Ch 7.1 Forming Ions.

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Presentation transcript:

Ch 7.1 Forming Ions

Review… Cations are Groups 1A, 2A, and 3A They have positive charges. Anions are Groups 5A, 6A, and 7A They have negative charges They end in “ide” Majority of elements in Groups 4A and 8A do not usually form ions. These are called monatomic ions!

Ions of Transition Metals Some transition metals in Groups 1B-8B form more than one cation with different charges. Examples: Iron forms Fe2+ and Fe3+ Two methods of naming: Stock System: Iron (II) ion and Iron (III) ion Classical name: Ferrous ion and Ferric ion *We will use the stock system.

Common Metal Ions with More than One Ionic Charge Symbol Stock Name Classical Name Cu+ Copper(I) ion Cuprous ion Cu2+ Copper(II) ion Cupric ion Fe2+ Iron(II) ion Ferrous ion Fe3+ Iron(III) ion Ferric ion *Hg22+ Mercury(I) ion Mercurous ion Hg2+ Mercury(II) ion Mercuric ion Pb2+ Lead(II) ion Plumbous ion Pb4+ Lead(IV) ion Plumbic ion

Common Metal Ions with More than One Ionic Charge Symbol Stock Name Classical Name Sn2+ Tin(II) ion Stannous ion Sn4+ Tin(IV) ion Stannic ion Cr2+ Chromium(II) ion Chromous ion Cr3+ Chromium(III) ion Chromic ion Mn2+ Manganese(II) ion Manganous ion Mn3+ Manganese(III) ion Manganic ion Co2+ Cobalt(II) ion Cobaltous ion Co3+ Cobalt(III) ion Cobaltic ion

Other Transition Metal Ions Some transition metals have only one charge and do not use a Roman numeral. Examples (Write on your periodic table!) Silver: Ag+ Cadmium: Cd2+ Zinc: Zn2+

Polyatomic Ions The names of most polyatomic anions end in “-ite” or “-ate” The “-ite” ending indicates one less oxygen atom than the “-ate” ending Remember to use parentheses if more than one is needed.

Writing a Formula Write the formula that will form between Ba and Cl Solution: 1. Write the positive ion of metal first, and then negative ion Ba2+ Cl 2. Do the charges equal zero? NO!! 3. Use Criss-Cross method – write subscripts Ba2+ Cl1 BaCl2 8

Learning Check Write the correct formula for the compounds containing the following ions: 1. Na+, S2- 2. Al3+, Cl- 3. Mg2+, N3- 4. Al3+, S2- 9

Solution 1. Na+, S2- Na2S 2. Al3+, Cl- AlCl3 3. Mg2+, N3- Mg3N2 4. Al3+, S2- Al2S3 10

Binary Compound: is composed of two elements and can be either ionic or molecular.

Naming Binary Ionic Compounds To name a binary ionic compound, place the cation name first, followed by the anion name. Remember the anion ends in “-ide” Examples: Cs2O NaBr CuO Cesium Oxide Sodium Bromide Copper(II) Oxide

Naming Compounds with Polyatomic Ions State the cation first and then the anion just as you did in naming binary ionic compounds. KNO3 Mg(ClO2)2 Potassium Nitrate Magnesium Chlorite

Naming Binary Molecular Compounds Binary molecular compound: must be composed of two nonmetals Use prefixes to indicate the number and kind of atom in the compound Use the following general format: 1st name: prefix + element name 2nd name: prefix + element name + “ide” If there is only 1 of the 1st element, no prefix.

Prefixes in Covalent Compounds pg 228 Number of atoms Prefix 1 mono- 6 hexa- 2 di- 7 hepta- 3 tri- 8 octa- 4 tetra- 9 nona- 5 penta- 10 deca-

Anyone want a cold glass of dihydrogen monoxide?

Examples Name the following CO Carbon Monoxide CO2 N2O Carbon Dioxide Cl2O8 Carbon Monoxide Carbon Dioxide Dinitrogen Monoxide Dichlorine Octoxide

Writing Formulas for Binary Molecular Compounds Use the prefixes in the name to tell you the subscript of each element in the formula. Then write the correct symbols for the two elements with the appropriate subscripts. The least electronegative element is written first Dinitrogen Tetroxide - N2O4

Examples Write formulas for the following: Nitrogen Monoxide Carbon Tetrachloride Diphosphorous Pentoxide NO CCl4 P2O5

Naming Acids An acid is a compound that contains one or more hydrogen atoms and produces hydrogen ions (H+) when dissolved in water. All acids begin with hydrogen General Format: HnX “X” represents a monatomic or polyatomic anion. “n” represents the number of hydrogen ions

3 Rules for Naming Common Acids If the name of “X” ends in -ate: ____________-ic acid If the name of “X” ends in -ite: ____________-ous acid If the name of “X” ends in -ide: hydro-__________-ic acid

Name these acids H2SO4 HCl H2S HNO3 HClO2 Sulfuric Acid Hydrochloric Acid Hydrosulfuric Acid Nitric Acid Chlorous Acid

Writing Formulas for Acids If the acid ends in –ic, then “X” ends in –ate If the acid ends in –ous, then “X” ends in –ite If the acid has hydro-______-ic, then “X” ends in –ide. The subscript on hydrogen is equal to the charge of “X”.

Write the Formula for the Following Acids Hydrobromic Acid Carbonic Acid Phosphoric Acid Sulfurous Acid HBr H2CO3 H3PO4 H2SO3

Homework 7.1 pg 251 #14, 15(a-d), 41

Ch 7.3 Using Chemical Formulas

The Mass of a Mole of an Element Molar Mass: is the atomic mass of an element expressed in grams/mole (g/mol). Carbon = 12.01 g/mol Hydrogen = 1.01 g/mol When dealing with molar mass, round off to two decimals. 12.011 g/mol  12.01 g/mol

The Mass of a Mole of a Compound You calculate the mass of a molecule by adding up the molar masses of the atoms making up the molecules. Example: H2O H = 1.01 g x 2 atoms = 2.02 g/mol O = 16.00 g x 1 atom = 16.00 g/mol Molar Mass of H2O = 2.02 g/mol + 16.00 g/mol = 18.02 g/mol This applies to both molecular and ionic compounds

Find the molar mass of PCl3 P = 30.97 g x 1 atom = 30.97 g/mol Cl = 35.45 g x 3 atoms = 106.35 g/mol PCl3 = 30.97 g + 106.35 g = 137.32 g/mol What is the molar mass of Sodium Hydrogen Carbonate (NaHCO3) ? Na = 22.99 g x 1 atom = 22.99 g/mol H = 1.01 g x 1 atom = 1.01 g/mol C = 12.01 g x 1 atom = 12.01 g/mol O = 16.00 g x 3 atoms = 48.00 g/mol NaHCO3 = 22.99 + 1.01 + 12.01 + 48.00 = 84.01 g/mol 137.5 g/mol 84.0 g NaHCO3

Converting Moles to Mass You can use the molar mass of an element or compound to convert between the mass of a substance and the moles of a substance. Mass (g) = # of moles x mass (g) 1 mole Example: If molar mass of NaCl is 58.44 g/mol, what is the mass of 3.00 mol NaCl? Mass of NaCl = 3.00 mol x 58.44g = 1 mol 175 g NaCl

Example 2: Moles to Mass What is the mass of 9.45 mol of Aluminum Oxide (Al2O3)? Find molar mass of Al2O3 = 101.96 g/mol Mass = 9.45 mol Al2O3 x 101.96 g Al2O3 1 mol Al2O3 = 964 g Al2O3

Converting Mass to Moles You can invert the conversion factor to find moles when given the mass. Moles = mass (g) x 1 mole mass (g) Example: If molar mass of Na2SO4 142.05 g/mol, how many moles is 10.0 g of Na2SO4? Moles of Na2SO4 = 10.0 g x 1 mol = 142.05 g = 0.0704 mol Na2SO4

Example 2: Mass to Moles How many moles are in 75.0 g of Dinitrogen Trioxide? Find molar mass of N2O3 = 76.02 g/mol Moles = 75.0 g N2O3 x 1 mole = 76.02 g N2O3 0.987 mol N2O3

Percent Composition Percent Composition: the relative amount of the elements in a compound. Also known as the percent by mass It can be calculated in two ways: Using Mass Data Using the Chemical Formula % mass of element= mass of element x100% mass of compound

Example When a 13.60 g sample of a compound containing Mg and O is decomposed, 5.40 g O is obtained. What is the % composition of this compound? Mass of compound: 13.60 g Mass of oxygen: 5.40 g O Mass of magnesium: 13.60 g - 5.40 g = 8.20 g Mg % Mg = 8.20 g Mg x 100% = 13.60 g % O = 5.40 g O x 100% = 60.3% 39.7%

Find the percent composition of Cu2S. Find mass of Cu and S Cu = 63.55 x 2 = 127.10 g S = 32.07 g Find mass of Cu2S 127.10 g + 32.07 g = 159.17 g % Composition Cu = 127.10 g x 100% = 159.17 g S = 32.07 g x 100% = 79.85% 20.15%

Find the percentage by mass of water in the hydrate Na2CO310H2O Find the mass of 10H2O H2O = 10 mol x 18.02 g/mol = 180.2 g Find the mass of Na2CO310H2O Na2CO3 = 1 mol x 105.99 g/mol =105.99 g 105.99 g + 180.2 g = 286.2 g % By Mass of Water H2O = 180.2 g H2O x 100% = 286.2 g Na2CO310H2O 62.96%

Homework 7.3 pg 253 #30-33

Ch 7.4 Determining Chemical Formulas

Empirical Formulas Empirical Formula: shows the smallest whole-number ratio of the atoms of the elements in a compound. Example: The Empirical Formula for Hydrogen Peroxide (H2O2) is HO with a 1:1 ratio. The Empirical Formula for Carbon Dioxide (CO2) is CO2 with a 1:2 ratio.

Determining the Empirical Formula of a Compound A compound is found to contain 25.9% Nitrogen and 74.1% Oxygen. What is the Empirical Formula of the compound? 25.9 g N x 1 mol N = 14.01 g N 74.1 g O x 1 mol O = 16.00 g O N1.85O4.63 = N1O2.5 = N2O5 1.85 mol N 4.63 mol O

Molecular Formulas Molecular Formula: tells the actual number of each kind of atom present in a molecule of a compound Example: The Molecular Formula for Hydrogen Peroxide is H2O2. The Molecular Formula for Carbon Dioxide is CO2 It is possible to find the Molecular Formula using the Empirical Formula if you know the molar mass of the compound.

Finding the Molecular Formula Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N Step 1: Find the empirical formula molar mass 12.01 + (4 x 1.01) + 14.01 = 30.06 g/mol Step 2: Divide molar mass by EF molar mass 60.0 g/mol = 1.99  2 30.06 g/mol Step 3: Multiply empirical formula by 2 CH4N x 2 = C2H8N2

Homework 7.4 pg 253 #36-38