Upcoming SlideShare
×

# Ch 1 06 the arithmetic of chemistry

1,908 views

Published on

1 Like
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

Views
Total views
1,908
On SlideShare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
48
0
Likes
1
Embeds 0
No embeds

No notes for slide

### Ch 1 06 the arithmetic of chemistry

1. 1. The Arithmetic of Chemistry Stoichiometry and Chemical Equations
2. 2. 1. FORMULA: Potassium bromide2. FORMULA: Iron (III) oxide3. NAME: BaCl24. NAME: (NH4)2S5. What IFA exist/s between ammonia and acetone?
3. 3. Micro World Macro World atoms & molecules gramsAtomic mass is the mass of an atom inatomic mass units (amu) By definition: 1 atom 12C “weighs” 12 amu On this scale 1H = 1.008 amu 16O = 16.00 amu 3.1
4. 4. Natural lithium is: 7.42% 6Li (6.015 amu) 92.58% 7Li (7.016 amu) Average atomic mass of lithium: 7.42 x 6.015 + 92.58 x 7.016 = 6.941 amu 100 3.1
5. 5. Average atomic mass (6.941)
6. 6. The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12C 1 mol = NA = 6.0221367 x 1023 Avogadro’s number (NA) 3.2
7. 7. The MoleTherefore: 1.00 mole of water = 6.02 x 1023 molecules of H2O 1.00 mole of NaCl = 6.02 x 1023 Na+ ions = 6.02 x 1023 Cl- ions
8. 8. eggsMolar mass is the mass of 1 mole of shoes in grams marbles atoms 1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g 1 12C atom = 12.00 amu 1 mole 12C atoms = 12.00 g 12C 1 mole lithium atoms = 6.941 g of Li For any element atomic mass (amu) = molar mass (grams) 3.2
9. 9. Mole to Mass RelationshipsBalanced chemical reaction gives molar ratios
10. 10. Amadeo Avogadro Jean Baptiste Perrin (1776-1856) ( 1870-1942) Jean Perrin originated the term “Avogadro’s Number” in 1909. Perrin won the Nobel Physics Prize in 1926
11. 11. One Mole of:C S HgCu Fe 3.2
12. 12. 1 12C atom 12.00 g 1.66 x 10-24 g x 23 12C atoms =12.00 amu 6.022 x 10 1 amu 1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu M = molar mass in g/mol NA = Avogadro’s number 3.2
13. 13. Do You Understand Molar Mass? How many atoms are in 0.551 g of potassium (K) ? 1 mol K = 39.10 g K 1 mol K = 6.022 x 1023 atoms K 1 mol K 6.022 x 1023 atoms K0.551 g K x x = 39.10 g K 1 mol K 8.49 x 1021 atoms K 3.2
14. 14. Molecular mass (or molecular weight) is the sum ofthe atomic masses (in amu) in a molecule. 1S 32.07 amu 2O + 2 x 16.00 amu SO2 SO2 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2 3.3
15. 15. Do You Understand Molecular Mass? How many H atoms are in 72.5 g of C3H8O ?1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O 1 mol C3H8O molecules = 8 mol H atoms 1 mol H = 6.022 x 1023 atoms H 1 mol C3H8O 8 mol H atoms 6.022 x 1023 H atoms72.5 g C3H8O x x x = 60 g C3H8O 1 mol C3H8O 1 mol H atoms 5.82 x 1024 atoms H 3.3
16. 16. Percent composition of an element in a compound = n x molar mass of element x 100% molar mass of compound n is the number of moles of the element in 1 mole of the compound 2 x (12.01 g) %C = x 100% = 52.14% 46.07 g 6 x (1.008 g) %H = x 100% = 13.13% 46.07 g 1 x (16.00 g) %O = x 100% = 34.73% 46.07 g C2H6O 52.14% + 13.13% + 34.73% = 100.0% 3.5
17. 17. A process in which one or more substances is changed into oneor more new substances is a chemical reactionA chemical equation uses chemical symbols to show whathappens during a chemical reaction3 ways of representing the reaction of H2 with O2 to form H2O reactants products 3.7
18. 18. How to “Read” Chemical Equations 2 Mg + O2 2 MgO2 atoms Mg + 1 molecule O2 makes 2 formula units MgO 2 moles Mg + 1 mole O2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO IS NOT 2 grams Mg + 1 gram O2 makes 2 g MgO 3.7
19. 19. Balancing Chemical Equations Must have the same number of each element on both sides of the equation
20. 20. Recall: The Laws of ChemicalCombination  Law of Conservation of Matter Matter is neither created nor destroyed in a chemical reaction.  Law of Definite Composition Elements combine in fixed amounts in a compound.  Law of Multiple Proportion If two elements A and B combine to form more than one compound, the masses of B that react with a given mass of A are in ratio of small whole numbers.
21. 21. Recall: Writing the Formulaof Compounds  Formulas for molecules: H2O, CO2, CH4, C6H12O6  Formulas for ions (charged particles): OH- , NO3- , PO43- , H+, NH4+  Formulas for ionic compounds NaOH, NH4NO3, FeCl3
22. 22. Balancing Chemical Equations1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.Ethane reacts with oxygen to form carbon dioxide and water C2H6 + O2 CO2 + H2O2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 2C2H6 NOT C4H12 3.7
23. 23. Balancing Chemical Equations 3. Start by balancing those elements that appear in only one reactant and one product. C2H6 + O2 CO2 + H2O start with C or H but not O2 carbon 1 carbon multiply CO2 by 2 on left on right C2H6 + O2 2CO2 + H2O 6 hydrogen 2 hydrogen multiply H2O by 3 on left on right C2H6 + O2 2CO2 + 3H2O 3.7
24. 24. Balancing Chemical Equations4. Balance those elements that appear in two or more reactants or products. C2H6 + O2 2CO2 + 3H2O multiply O2 by 7 2 2 oxygen 4 oxygen + 3 oxygen = 7 oxygen on left (2x2) (3x1) on right C2H6 + 7 O2 remove fraction 2CO2 + 3H2O 2 multiply both sides by 2 2C2H6 + 7O2 4CO2 + 6H2O 3.7
25. 25. Balancing Chemical Equations5. Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C2H6 + 7O2 4CO2 + 6H2O 12C (2 xx2) 14 H (2 6) 4 O (7 2) 1412 H (6 2 + 6) O (4 C x 2) 4x Reactants Products 4C 4C 12 H 12 H 14 O 14 O 3.7
26. 26. Mass Changes in Chemical Reactions1. Write balanced chemical equation2. Convert quantities of known substances into moles3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity4. Convert moles of sought quantity into desired units 3.8
27. 27. Methanol burns in air according to the equation 2CH3OH + 3O2 2CO2 + 4H2O If 209 g of methanol are used up in the combustion, what mass of water is produced?grams CH3OH moles CH3OH moles H2O grams H2O molar mass coefficients molar mass CH3OH chemical equation H2O 1 mol CH3OH 4 mol H2O 18.0 g H2O 209 g CH3OH x x x = 32.0 g CH3OH 2 mol CH3OH 1 mol H2O 235 g H2O 3.8
28. 28. Limiting Reagents 6 green used up red left over 3.9
29. 29. Assembling a Handout
30. 30. Assembling a Handout
31. 31. Assembling a HandoutWhat is the limiting page?
32. 32. Assembling a HandoutWhat is the limiting page?
33. 33. Assembling a HandoutWhat is the limiting page?
34. 34. Do You Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe2O3 2Al + Fe2O3 Al2O3 + 2Fe Calculate the mass of Al2O3 formed.g Al mol Al mol Fe2O3 needed g Fe2O3 needed ORg Fe2O3 mol Fe2O3 mol Al needed g Al needed 1 mol Al 1 mol Fe2O3 160. g Fe2O3 124 g Al x x x = 367 g Fe2O3 27.0 g Al 2 mol Al 1 mol Fe2O3 Start with 124 g Al need 367 g Fe2O3 Have more Fe2O3 (601 g) so Al is limiting reagent 3.9
35. 35. Use limiting reagent (Al) to calculate amount of product that can be formed. g Al mol Al mol Al2O3 g Al2O3 2Al + Fe2O3 Al2O3 + 2Fe 1 mol Al 1 mol Al2O3 102. g Al2O3124 g Al x x x = 234 g Al2O3 27.0 g Al 2 mol Al 1 mol Al2O3 3.9
36. 36. Theoretical Yield is the amount of product that wouldresult if all the limiting reagent reacted.Actual Yield is the amount of product actually obtainedfrom a reaction. Actual Yield % Yield = x 100 Theoretical Yield 3.10
37. 37. Test yourself: Limiting reagent In the Haber process, ammonia (NH3) is produced from hydrogen gas and nitrogen gas. If I have 40.0g hydrogen gas and 30.0g nitrogen gas, what will be the mass of ammonia I produce? What amount (in moles) of reactants will be used up?
38. 38. Some General Types of Simple Chemical Reactions1. Combination/Addition Reactions2. Decomposition Reactions3. Single replacement/displacement/substitution4. Double replacement/displacement5. Neutralization6. Combustion
39. 39. 1. Combination Reactions Two or more substances (elements or compounds) react to produce one compound Also known as “synthesis” or “addition” A + Z  AZ  Metal oxygen metal oxide  Nonmetal oxygen nonmetal oxide  Metal nonmetal salt  Water metal oxide base  Metal oxide nonmetal oxide salt
40. 40. Test Yourself: Combination1. Magnesium metal and oxygen react to form a metal oxide. Balance the equation.2. Sulfur trioxide and sulfur dioxide can be formed (in separate reactions) from sulfur and oxygen gas.3. Carbon and oxygen combine to form two different species, depending on the amount of oxygen.4. Water and sulfur trioxide form sulfuric acid.
41. 41. Stoichiometry: Combination Working at 273.15K and 1atm (STP), I have 10.0g of carbon and 56L of oxygen. Under these conditions, I know 1mol of any gas has a volume of 22.4L (Molar volume at STP). What is the limiting reagent? If carbon monoxide made sure to be formed, what will be its volume? If the reaction takes place in a giant balloon, what will be the final volume of the balloon (considering the conditions are kept constant).
42. 42. 2. Decomposition ONE substance reacts to form two or more substances (elements or compounds). AZ  A + Z  Compound compound/element oxygen  Metal carbonate metal oxide carbon dioxide  Hydrated salt salt water  Compound compound/element water
43. 43. Test yourself: decomposition1. Mercury (II) oxide decomposes into mercury and oxygen.2. Potassium nitrate decomposes into potassium nitrite and oxygen.3. Hydrogen peroxide decomposes into hydrogen and oxygen.4. Water electrolysis creates hydrogen and oxygen.
44. 44. Stoichiometry: Decomposition Sodium azide produces nitrogen gas and sodium metal. At standard temperature and pressure, what will be the volume of nitrogen gas produced from 60.0g sodium azide?
45. 45. 3. Single Replacement Also known as single displacement or substitution. A metal (A) replaces a metal ion (B=metal) in its salt or a hydrogen ion (B=H) in an acid  A + BZ  AZ +B  Ex. Fe(s) + CuSO4(aq)  FeSO4(aq) + Cu(s)  Ex. Sn(s) + HCl(aq)  SnCl2(aq) + H2(g) A nonmetal (X) replacing another nonmetal (Z) in its salt (B=metal) or acid (B=H)  Cl2(g) + NaBr (aq)  _____________________________  Bromine + Potassium Iodide ______________________
46. 46. 4. Double Replacement Also known as “Double displacement”, “Metathesis”, or “Double decomposition.” Two compounds are involved with the cation of one compound EXCHANGING with the cation of another compound. AX + BZ  AZ + BX These reactions proceed if one of the ff. is satisfied: 1. An insoluble/slightly soluble product is formed (PRECIPITATE formation) 2. A weakly ionized species is produced. The most common species of this type is water. 3. A gas is produced as a product.
47. 47. Test yourself: metathesis1. Insoluble silver chloride is produced in the double displacement of silver nitrate and hydrochloric acid.2. Nickel (II) nitrate reacts with sodium hydroxide in a metathesis reaction.3. Double decomposition occurs with the addition of table salt to an aqueous solution of silver nitrate.
48. 48. 5. Neutralization Reaction of an acid and a base that usually produces a salt and water. HX + MOH  MX + HOH  Hydrochloric acid and sodium hydroxide.  Hydrochloric acid and magnesium hydroxide.  Sulfuric acid and Barium hydroxide To be discussed in a separate lesson
49. 49. 6. Combustion Reactions involving oxygen (metal +oxygen), (nonmetal + oxygen), (organic compounds + oxygen) are sometimes called combustion reactions. Energy is given off in combustion reactions. For organic compounds, water and carbon dioxide are usual byproducts. Ex. Metabolism of food, fuel combustion:  Sucrose (C22H12O11) combusts  Combustion of propane  Combustion of hydrogen