1 / 33

A. Naming Elements and Compounds 1. Polyatomic Molecular Elements

Chem 20 Review. A. Naming Elements and Compounds 1. Polyatomic Molecular Elements. elements that exist in form instead of form. molecular. atomic. halogens, hydrogen, phosphorus, sulfur, oxygen, nitrogen. ***SHPON (8, 2, 4, 2, 2). 2. Molecular Compounds.

sophia-parsons
Download Presentation

A. Naming Elements and Compounds 1. Polyatomic Molecular Elements

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chem 20 Review A. Naming Elements and Compounds 1. Polyatomic Molecular Elements • elements that exist in form instead of form molecular atomic • halogens, hydrogen, phosphorus, sulfur, oxygen, nitrogen ***SHPON (8, 2, 4, 2, 2)

  2. 2. Molecular Compounds non-metals only • use prefixes to indicate the number of atoms in the molecule eg) carbon dioxide = dinitrogen monoxide = CCl4 = CO2 N2O carbon tetrachloride • there are many molecular compounds that still use their common names or IUPAC names (organic molecules) eg) hydrogen peroxide, glucose, ammonia, sucrose

  3. 3. Acids • see data booklet and Nelson pg 553 Rules 1. hydrogen becomes acid 2. hydrogen becomes acid 3. hydrogen becomes acid _____ide hydro____ic _____ate _______ic _____ite ______ous

  4. Try These: 1. hydrogen iodide = 2. hydrogen phosphate = 3. hydrogen nitrite = 4. hydrogen sulphite = hydroiodic acid phosphoric acid nitrous acid sulphurous acid

  5. 4. Ionic Compounds • name in full then name the with or use cation (positive ion) anion (negative ion) the “ide” ending the polyatomic ion name eg) NaCl = Mg3(PO4)2 = sodium chloride magnesium phosphate when writing formulas, look up the symbol for each ion then balance the charges using subscripts

  6. hydrated ionic compounds contain water in their atomic structures • indicated by where x is the number of water molecules “xH2O” eg) CuSO45H2O = copper (II) sulphate pentahydrate

  7. Your Assignment: pgs 1-3 in workbook

  8. B. Bonding • recall types of intermolecular forces • London Dispersion – occurs between all molecules and is the attraction of the electrons in one molecule to the protons in another molecule

  9. Dipole-Dipole – occurs only between polar molecules and is the electrostatic attraction of the polar ends of molecules • Hydrogen Bonding – occurs only when H is bonded to O, F or N and is the attraction of the H to the O, F or N in another molecule

  10. C. Chemical Reactions 1. Classification of Reactions • clues to a chemical reaction: 1. colour change 2. gas produced 3. energy change 4. precipitate formed • conservation laws: 1. mass - number and kind of atom (balancing) 2. energy (1st Law of Thermo)

  11. types of reactions: 1. composition/formation 2. decomposition 3. single replacement 4. double replacement 5. hydrocarbon combustion 6. other

  12. 2. Energy in Chemical Reactions • endothermic = energy is absorbed • exothermic = energy is released • bond energy = energy released when bonds are formed or energy absorbed when bonds are broken

  13. 3. States of Matter • gives the states of elements at room temperature periodic table • all ionic compounds at room temperature (by themselves) are solid • molecular compounds can be at room temperature solids, liquids or gases • acids are assumed to be aqueous • ionic compounds may be when mixed with water aqueous or solid

  14. Your Assignment: pgs 4-5 in workbook

  15. D. Significant Digits • represents the degree of accuracy of using measured values • two different rules are used: 1. Addition/Subtraction: add or subtract then round to the lowest number of decimal places 2. Multiplication/Division: multiply or divide then round to the lowest number of sig digs

  16. E. Solutions and Gases 1. Preparation of Solutions • concentration is most commonly measured in mol/L • mol/L can also be expressed as “molarity” or M eg) 0.300 mol/L = 0.300 M • solubility = the concentration of solute in a saturated solution at a given temperature • solubility is measured in g/100 mL

  17. molar solubility is measured in mol/L • to determine the solubility of an ionic compound, check the solubility table in the Data Booklet

  18. Steps for Solution Preparation 1. Calculate the mass of the solute required to achieve a specific concentration and volume. 2. Measure mass. 3. Dissolve the solute in half of the volume of solvent. 4. Transfer solution to a volumetric flask. 5. Bring solution up to final volume and mix by inverting.

  19. 2. Dilution of Solutions • when a solution is diluted, only the amount of is increased solvent (usually water) • the remains constant number of moles ViCi = VfCf

  20. Your Assignment: 1. pgs 6-8 in workbook

  21. 3. Types of Solutes • nonelectrolytes: substances that dissolve to yield solutions that do not conduct electricity eg) molecular compounds • electrolytes: substances that dissolve to yield solutions that conduct electricity eg) ionic compounds and acids • dissociation of solutes in water: 1. electrolytes: 2. non-electrolytes: Na2CO3(s) 2Na+(aq) + CO32-(aq) C6H12O6(s) C6H12O6(aq)

  22. 4. Determining Ionic Concentrations • you can use the concentration of a solute to determine the ion concentrations once it is dissolved in water (dissociated) Steps 1. Write a balanced dissociation equation. 2. Write down concentration given. 3. Determine the ion concentrations using the mole ratio.

  23. Example What is the concentration of each ion in a 0.23 mol/L solution of aluminum sulphate? 1 Al2(SO4)3(s) 2 Al3+(aq) + 3 SO42-(aq) C = 0.23 mol/L C = 0.23 mol/L x 2 1 C = 0.23 mol/L x 3 1 = 0.46 mol/L = 0.69 mol/L

  24. Your Assignment: pg 9 in workbook

  25. 5. Non Ionic vs. Net Ionic Equations • net ionic reactions are used to show only the reacting ions…spectator ions (non-reacting) are omitted • write the non-ionic reaction, the total ionic reaction and the net ionic reaction

  26. Example What is the net ionic reaction for the reaction of bromine and sodium iodide? Non Ionic: Total Ionic: Net Ionic: Br2(l) + NaI(aq) 2 I2(s) + NaBr(aq) 2 Br2(l) + Na+(aq) 2 + I–(aq)  2 I2(s) + Na+(aq) 2 + Br–(aq) 2 Br2(l) + 2I–(aq)  I2(s) + 2Br–(aq)

  27. Your Assignment: pg 10 in workbook

  28. 6. Gases • Ideal Gas Law: PV = nRT • STP = 273.15 K, 101.325 kPa, 22.4 L/mol • SATP = 298.15 K, 100 kPa, 24.8 L/mol • conversion to Kelvin = 273.15 + xC

  29. F. Stoichiometry Steps 1. Write a balanced chemical (or net ionic) equation. 2. Write down the given information. 3. Find moles of given using n = m, C = n , PV = nRT M V 4. Find moles of wanted using the mole ratio: wanted/given. 5. Find answer to question.

  30. Example 1 If 5.00 g of sodium reacts with excess chlorine gas, how much sodium chloride is produced? 2Na(s) + Cl2(g) 2 NaCl(aq) m = 5.00 g M = 22.99 g/mol x g M = 58.44 g/mol n = m M = 5.00g 22.99 g/mol = 0.217… mol n = 0.217… mol x 2 2 m = nM = (0.217… mol)(58.44 g/mol) = 12.7 g

  31. Example 2 Liquid bromine is added to 500 mL of a solution containing 0.150 mol/L iodide ion. What mass of iodine will be produced? 2I-(aq) + Br2(l) 2Br-(aq) + I2(s) V = 500 mL = 0.500 L C = 0.150 mol/L x g M = 253.80 g/mol n = 0.0750 mol x 1 2 = 0.0375 mol n = CV = (0.150 mol/L)(0.500 L) = 0.0750 mol m = nM = (0.0375mol)(253.80 g/mol) = 9.52 g

  32. Example 3 If 5.0 g of sodium reacts with 5.0 g of chlorine, how much sodium chloride is produced? 2Na(s) + Cl2(g) 2NaCl(aq) x g m = 5.0 g M = 22.99 g/mol m = 5.0 g M = 70.90 g/mol M = 58.44 g/mol n = m M = 5.0 g 22.99 g/mol = 0.217…mol n = m M = 5.0 g 70.90 g/mol = 0.0705…mol n =0.0705 molx 2 1 = 0.141…mol m = nM = (0.141..mol)(58.44 g/mol) = 8.2 g  2 = 0.108…mol  1 = 0.0705…mol Cl2(g) is limiting

  33. Your Assignment: pgs 11-13 in workbook

More Related