.Find the molar mass of Cu 2 S: 2.Find the percentage by mass of each element: Solution: Cu 2 = 63.5 x 2 = g Chapter 7 – Section 3: Using Chemical Formulas S = g g % Cu = g g x 100 = 79.8% Cu % S = 32.1 g g x 100 = 20.2% S