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Mass Relationships in Chemical Reactions

Mass Relationships in Chemical Reactions. Micro World atoms & molecules. Macro World grams. Atomic mass is the mass of an atom in atomic mass units (amu). By definition: 1 atom 12 C “weighs” 12 amu. On this scale 1 H = 1.008 amu 16 O = 16.00 amu. 3.1.

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Mass Relationships in Chemical Reactions

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  1. Mass Relationships in Chemical Reactions

  2. Micro World atoms & molecules Macro World grams Atomic mass is the mass of an atom in atomic mass units (amu) By definition: 1 atom 12C “weighs” 12 amu On this scale 1H = 1.008 amu 16O = 16.00 amu 3.1

  3. (7.42% x 6.015) + (92.58% x 7.016) 100 Natural lithium is: 7.42% 6Li (6.015 amu) 92.58% 7Li (7.016 amu) Average atomic mass of lithium: = 6.941 amu 3.1

  4. Dozen = 12 Pair = 2 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

  5. eggs shoes Molar mass is the mass of 1 mole of 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

  6. One Mole of: S C Hg Cu Fe 3.2

  7. Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. 1 X 32.07 g 1S + 2 x 16.00 g 2O SO2 64.07 g/mole For all practical purposes think of amu as grams What is the molecular weight ofSO2 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2 3.3

  8. Molecular weights (covalent compounds) and formula weights (ionic compounds) are found exactly the same way Try a few Find the formula weight for NaCl Find the molecular weight for H2SO4

  9. I STOICHIOMETRY A. Relationship between balanced chemical equations andmass or volume 1. Mass - mass problems a. The first thing to do is balanced the chemical equations i. this will allow you to find the mole ratio between the reactants andproducts

  10. 2H2 + O2 --> 2H2O 2 moles 1 mole 2 moles 4.04 g 32.0 g 36.0 g

  11. ex. How many grams of KCl are produced when 200.0 g of KClO3 decomposes? Oxygen is the other product 1st write the balanced equation 2moles 2moles 3moles 2nd find mole ratio 2KClO3 ---> 2KCl + 3 O2 122.6g/m 74.55g/m 32.00g/m 200.0 g x 3th write down given/asked 4th find mwt or fwt

  12. 5th solve the problem a. start with the given b. change grams of given to moles of given using themolecular or formula weight c. change the moles of given to moles of asked using themole ratio d. change the moles of asked to grams of asked using themolecular or formula weight

  13. M.W (1 mole KClO3) (200.0 g KClO3) (2 moles KCl) ( 74.55 g KCl) ( 122.6 g KClO3 ) (2 mole KClO3) (1 mole KCl) Given mole ratio M.W Your answer will come out to be in g KCl Remember your sig figs 243.2 g KCl

  14. You try one How many grams of NaOH is required to completely neutralize 155 grams of HCl? The product of this reaction is HOH and NaCl.

  15. 1 mole 1 mole 1 mole 1 mole NaOH + HCl ===> HOH + NaCl X g 155 g HCl Na = 1 x 23.0 g = 23.0 g H = 1 x 1.01 g = 1.01 g Cl =1 x 35.5 g = 35.5 g O = 1 x 16.0 g = 16.0 g H = 1.01 g = 1.01 g 170. g NaOH 40.0 g/mole 36.5 g/mole 1 mole NaOH 40.0 g NaOH ( ( ) 1 mole HCl ) ( ) ( ) 155 g HCl 1 mole HCl 1 mole NaOH 36.5 g HCl

  16. Try another How many grams of Potassium is required to produce 200. gof KOH when it is involved in a single replacement reaction with water? Hydrogen is the other product 2 moles 1 moles 2 moles 2 moles 2 K + HOH ===> KOH + H2 2 2 200. g KOH X g 39.1 g/mole 56.1 g/mole 39.1g K ) ) ( 2 moles K ) ( 1 mole KOH ) ( ( 200.g KOH 1 mole K 56.1 g KOH 2 moles KOH

  17. grams CH3OH moles CH3OH moles H2O grams H2O 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? molar mass H2O molar mass CH3OH coefficients chemical equation 18.0 g H2O ( ) ( 4 mol H2O ) ) ( 1 mol CH3OH ( ) = 209 g CH3OH 1 mol H2O 2 mol CH3OH 32.0 g CH3OH 235 g H2O 3.8

  18. If 120. grams of mercuric oxide are decomposed, what mass of oxygen is produced? • HgO ==> Hg + O2 2 mole 1mole 2 mole 2 2 x 120.g O = 2 x 16.0 g = 32.0 g/mole Hg = 1 x201 = 201g O = 1 x 16.0 = 16.0 g 217 g/mole 1 mole HgO ) 32.0 g O2 ( ) ( ( 1 mole O2 ) ( ) 120. gHgO 2 moles HgO 1 mole O2 217 g HgO = 8.85 g O2

  19. When 70.0 grams of hydrogen peroxide are decomposed, how many grams of oxygen are released? • H2O2===> H2O + O2 2 moles 2 moles 1 moles 2 2 70.0 g x 34.0 g/mole H2O2 32.0 g/mole O2 ) ( ) 1 mole H2O2 ) ( 1 mole O2 ) ( 32.0 g O2 ( 70.0 g H2O2 2 moles H2O2 1mole O2 34.0 g H2O2 = 32.9 g O2

  20. If 40.00 grams of aluminum react with hydrochloric acid (HCl), how many grams of hydrogen are produced? • Al + HCl ===> AlCl3 + H2 2 moles Al 6 moles HCl 2 moles AlCl3 3 moles H2 2 6 2 3 x g 40.00 g Al 2.016 g/mole 26.98 g /mole Al 1 mole Al ) ) ( 3 moles H2 ) ( 2.016 g H2 ( ) ( 40.00 g Al 1 mole H2 26.98 g Al 2 moles Al

  21. In the reduction of 84.0 grams of copper II oxide, how many grams of hydrogen enter into the reaction? • CuO + H 2 ===> Cu + H2O 1 mole 1 mole 1 mol 1mole 84.0 g CuO X g H2 79.4 g/mole 2.00 g/mole H2 2.00 g H2 ( ) ( ) ) ) ( 1 mole CuO ( 1 mole H2 84.0 g CuO 1 mole H2 79.4 g CuO 1 mole CuO

  22. 5. How many grams of potassium chlorate are required in the preparation of 90.0 grams of oxygen? KClO3 ===> KCl + O2

  23. In the electrolysis of 144 grams of water, how many grams of hydrogen are produced? • H2O ===> H2 + O2

  24. Mass Changes in Chemical Reactions • Write balanced chemical equation • Convert quantities of known substances into moles • Use coefficients in balanced equation to calculate the number of moles of the sought quantity • Convert moles of sought quantity into desired units 3.8

  25. 6 green used up 6 red left over Limiting Reagents 3.9

  26. Method 1 • Pick A Product • Try ALL the reactants • The lowest answer will be the correct answer • The reactant that gives the lowest answer will be the limiting reactant

  27. Limiting Reactant: Method 1 • 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl2 2 AlCl3 • Start with Al: • Now Cl2: 10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g AlCl3 27.0 g Al 2 mol Al 1 mol AlCl3 = 49.4g AlCl3 35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g AlCl3 71.0 g Cl2 3 mol Cl2 1 mol AlCl3 = 43.9g AlCl3

  28. Method 2 • Convert one of the reactants to the other REACTANT • See if there is enough reactant “A” to use up the other reactants • If there is less than the GIVEN amount, it is the limiting reactant • Then, you can find the desired species

  29. 5.0 g of H2 reacts with 10.0 g of O2. How many grams of H2O is produced? 2 moles 1moles 2 moles 2 H2 + O2 H2O 2 5.0 g 10.0 g x 2.0 g/mole 32.0g/mole 18 .0g/mole 1 mole H2 1 mole O2 ( ( ) ( 32 g O2 ) ) ( ) 5.0 g = 40. g O2 2.0g H2 2mole H2 1 mole O2 2 mole H2O ) ( 1 mole O2 ) ( ( 18.0 g H2O ) = 11.2 g H2O ( 10.0 g ) 32.0g O2 1mole O2 1 mole H2

  30. Do You Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe2O3 2Al + Fe2O3 Al2O3 + 2Fe

  31. g Al mol Al mol Fe2O3 needed g Fe2O3 needed OR g Fe2O3 mol Fe2O3 mol Al needed g Al needed Calculate the mass of Al2O3 formed. 2Al + Fe2O3 Al2O3 + 2Fe ( ) ( ) ) ( ) 1 mol Al ( 1 mol Fe2O3 160. g Fe2O3 = 367 g Fe2O3 124 g Al 1 mol Fe2O3 27.0 g Al 2 mol Al 3.9

  32. need 367 g Fe2O3 Start with 124 g Al Have more Fe2O3 (601 g) so Al is limiting reagent

  33. 2Al + Fe2O3 Al2O3 + 2Fe g Al mol Al mol Al2O3 g Al2O3 Use limiting reagent (Al) to calculate amount of product that can be formed. ( ) ( ) ( ) 1 mol Al 1 mol Al2O3 102. g Al2O3 ( ) = 234 g Al2O3 124 g Al 27.0 g Al 2 mol Al 1 mol Al2O3 3.9

  34. % Yield = x 100 Actual Yield Theoretical Yield Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. 3.10

  35. 2 x (12.01 g) 6 x (1.008 g) 1 x (16.00 g) n x molar mass of element %C = %H = %O = x 100% = 34.73% x 100% = 13.13% x 100% = 52.14% x 100% 46.07 g 46.07 g 46.07 g molar mass of compound C2H6O Percent composition of an element in a compound = n is the number of moles of the element in 1 mole of the compound 52.14% + 13.13% + 34.73% = 100.0% 3.5

  36. You try two Find percent composition of H2O H3PO4

  37. To obtain an Empirical Formula 1. Determine the mass in grams of each element present, if necessary. 2. Calculate the number of moles of each element. 3. Divide each by the smallest number of moles to obtain the simplest whole number ratio. • If whole numbers are not obtained* in step 3), multiply through by the smallest number that will give all whole numbers *Be careful! Do not round off numbers prematurely

  38. e.g. problem Find empirical formula for a compound found to contain 60.8 % Na 28.60 % B 10.60 % H

  39. 2.644 moles 60.80 % Na = 60.80 g Na ( ) ( 1 mole Na ) 1 = = 23.00 g ) ( 2.644 moles 28.60 % B = 28.60g B ( ) ( 1 mole B ) 2.646 moles = = 1 ( 10.81 g B ) 2.644 moles 10.60 % H = ( 10.60g H ) ( 1 mole H ) 10.52 moles = 4 = ( 1.008 g H ) 2.644 moles Na1 B1 H4

  40. Stop both 6th and 8th

  41. Empirical Formula from % Composition A substance has the following composition by mass: 60.80 % Na ; 28.60 % B ; 10.60 % H What is the empirical formula of the substance? Consider a sample size of 100 grams This will contain 60.80 g of Na , 28.60 grams of B and 10.60 grams H Determine the number of moles of each Determine the simplest whole number ratio

  42. You try one A compound was found to contain 11 % Hydrogen and 89 % O. What is the empirical formula?

  43. Types of Formulas • Empirical Formula The formula of a compound that expresses the smallest whole number ratio of the atoms present. Ionic formula are always empirical formula • Molecular Formula The formula that states the actual number of each kind of atom found in one molecule of the compound.

  44. require mole ratios so convert grams to moles A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine a formula for this substance. ( ) ( ) 1 mole N2 0.167 moles of N moles of N = 2.34 g N = 14.01g N2 0.167 moles of N ) ( ) ( 1 mole O2 0.334moles of O moles of O = 5.34g O = 16.0 g O2 0.167 moles of N NO2

  45. Calculation of the Molecular Formula From the previous problem, it was found a compound has an empirical formula of NO2. The colourless liquid, used in rocket engines has a molar mass of 92.0 g/mole. What is the molecular formula of this substance? empirical formula mass: 14.01+2 (16.00) = 46.01 g/mol n = molar mass = 92.0 g/mol emp. f. mass 46.01 g/mol n = 2 So the molecular formula is twice the size of the empirical formula NO2 = N2O4

  46. 8th period start here

  47. 6th period start here

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