C h a p t e rC h a p t e r C h a p t e rC h a p t e r Chemistry, 5 th Edition McMurry/Fay Chemistry, 5 th Edition McMurry/Fay 3 3 Formulas, Equations, and Moles

2 The Structure of Atoms Atomic Mass Unit 1 amu =1/12 of the mass of on atom of Carbon-12 1 amu = x g

3 Mass:proton = amu neutron = amu electron = C atom = amu 13 C atom = amu Atomic and Molecular Mass

4 The atomic masses as tabulated in the periodic table are the averages of the naturally occurring isotopes. Mass of C = average of 12 C and 13 C = x 12 amu x amu = amu

5 Atomic and Molecular Mass The mass of a molecule is just the sum of the masses of the atoms making up the molecule. m(C 2 H 4 O 2 ) = 2·m C + 4·m H + 2·m O = 2·(12.01) + 4·(1.01) + 2·(16.00) = amu

6 Avogadro and the Mole One mole of a substance is the gram mass value equal to the amu mass of the substance. One mole of any substance contains 6.02 x units of that substance. Avogadro’s Number (N A, x ) is the numerical value assigned to the unit, 1 mole.

7 Avogadro and the Mole

8 Methionine, an amino acid used by organisms to make proteins, is represented below. Write the formula for methionine and calculate its molar mass. (red = O; gray = C; blue = N; yellow = S; ivory = H)

9 Avogadro and the Mole The Mole: Allows us to make comparisons between substances that have different masses.

10 Balancing Chemical Equations A balanced chemical equation represents the conversion of the reactants to products such that the number of atoms of each element is conserved. reactants  products limestone  quicklime + gas Calcium carbonate  calcium oxide + carbon dioxide CaCO 3 (s)  CaO(s) + CO 2 (g)

11 Balancing Chemical Equations CaCO 3 (s)  CaO(s) + CO 2 (g) The letters in parentheses following each substance are called State Symbols (g) → gas(l) → liquid(s) → solid(aq) → aqueous

12 Balancing Chemical Equations A balanced equation MUST have the same number of atoms of each element on both sides of the equation. H 2 + O 2 → H 2 ONot Balanced H 2 + ½O 2 → H 2 OBalanced 2H 2 + O 2 → 2H 2 OBalanced

13 Balancing Chemical Equations The numbers multiplying chemical formulas in a chemical equation are called: Stoichiometric Coefficients (S.C.) 2H 2 + O 2 → 2H 2 OBalanced Here 2, 1, and 2 are stoichiometric coefficients.

14 Balancing Chemical Equations Hints for Balancing Chemical Equations: 1) Save single element molecules for last. 2) Try not to change the S.C. of a molecule containing an element that is already balanced. 3) If possible, begin with the most complex molecule that has no elements balanced.

15 Balancing Chemical Equations Hints for Balancing Chemical Equations: 4) Otherwise, trial and error!!

16 Balancing Chemical Equations Example 1:CH 4 + O 2 → CO 2 + H 2 O Balance O 2 last C is already balanced Start by changing S.C. of H 2 O to balance H CH 4 + O 2 → CO 2 + 2H 2 O

17 Balancing Chemical Equations Example 1: CH 4 + O 2 → CO 2 + 2H 2 O Now C and H are balanced Balance O by changing the S.C. of O 2 CH 4 + 2O 2 → CO 2 + 2H 2 O BALANCED!

18 Balancing Chemical Equations Example 2: B 2 H 6 + O 2 → B 2 O 3 + H 2 O Balance O last B is already balanced Start by changing S.C. of H 2 O: B 2 H 6 + O 2 → B 2 O 3 + 3H 2 O

19 Balancing Chemical Equations Example 2: B 2 H 6 + O 2 → B 2 O 3 + 3H 2 O B and H are balanced Balance O by changing S.C. of O 2 B 2 H 6 + 3O 2 → B 2 O 3 + 3H 2 O BALANCED!

20 Balancing Chemical Equations Example 3:MnO 2 + KOH + O 2 → K 2 MnO 4 + H 2 O Balance O last Mn is already balanced Change S.C. of KOH to balance K MnO 2 + 2KOH + O 2 → K 2 MnO 4 + H 2 O

21 Balancing Chemical Equations Example 3:MnO 2 + 2KOH + O 2 → K 2 MnO 4 + H 2 O Mn, K, and H are balanced (H was balanced by chance) Balance O MnO 2 + 2KOH + ½O 2 → K 2 MnO 4 + H 2 O or 2MnO 2 + 4KOH + O 2 → 2K 2 MnO 4 + 2H 2 O

22 Balancing Chemical Equations Example 4:NaNO 2 + H 2 SO 4 → NO + HNO 3 + H 2 O + Na 2 SO 4 Hard one (no single element molecules) S is balanced Start with NaNO 2 to balance Na 2NaNO 2 + H 2 SO 4 → NO + HNO 3 + H 2 O + Na 2 SO 4

23 Balancing Chemical Equations Example 4:2NaNO 2 + H 2 SO 4 → NO + HNO 3 + H 2 O + Na 2 SO 4 S, Na, and N are balanced Cannot balance H without changing S.C. for H 2 SO 4 ! Boo!Option 1: trial and error Option 2: Go on to next problem!

24 Balancing Chemical Equations Balance the following equations: C 6 H 12 O 6 → C 2 H 6 O + CO 2 Fe + O 2 → Fe 2 O 3 NH 3 + Cl 2 → N 2 H 4 + NH 4 Cl KClO 3 + C 12 H 22 O 11 → KCl + CO 2 + H 2 O

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26 Balancing Chemical Equations Balance the following equations: C 6 H 12 O 6 → 2C 2 H 6 O + 2CO 2 Fe + O 2 → Fe 2 O 3 NH 3 + Cl 2 → N 2 H 4 + NH 4 Cl KClO 3 + C 12 H 22 O 11 → KCl + CO 2 + H 2 O

27 Balancing Chemical Equations Balance the following equations: C 6 H 12 O 6 → 2C 2 H 6 O + 2CO 2 4Fe + 3O 2 → 2Fe 2 O 3 (balance O first) NH 3 + Cl 2 → N 2 H 4 + NH 4 Cl KClO 3 + C 12 H 22 O 11 → KCl + CO 2 + H 2 O

28 Balancing Chemical Equations Balance the following equations: C 6 H 12 O 6 → 2C 2 H 6 O + 2CO 2 4Fe + 3O 2 → 2Fe 2 O 3 (balance O first) NH 3 + Cl 2 → N 2 H 4 + NH 4 Cl N:H is 1:3 on left, must get 1:3 on right!

29 Balancing Chemical Equations NH 3 + Cl 2 → N 2 H 4 + NH 4 Cl N:H is 1:3 on left, must get 1:3 on right! 4NH 3 + Cl 2 → N 2 H 4 + 2NH 4 Cl

30 Balancing Chemical Equations Balance the following equations: C 6 H 12 O 6 → 2C 2 H 6 O + 2CO 2 4Fe + 3O 2 → 2Fe 2 O 3 4NH 3 + Cl 2 → N 2 H 4 + 2NH 4 Cl KClO 3 + C 12 H 22 O 11 → KCl + CO 2 + H 2 O (tough!)

31 Balancing Chemical Equations Balance the following equations: KClO 3 + C 12 H 22 O 11 → KCl + CO 2 + H 2 O balance C KClO 3 + C 12 H 22 O 11 → KCl + 12CO 2 + H 2 O balance H KClO 3 + C 12 H 22 O 11 → KCl + 12CO H 2 O balance O 8KClO 3 + C 12 H 22 O 11 → KCl + 12CO H 2 O

32 Balancing Chemical Equations Balance the following equations: 8KClO 3 + C 12 H 22 O 11 → KCl + 12CO H 2 O balance K (and hope Cl is balanced) 8KClO 3 + C 12 H 22 O 11 → 8KCl + 12CO H 2 O Balanced!

33 Balancing Chemical Equations Write a balanced equation for the reaction of element A (red spheres) with element B (green spheres) as represented below:

34 Avogadro and the Mole

35 Stoichiometry Stoichiometry: Relates the moles of products and reactants to each other and to measurable quantities.

36 Stoichiometry Aqueous solutions of NaOCl (household bleach) are prepared by the reaction of NaOH with Cl 2 : 2 NaOH(aq) + Cl 2 (g)  NaOCl(aq) + NaCl(aq) + H 2 O(l) How many grams of NaOH are needed to react with 25.0 g of Cl 2 ?

37 Stoichiometry 2 NaOH + Cl 2 → NaOCl + NaCl + H 2 O 25.0 g Cl 2 reacts with ? g NaOH

38 Avogadro and the Mole Calculate the molar mass of the following: Fe 2 O 3 (Rust) C 6 H 8 O 7 (Citric acid) C 16 H 18 N 2 O 4 (Penicillin G) Balance the following, and determine how many moles of CO will react with moles of Fe 2 O 3. Fe 2 O 3 (s) + CO(g) Fe(s) + CO 2 (g)

39 Avogadro and the Mole Fe 2 O 3 + CO → Fe + CO 2 Balance (not a simple one) Save Fe for last C is balanced, but can’t balance O In the products the ratio C:O is 1:2 and can’t change Make the ratio C:O in reactants 1:2 Fe 2 O 3 + 3CO → 2Fe + 3CO 2

40 Avogadro and the Mole Fe 2 O 3 + 3CO → 2Fe + 3CO 2

41 Stoichiometry

42 Stoichiometry Aspirin is prepared by reaction of salicylic acid (C 7 H 6 O 3 ) with acetic anhydride (C 4 H 6 O 3 ) to form aspirin (C 9 H 8 O 4 ) and acetic acid (CH 3 CO 2 H). Use this information to determine the mass of acetic anhydride required to react with 4.50 g of salicylic acid. How many grams of aspirin will result? How many grams of acetic acid will be produced as a by-product?

43 Stoichiometry Salicylic acid + Acetic anhydride → Aspirin + acetic acid C 7 H 6 O 3 + C 4 H 6 O 3 → C 9 H 8 O 4 + CH 3 CO 2 H C 7 H 6 O 3 + C 4 H 6 O 3 → C 9 H 8 O 4 + C 2 H 4 O 2 Balanced! Equal # moles for all

44 Stoichiometry 4.50 g Salicylic acid (C 7 H 6 O 3 ) = ? moles MW C 7 H 6 O 3 = 7 x x x = g/mole

45 Stoichiometry Since all compounds have the same S.C., there must be moles of all 4 of them involved in the reaction. g Aspirin (C 9 H 8 O 4 ) = moles x MW Aspirin =.0326 x [9x x x16.00] =.0326 mole x g/mole 5.87 g Aspirin

46 Stoichiometry Yields of Chemical Reactions: If the actual amount of product formed in a reaction is less than the theoretical amount, we can calculate a percentage yield.

47 Stoichiometry Dichloromethane (CH 2 Cl 2 ) is prepared by reaction of methane (CH 4 ) with chlorine (Cl 2 ) giving hydrogen chloride as a by-product. How many grams of dichloromethane result from the reaction of 1.85 kg of methane if the yield is 43.1%?

48 Stoichiometry CH 4 + Cl 2 → CH 2 Cl 2 + HCl Balance CH 4 + 2Cl 2 → CH 2 Cl 2 + 2HCl 1.85 kg CH 4 = ? moles CH 4

49 Stoichiometry CH 4 + 2Cl 2 → CH 2 Cl 2 + 2HCl 1.85 kg CH 4 = ? moles CH 4 MW CH 4 = 1x x1.008 = g/mole

50 Stoichiometry CH 4 + 2Cl 2 → CH 2 Cl 2 + 2HCl 115 moles CH 4 in theory we should produce: 115 moles of CH 2 Cl 2 and 230 moles of HCl And use up 230 moles of Cl 2

51 Stoichiometry CH 4 + 2Cl 2 → CH 2 Cl 2 + 2HCl 115 moles of CH 2 Cl 2 = ? g MW CH 2 Cl 2 = x x35.45 = moles x (84.03 g/mole) = 9770 g

52 Stoichiometry CH 4 + 2Cl 2 → CH 2 Cl 2 + 2HCl Expect 9770 g CH 2 Cl 2 but the yield is 43.1% So we produced just x 9770 g 4.21 kg CH 2 Cl 2

53 Stoichiometry CH 4 + 2Cl 2 → CH 2 Cl 2 + 2HCl Suppose the reaction went to completion (100% yield) Is mass conserved?

54 Stoichiometry CH 4 + 2Cl 2 → CH 2 Cl 2 + 2HCl Start with 115 moles CH 4 and 230 moles Cl 2 total mass = 115x x70.90 = = only 3 sig. figs. → 18.2 kg

55 Stoichiometry CH 4 + 2Cl 2 → CH 2 Cl 2 + 2HCl End with 115 moles CH 2 Cl 2 and 230 moles HCl total mass = 115x x36.46 = = only 3 sig. figs → 18.2 kg