1. Mass Relationships in Chemical Reactions

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

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

4. 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 12 C 3.2 1 mol = N A = 6.0221367 x 10 23 Avogadro’s number (N A )

5. Molar mass is the mass of 1 mole of in grams eggs shoes marbles atoms 1 mole 12 C atoms = 6.022 x 10 23 atoms = 12.00 g 1 12 C atom = 12.00 amu 1 mole 12 C atoms = 12.00 g 12 C 1 mole lithium atoms = 6.941 g of Li For any element atomic mass (amu) = molar mass (grams) 3.2

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

7. Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. 1S 1 X32.07 g 2O + 2 x 16.00 g SO 2 64.07 g/mole 1 molecule SO 2 = 64.07 amu 1 mole SO 2 = 64.07 g SO 2 3.3 What is the molecular weight of SO 2 For all practical purposes think of amu as grams

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 H 2 SO 4

9. I STOICHIOMETRY A.Relationship between balanced chemical equations and mass 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 and products

10. 2H 2 + O 2 --> 2H 2 O 2 moles1 mole2 moles 4.04 g32.0 g36.0 g

11. ex.How many grams of KCl are produced when 200.0 g of KClO 3 decomposes? Oxygen isthe other product 1st write the balanced equation 2KClO 3 ---> 2KCl + 3 O 2 2nd find mole ratio 2moles2moles3moles 4th find mwt or fwt 122.6g/m74.55g/m 32.00g/m 3th write down given/asked 200.0 g x

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

13. (200.0 g KClO 3 ) ( 122.6 g KClO 3 ) (1 mole KClO 3 ) (2 mole KClO 3 ) (2 moles KCl) (1 mole KCl) ( 74.55 g KCl) Your answer will come out to be in g KCl Remember your sig figs 243.2 g KCl Given M.W mole ratio M.W

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. NaOH + HCl ===> HOH + NaCl 1 mole X g155 g HCl Na = 1 x 23.0 g = 23.0 g O = 1 x 16.0 g = 16.0 g H = 1.01 g = 1.01 g 40.0 g/mole H = 1 x 1.01 g = 1.01 g Cl =1 x 35.5 g = 35.5 g 36.5 g/mole 155 g HCl () ( ) 36.5 g HCl 1 mole HCl () 1 mole NaOH () 40.0 g NaOH 170. g NaOH

16. Try another How many grams of Potassium is required to produce 200. g of KOH when it is involved in a single replacement reaction with water? Hydrogen is the other product K + HOH ===> KOH + H 2 222 2 moles 1 moles X g 200. g KOH 39.1 g/mole56.1 g/mole 200.g KOH ( )() 56.1 g KOH 1 mole KOH ( ) 2 moles KOH 2 moles K ( ) 39.1g K 1 mole K

17. Methanol burns in air according to the equation 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH 3 OHmoles CH 3 OHmoles H 2 Ograms H 2 O molar mass CH 3 OH coefficients chemical equation molar mass H 2 O 209 g CH 3 OH 1 mol CH 3 OH 32.0 g CH 3 OH 4 mol H 2 O 2 mol CH 3 OH 18.0 g H 2 O 1 mol H 2 O = 235 g H 2 O 3.8 ( ( ) )( )( )

18. 1.If 120. grams of mercuric oxide are decomposed, what mass of oxygen is produced? HgO ==> Hg + O 2 120.g x 22 2 mole 1mole Hg = 1 x201 = 201g O = 1 x 16.0 = 16.0 g 217 g/mole O = 2 x 16.0 g = 32.0 g/mole 120. gHgO ()((( ) )) 217 g HgO 1 mole HgO 2 moles HgO 1 mole O 2 32.0 g O 2 = 8.85 g O 2

19. 2.When 70.0 grams of hydrogen peroxide are decomposed, how many grams of oxygen are released? H 2 O 2 ===> H 2 O + O 2 22 70.0 gx 2 moles 1 moles 34.0 g/mole H 2 O 2 32.0 g/mole O 2 70.0 g H 2 O 2 () ( ) 34.0 g H 2 O 2 1 mole H 2 O 2 () 2 moles H 2 O 2 1 mole O 2 ( ) 32.0 g O 2 = 32.9 g O 2

20. 3.If 40.00 grams of aluminum react with hydrochloric acid (HCl), how many grams of hydrogen are produced? Al + HCl ===> AlCl 3 + H 2 6223 2 moles Al6 moles HCl2 moles AlCl 3 3 moles H 2 40.00 g Al x g 26.98 g /mole Al 2.016 g/mole 40.00 g Al ()( ) 26.98 g Al 1 mole Al () 2 moles Al 3 moles H 2 ( ) 1 mole H 2 2.016 g H 2

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

22. 5.How many grams of potassium chlorate are required in the preparation of 90.0 grams of oxygen? KClO 3 ===> KCl + O 2

23. 6.In the electrolysis of 144 grams of water, how many grams of hydrogen are produced? H 2 O ===> H 2 + O 2

24. 1.Write balanced chemical equation 2.Convert quantities of known substances into moles 3.Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4.Convert moles of sought quantity into desired units Mass Changes in Chemical Reactions 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 Cl 2  2 AlCl 3 Start with Al: Now Cl 2 : 10.0 g Al 1 mol Al 2 mol AlCl 3 133.5 g AlCl 3 27.0 g Al 2 mol Al 1 mol AlCl 3 = 49.4g AlCl 3 35.0g Cl 2 1 mol Cl 2 2 mol AlCl 3 133.5 g AlCl 3 71.0 g Cl 2 3 mol Cl 2 1 mol AlCl 3 = 43.9g AlCl 3

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 H 2 reacts with 10.0 g of O 2. How many grams of H 2 O is produced? H 2 + O 2  H 2 O22 5.0 g10.0 g 2 moles1moles2 moles x 2.0 g/mole32.0g/mole18.0g/mole () (( ) ) 5.0 g 2.0g H 2 1 mole H 2 2mole H 2 1 mole O 2 () 32 g O 2 = 40. g O 2 10.0 g ( ) ()(( ) ) 32.0g O 2 1 mole O 2 2 mole H 2 O 1 mole H 2 18.0 g H 2 O = 11.2 g H 2 O

30. Do You Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe

31. Calculate the mass of Al 2 O 3 formed. g Almol Almol Fe 2 O 3 neededg Fe 2 O 3 needed OR g Fe 2 O 3 mol Fe 2 O 3 mol Al neededg Al needed 124 g Al 1 mol Al 27.0 g Al 1 mol Fe 2 O 3 2 mol Al 160. g Fe 2 O 3 1 mol Fe 2 O 3 = 367 g Fe 2 O 3 3.9 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe ()() ( )()

32. Have more Fe 2 O 3 (601 g) so Al is limiting reagent Start with 124 g Al need 367 g Fe 2 O 3

33. Use limiting reagent (Al) to calculate amount of product that can be formed. g Almol Almol Al 2 O 3 g Al 2 O 3 124 g Al 1 mol Al1 mol Al 2 O 3 2 mol Al 102. g Al 2 O 3 1 mol Al 2 O 3 = 234 g Al 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe 3.9 () ( 27.0 g Al )()()

34. 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. % Yield = Actual Yield Theoretical Yield x 100 3.10

35. Percent composition of an element in a compound = n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound C2H6OC2H6O %C = 2 x (12.01 g) 46.07 g x 100% = 52.14%H = 6 x (1.008 g) 46.07 g x 100% = 13.13%O = 1 x (16.00 g) 46.07 g x 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.0% 3.5

36. You try two Find percent composition of H 2 O H 3 PO 4

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. 4.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. 60.80 % Na = 60.80 g Na ()(1 mole Na) ) ( 23.00 g = 2.644 moles 28.60 % B = 28.60g B(( ( )) )10.81 g B 1 mole B = 2.646 moles 10.60 % H =(( ( )) ) 10.60g H 1.008 g H 1 mole H = 10.52 moles 2.644 moles = = = 1 1 4 Na 1 B1B1 H4H4

40. Stop both 6th and 8th

41. Empirical Formula from % Composition Determine the simplest whole number ratio Determine the number of moles of each This will contain 60.80 g of Na, 28.60 grams of B and 10.60 grams H Consider a sample size of 100 grams 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?

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. A. 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. 0.167 moles of N require mole ratios so convert grams to moles moles of N = moles of O = ( 2.34 g N )() 14.01g N 2 1 mole N 2 0.334moles of O ( )() 5.34g O 16.0 g O 2 1 mole O 2 = = 0.167 moles of N NO 2

45. Calculation of the Molecular Formula From the previous problem, it was found a compound has an empirical formula of NO 2. 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 NO 2 = N 2 O 4

46. 8 th period start here

47. 6 th period start here