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Unit 6 Chapter 12 Chemical Quantities or. Stoichiometry stoichiometryusing balanced chemical equations to obtain info.

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Presentation on theme: "Unit 6 Chapter 12 Chemical Quantities or. Stoichiometry stoichiometryusing balanced chemical equations to obtain info."— Presentation transcript:

1 Unit 6 Chapter 12 Chemical Quantities or

2 Stoichiometry stoichiometryusing balanced chemical equations to obtain info

3 12.1 Counting Particles of Matter Particles of matter are too small and numerous to count SI unit of chemical quantity = the mole (abbreviated mol) nThe mole is a counting unit

4 How do you measure how much? How do you measure how much? You can measure mass = grams or volume = liters or you can count pieces = MOLES

5 Moles 1 mole = 6.02 x representative particles n Treat it like a very large dozen n 6.02 x is called Avogadro's number

6 Representative particles = The smallest pieces of a substance representative particles = n ATOMS n IONS n MOLECULES n FORMULA UNITS

7 Representative particles n For a molecular (covalent) compounds it is a molecule compound with all nonmetals CO, BF 3, Cl 2 CO, BF 3, Cl 2 1 mole = 6.02 x molecules

8 Representative particles For an element it is an atom –Unless it is diatomic one symbol, no charge: Br,Cs 1 mole = 6.02 x atoms

9 Representative particles For an ionic compound it is a formula unit compound with metal and nonmetal - KI, Na 2 SO 4 1 mole = 6.02 x formula units

10 Representative particles For ions it is 1 mol ions one symbol with charge (monatomic) or more than one symbol with charge (polyatomic: Na+, N 3 -, (C 2 H 3 O 2 ) – 1 mole ions = 6.02 x ions

11 Molar Mass = The mass of 1 mole of an element in grams. n We can make conversion factors from these. –Example - We can write this as g C = 1 mol n To change grams of a compound to moles of a compound. n Or moles to grams

12 Molar Masses n the atomic masses on the periodic table –have a unit of amu (atomic mass unit) GAM = gram atomic mass = the atomic mass (listed on the periodic table) written in grams n 1 atom Xe = u n GAM of Xe = g

13 molar massthe mass, in g, of 1 mole of a substance Add up the gram atomic mass of all elements in compound to calculate molar mass of a substance

14 Find the molar mass of methane, CH 4. CH 4 = 1(12.0) + 4(1.0) = = 16.0 g Find the molar mass of calcium hydroxide, Ca(OH) 2. n Ca(OH) 2 = 1(40.1) + 2(16.0) + 2(1.0) = 74.1 g

15 1. What is the molar mass of 1 mole of phosphorus? 2. What is the molar mass of a molecule of H 2 ? 3. What is the molar mass of a formula unit of MgCl 2 ? g g 3.94 g

16 Molar Volume: volume-to-mole and mole-to-volume conversions At STP, all gases occupy the same amount of space: MOLAR VOLUME of any gas at STP: 22.4 L = 1 mol

17 dimensional analysis = using the units (dimensions) to solve problems steps for success: 1) identify unknown (read carefully) 2) identify known (read carefully) Play checkers with the units, moving them diagonally, canceling when appropriate. All units should cancel except those of the desired answer. 3) plan solution 4) calculate 5) check (sig.figs., units, and math)

18 CONVERSION FACTOR SUMMARY: 6.02 x 10^ 23 representative particles 1 MOLE & 1 MOLE 6.02 x 10^ 23 representative particles

19 CONVERSION FACTOR SUMMARY: MOLAR MASS (g) & 1 MOLE 1 MOLE MOLAR MASS (g) for a gas at STP: 22.4 L & 1 MOLE 1 MOLE 22.4 L

20 Calculation question How many molecules of CO 2 are the in 4.56 moles of CO 2 ? 4.56 moles CO 2 X 6.02 x molecules CO mole CO 2 = 27.5 x molecules CO 2

21 For example n How many moles is 5.69 g of NaOH? 5.69 g NaOH x 1 mole= mol NaOH gNaOH need to change grams to moles for NaOH l 1mole Na = 22.99g 1 mol O = g 1 mole of H = 1.01 g l 1 mole NaOH = g

22 Examples n How much would 2.34 moles of carbon weigh? 2.34 mol C X g C 1 1mol C = g C

23 Calculation question How many moles of salt is 5.87 x formula units? 5.87 X formula units NaCl X 1 mole NaCl_ x f.u. NaCl = moles NaCl

24 Examples n How many moles of magnesium in 4.61 g of Mg? 4.61 g Mg X _1 mol Mg g Mg = mol Mg

25 example What is the volume, in L, of mol of NO 2 gas at STP? mol NO 2 x 22.4 L NO 2 = 11.1 L NO mol NO 2

26 How many moles are found in 84 L of neon gas at STP? 84 L Ne x 1 mol Ne = 3.8 mol Ne L Ne

27 There are many types of mole problems: 1 step: r.p. mol & mol r.p. mass mol & mol mass 2 step: mass r.p. & r.p. mass mass volume & volume mass r.p. volume & volume r.p.

28 Moles of Compounds 1 mole of a compound contains as many moles of each element as are indicated by the subscripts in the formula for the compound. Example: 1 mole of ammonia (NH 3 ) has 1 mole of nitrogen atoms and 3 moles of hydrogen atoms.

29 Percent composition composition of a compound is the percent by mass of each element in the compound. n Find the mass of each element n divide by the total mass of compound.

30 Determine the percent composition of calcium chloride (CaCl 2 ). Step 1: determine the molar masses of each element in compound and the compound Ca = g Cl 2 = 2 x g = g CaCl 2 = g

31 Step 2: divide the molar mass of each element by the molar mass of the compound and times by 100 ****make sure they add up to 100

32 Empirical Formula n The lowest whole number ratio of elements in a compound. n CH 2 nH2OnH2OnH2OnH2O

33 You can determine the empirical formula from percent composition and mole ratios *** percent means parts per hundre d nassume you have 100 g of the compound nStep 1: change the percent sign to g (grams) Calculate the empirical formula of a compound composed of % C, % H, and %N

34 Step 2:calculate the number of moles for each element by converting grams to moles using 1 mole = molar mass g C x 1mol C = mole C g C g H x 1mol H = 16.1 mole H g H g N x 1mol N = mole N g N

35 Step 3: divide each number of moles by the smallest number of moles mole C = 1 mol C mole H = 5 mol H mole N = 1 mol N So empirical formula is CH 5 N

36 Determine the empirical formula of a compound with percent composition of 38.4 % Mn, 16.8 % C, 44.7 % O 38.4 g Mn X 1mol Mn = mole Mn 1 55 g Mn 16.8 g C X 1mol C = mole C 1 12 g C 45.7 g O X 1mol O = mole O 1 16 g O

37 0.699 mole Mn = 1 mol Mn mole C = 2 mol C mole O = 4 mol O So formula is MnC 2 O 4

38 For many compounds, the empirical formula is not the true formula A molecular formula tells the exact number of atoms of each element in a molecule

39 Example: acetic acid molecular formula = C 2 H 4 O 2 empirical formula = CH 2 O The molecular formula for a compound is always a whole-number multiple of the empirical formula.

40 To determine molecular formula you need to know the molar mass of the compound n Divide the actual molar mass by the molar mass of the empirical formula. n Multiply the empirical formula by this number. molar mass of compound molar mass of empirical formula

41 Example: A compound has an empirical formula of ClCH 2 and a molar mass of g/mol. What is its molecular formula? molar mass of ClCH 2 = 49.0 g/mol = X (ClCH 2 ) Empirical formula = ClCH 2 Molecular formula = Cl 2 C 2 H 4

42 Example n n A compound has an empirical formula of CH 2 O and a molar mass of g/mol. What is its molecular formula?

43 Example n n Ibuprofen is % C, 8.80 % H, % O, and has a molar mass of about 207 g/mol. What is its molecular formula?

44 Unit 6 Chemical Quantities Continued

45 remember our friend the mole? n Recall that the mole should be treated like a dozen n 1 dozen = 12 pieces & n 1 mole = pieces (or 6.02 X r.p)

46 remember our friend the mole? You still need to know how and when to use: Avogadros number (6.02 x 1023), representative particles Avogadros number (6.02 x 1023), representative particles Molar mass, grams in one mole, Molar mass, grams in one mole, Molar volume of a gas at STP (22.4 L) Molar volume of a gas at STP (22.4 L)

47 Conversion factor review 1 mole = 6.02 x 1023 r.p 1 mole = molar mass (g) 1 mole = 22.4 L

48 So what about this silly mole? Our useful friend the mole allows us to do calculations called stoichiometryusing balanced chemical equations to obtain info ex: 2C + O 2 2CO

49 Mole - Mole (MOL – MOL) Conversions n new conversion factor – # of mol A = # of mol B # = coefficients in balanced equation

50 I.Mole - Mole (MOL – MOL) Conversions A. the most important, most basic stoich calculation B. uses the coefficients of a balanced equation to compare the amounts of reactants and products C. coefficients are mole ratios

51 D. the way to go from substance A to substance B E. mol – mol is the only time the mole number in the conversion is not automatically 1. (Avogadros #, molar mass, and 22.4 L are all = to 1 mol) MOL – MOL : # mol A# mol B # mol B # mol A # mol B # mol A # = coefficients

52 MOL – MOL Conversions # mol A# mol B # mol A# mol B # mol B # mol A # mol B # mol A ex: How many moles of carbon monoxide are produced when mol of oxygen reacts with carbon? 2C + O 2 2CO ***always start with a balanced chemical equation

53 How many moles of carbon monoxide are produced when mol of oxygen reacts with carbon? 2C + O 2 2CO Step 1; determine given and unknown given = mol O 2 unknown = ? mol CO

54 How many moles of carbon monoxide are produced when mol of oxygen reacts with carbon? 2C + O 2 2CO Step 2; setup conversion using mole ratios (coefficients) in balanced equation mol O 2 X 2 mol CO = 1.5 mol CO 1 1 mol O mol O 2

55 4Al + 3O 2 2Al 2 O 3 Find the number of moles of the reactants, given mol of product is formed. 4Al + 3O 2 2Al 2 O 3 Step 1; determine given and unknown given = mol Al 2 O 3 unknown = ? mol Al = ? mol O 2 = ? mol O 2

56 Step 2; setup conversion using mole ratios (coefficients) in balanced equation mol Al 2 O 3 X 4 mol Al = 1.32 mol Al 1 2 mol Al 2 O mol Al 2 O mol Al 2 O 3 X 3 mol O 2 = mol O mol Al 2 O mol Al 2 O 3 4Al + 3O 2 2Al 2 O 3

57 II. MASS – MASS Conversions – Using molar mass in stoich problems to predict masses of reactants and/or products A. a balanced chemical equation can be used to compare masses of reactants and products B. mass – mass cannot change which substance you are dealing with; only mol – mol can do that

58 MASS – MASS Conversions GIVEN g A X 1 mol A X # mol B X PT g B 1 PT g A # mol A 1 mol B 1 PT g A # mol A 1 mol B PT = periodic table, molar mass # = coefficients, chemical equation

59 How many grams of hydrochloric acid are made from the reaction of g of hydrogen gas with excess chlorine gas? H 2 + Cl 2 2HCl H 2 + Cl 2 2HCl Known = 0.500g H 2 Unknown = ? g HCl

60 0.500 g H 2 X 1 mol H 2 X 2 mol HCl X 36.5 gHCl g H 2 1 mol H 2 1 mol HCl g H 2 1 mol H 2 1 mol HCl = 18 g HCl H 2 + Cl 2 2HCl GIVEN g A X 1 mol A X # mol B X PT g B 1 PT g A # mol A 1 mol B 1 PT g A # mol A 1 mol B

61 Calculate the numbers of grams of oxygen formed when 25.0 g of sodium nitrate decomposes into sodium nitrite and oxygen. 2NaNO 3 2NaNO 2 + O 2 2NaNO 3 2NaNO 2 + O g NaNO 3 X 1 mol NaNO 3 X 1 mol O 2 X 32.0 g O g NaNO 3 2 mol NaNO 3 1 mol O g NaNO 3 2 mol NaNO 3 1 mol O 2 = 20.3 g O 2

62 mole – mass (mass – mole) calculationsmole – mass (mass – mole) calculations MASS – MOLE: GIVEN g A X 1 mol A X # mol B PT g A # mol A PT g A # mol A MOLE – MASS: GIVEN mol A X # mol B X PT g B # mol A 1 mol B # mol A 1 mol B PT = periodic table, molar mass # = coefficients

63 How many g of water are produced from the complete combustion of mol of C 2 H 2 ? 2C 2 H 2 + 5O 2 4CO 2 + 2H 2 O 2C 2 H 2 + 5O 2 4CO 2 + 2H 2 O mol C 2 H 2 X 2 mol H 2 O X 18.0 g H 2 O 1 2 mol C 2 H 2 1 mol H 2 O 1 2 mol C 2 H 2 1 mol H 2 O = 12.3 g H 2 O

64 Using the equation 2C 2 H 2 + 5O 2 4CO 2 + 2H 2 O how many moles of O 2 would be needed to produce g of CO 2 ? g CO 2 X 1 mol CO 2 X 5 mol O g CO 2 4 mol CO g CO 2 4 mol CO 2 = 1.59 mol O 2

65 mass – volume (volume – mass) calculations –mass – volume (volume – mass) calculations – MASS – VOLUME: STP)MASS – VOLUME: STP) GIVEN g A X 1 mol A X # mol B X 22.4 L B 1 PT g A # mol A 1 mol B 1 PT g A # mol A 1 mol B VOLUME – MASS: STP) GIVEN L A X 1 mol A X # mol B X PT g B L A # mol A 1 mol B L A # mol A 1 mol B PT = periodic table, molar mass # = coefficients

66 How many L of hydrogen are produced from the decomposition of 3.50 g of water at STP? 2H 2 O 2H 2 + O g H 2 O X 1 mol H 2 O X 2 mol H 2 X 22.4 L H g H 2 O 2 mol H 2 O 1 mol H g H 2 O 2 mol H 2 O 1 mol H 2 = 4.36 L H 2

67 Using the equation 2C 2 H 2 + 5O 2 4CO 2 + 2H 2 O 2C 2 H 2 + 5O 2 4CO 2 + 2H 2 O how many liters of water vapor are produced when 5.02 g of C 2 H 2 undergoes complete combustion? 5.02 g C 2 H 2 x 1 mol C 2 H 2 x 2 mol H 2 O x 22.4 L H 2 O g C 2 H 2 2 mol C 2 H 2 1 mol H 2 O g C 2 H 2 2 mol C 2 H 2 1 mol H 2 O = 4.32 L H 2 O

68 volume – volume calculations GIVEN L A x 1 mol A x # mol B x 22.4 L B L A # mol A 1 mol B L A # mol A 1 mol B # = coefficients (SHORT CUT: compare coefficients!)

69 How many L of sulfur trioxide are produced from the reaction of 36.1 L of oxygen with sulfur dioxide at STP? 2SO 2 + O 2 2SO 3 2SO 2 + O 2 2SO L O 2 x 1 mol O 2 x 2 mol SO 3 x 22.4 L SO L O 2 1 mol O 2 1 mol SO L O 2 1 mol O 2 1 mol SO 3 = 72.2 L SO 3 SHORTCUT: coefficient of O 2 = 1 coefficient of SO 3 = 2 so 36.1 L x 2 = 72.2 L SO 3

70 How many liters of CO 2 are produced from L of HCl reacting with NaHCO 3 ? NaHCO 3 + HCl NaCl + CO 2 + H 2 O L HCl x 1 mol HCl x 1 mol CO 2 x 22.4 L CO L HCl 1 mol HCl 1 mol CO L HCl 1 mol HCl 1 mol CO 2 = L CO 2 = L CO 2 SHORTCUT: coefficients = 1 mol HCl to 1 mol CO 2. so L HCl = L CO 2

71 mass –particle (particle – mass) calculations MASS – PARTICLE: GIVEN g A x 1 mol A x # mol B x 6.02 x r.p. B 1 PT g A # mol A 1 mol B 1 PT g A # mol A 1 mol B PARTICLE – MASS: GIVEN r.p. A x 1 mol A x # mol B x PT g B x r.p. A # mol A 1 mol B x r.p. A # mol A 1 mol B PT = periodic table, molar mass # = coefficients

72 How many molecules of NH 3 are produced from reacting 2.07 g of H 2 with excess N 2 ? N 2 + 3H 2 2NH 3 N 2 + 3H 2 2NH g H 2 x 1 mol H 2 x 2 mol NH 3 x 6.02 x NH g H 2 3 mol H 2 1 mol NH g H 2 3 mol H 2 1 mol NH 3 = 4.2 x molecules NH 3

73 Limiting Reactants Rarely are the reactants in a chemical reaction present in the exact mole ratios specified in the balanced equation. Usually, one or more of the reactants are present in excess, and the reaction proceeds until all of one reactant is used up.

74 The left-over reactants are called excess reactants. The reactant that is used up is called the limiting reactant.

75 Limiting Reactant To figure out the limiting reactant: You must do the stoichiometric calculation with all reactants. The one that produces the least amount of product is the limiting reactant.

76 Ex:4Al + 3O 2 2Al 2 O 3 What is the limiting reactant when 35 g of aluminum reacts with 35 g of oxygen? How much aluminum oxide is formed in this reaction? 35 g Al X 1 mol Al X 2 mol Al 2 O 3 = 1 27 g Al 4 mol Al 1 27 g Al 4 mol Al 35 g O 2 X 1 mol O 2 X 2 mol Al 2 O 3 = 1.1 mol Al 2 O g O 2 4 mol O g O 2 4 mol O 2 Aluminum is limiting reactant 0.65 mol Al 2 O 3 X 162 g Al 2 O 3 = g Al 2 O mol Al 2 O mol Al 2 O mol Al 2 O 3

77 Percent Yield theoretical yield amount of product predicted by the math (theory) actual yield amount of product obtained in lab

78 percent yield percentage of product recovered; comparison of actual and theoretical yields % YIELD = ACTUAL YIELD X 100 THEORETICAL YIELD THEORETICAL YIELD

79 35.0 g of product should be recovered from an experiment. A student collects 22.9 g at the end of the lab. What is the percent yield? 22.9 g X 100 = 65.4% 35.0 g

80 What is the percent yield if 2.89 g of NaCl is produced when 1.99 g of HCl reacts with excess NaOH? Water is the other product. HCl + NaOH NaCl + H 2 O HCl + NaOH NaCl + H 2 O Actual yield = 2.89 g NaCl Theoretical yield = ? 1.99 g HCl x 1 mol HCl x 1 mol NaCl x 58.5 g NaCl g HCl 1 mol HCl 1 mol NaCl g HCl 1 mol HCl 1 mol NaCl = THEORETICAL YIELD = 3.19 g NaCl % YIELD = 2.89 g NaCl X 100 = 90.6% 3.19 g NaCl 3.19 g NaCl

81 n The End


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