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Chapter 9 Chemical Quantities Chemistry 101. Formula and Molecule Ionic & covalent compounds  Formulaformula of NaCl Covalent compounds  Molecule molecule.

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Presentation on theme: "Chapter 9 Chemical Quantities Chemistry 101. Formula and Molecule Ionic & covalent compounds  Formulaformula of NaCl Covalent compounds  Molecule molecule."— Presentation transcript:

1 Chapter 9 Chemical Quantities Chemistry 101

2 Formula and Molecule Ionic & covalent compounds  Formulaformula of NaCl Covalent compounds  Molecule molecule of H 2 O Formula Weight of NaCl: 23 amu Na amu Cl = 58.5 amu NaCl Molecular Weight of H 2 O: 2 (1 amu H) + 16 amu O = 18 amu H 2 O

3 Mole Mole: formula weight of a substance (in gram). 12g of C = 1 mol C 23g of Na = 1 mol Na 58.5 g of NaCl = 1 mol NaCl 18 g of H 2 O = 1 mol of H 2 O

4 Avogadro’s number (6.02×10 23 ): number of formula units in one mole. 1 mole of apples = 6.02×10 23 apples 1 mole of A atoms = 6.02×10 23 atoms of A 1 mole of A molecules = 6.02×10 23 molecules of A 1 mole of A ions = 6.02×10 23 ions of A Molar mass (g/mol): mass of 1 mole of substance (in gram) (Formula weight) molar mass of Na = 23 g/mol molar mass of H 2 O = 18 g/mol

5 Relationships between amounts of substances in a chemical reaction. Look at the Coefficients! Stoichiometry 2H 2 O (l)  2H 2 (g) + O 2 (g) 2 moles 1 mole liters 1 liter 2 particles 1 particle 2 grams 1 gram 2  6.02  molecules 1  6.02  molecules

6 A mole B mole mass volume Particle (atom) (molecule) (ion) Particle (atom) (molecule) (ion) mass CH 4 + 2O 2  CO 2 + 2H 2 O 1 step: use coefficient in the balanced equation

7 1 Step Mole A  Mole B Volume A  Volume B # of Particles A  # of Particles B A mole B mole mass volume Particle (atom) (molecule) (ion) Particle (atom) (molecule) (ion) mass

8 CH 4 + 2O 2  CO 2 + 2H 2 O 23 mole CH 4 = ? moles H 2 O 23 mole CH 4 ( 2 moles H 2 O 1 mole CH 4 ) = 46 moles H 2 O A B 10 cc O 2 = ? cc CO 2 10 cc O 2 ( 1 cc CO 2 2 cc O 2 ) = 5 cc CO 2 A B 2  molecules H 2 O = ? molecules O 2 2  molecules H 2 O ( ) = 2  molecules O 2 2  (6.02  molecules H 2 O) 2  (6.02  molecules O 2 ) A B

9 2 Steps Mole A  Volume B Mass A  Mole B or Volume A # of Particles A  Mole B or Volume A A mole B mole mass volume Particle (atom) (molecule) (ion) Particle (atom) (molecule) (ion) mass

10 CH 4 + 2O 2  CO 2 + 2H 2 O 32 g CH 4 = ? moles CO 2 32 g CH 4 ( 1 mole CH 4 16 g CH 4 ) = 2.0 mole CO 2 1 mole CO 2 1 mole CH 4 )( B A 40. g CH 4 = ? L CH g CH 4 ( 1 mole CH 4 16 g CH 4 ) = 56 L CH L CH 4 1 mole CH 4 )( STP: 1 mole of substance (gas) = 22.4 L = cc (cm 3 or mL) AA 5 moles CO 2 = ? molecules O 2 5 moles CO 2 ( 2 mole O 2 ) = 6  molecules O 2 1 mole CO 2 1 mole O 2 )( B A 6.02  molecules O 2

11 3 Steps Mass A  Mass B Mass A  Volume B or # of Particles B # of Particles A  Volume B A mole B mole mass volume Particle (atom) (molecule) (ion) Particle (atom) (molecule) (ion) mass

12 CH 4 + 2O 2  CO 2 + 2H 2 O 46.0 g CH 4 = ? g H 2 O B A 46.0 g CH 4 ( 1 mole CH 4 16 g CH 4 )( ) = 104 g H 2 O 2 mole H 2 O 1 mole CH 4 )( 1 mole H 2 O 18 g H 2 O

13 Limiting Reagents N 2 (g) + O 2 (g)  2NO(g) 1 mole 2 moles 1 mole 4 moles 0 mole 3 moles 2 moles Before reaction: After reaction: Stoichiometry:

14 Limiting Reagents N 2 (g) + O 2 (g)  2NO(g) 1 mole 2 moles 1 mole 4 moles 0 mole 3 moles 2 moles Before reaction: After reaction: Stoichiometry: Used up first Left over

15 Limiting Reagents N 2 (g) + O 2 (g)  2NO(g) 1 mole 2 moles 1 mole 4 moles Limiting reagent 0 mole 3 moles 2 moles Before reaction: After reaction: Stoichiometry: Used up first Left over

16 Limiting Reagents Limiting reagents can control a reaction: N 2 (g) + O 2 (g)  2NO(g) Limiting reagent: is the reactant that is used up first.

17 To solve these problems, follow these steps: –Convert reactant A to moles of C (or D). –Convert reactant B to moles of C (or D). –Compare moles of C (or D) produced by A and B. Mole B Mole C Mole AMass A Mass B Mole CMass C Compare: lesser amount results from limiting reactant. Limiting Reagents

18 2 Al(s) + 3 Cl 2 (g)  2 AlCl 3 (s) A chemist combines 10.0 g of each reactant. Which is the limiting reagent? 10.0 g Al xx= mol AlCl 3 1 mol Al g Al 2 mol AlCl 3 2 mol Al 10.0 g Cl 2 xx= mol AlCl 3 1 mol Cl g Cl 2 2 mol AlCl 3 3 mol Cl 2 The smallest number is the result of the limiting reactant. So, Cl 2 is the limiting reactant. Example:

19 Limiting Reagents 2. What mass of product is produced (in g)? 2 Al(s) + 3 Cl 2 (g)  2 AlCl 3 (s) 3.What species are present in the final mixture? AlCl 3 (s) and some left-over Al(s) are present mol AlCl 3 x = g AlCl g AlCl 3 1 mol AlCl 3

20 4. How much Al(s) remains? 2 Al(s) + 3 Cl 2 (g)  2 AlCl 3 (s) 10.0 g original – 2.54 g used = 7.46 g Al(s) remains x = 2.54 g Al (reacted) 1 mol Cl g Cl 2 2 mol Al 3 mol Cl 2 x g Al 1 mol Al Limiting Reagents 10.0 g Cl 2 x Limiting reagent

21 C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(g) 1.A chemist combines 10.0 g of each reactant. Which is the limiting reagent? 2.What mass of each product is produced (in g)? 3.What species are present in the final mixture? Limiting Reagents Example:

22 Limiting Reagents C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(g) 1.A chemist combines 10.0 g of each reactant. Which is the limiting reagent? 10.0 g C 3 H 8 xx= mol CO 2 1 mol C 3 H g C 3 H 8 3 mol CO 2 1 mol C 3 H g O 2 xx= mol CO 2 1 mol O g O 2 3 mol CO 2 5 mol O 2 The smallest number is the result of the limiting reactant. So, O 2 is the limiting reactant.

23 Limiting Reagents C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(g) 2.What mass of each product is produced (in g)? 3.What species are present in the final mixture? CO 2 (g), H 2 O(g) and some remaining C 3 H 8 (g) mol CO 2 x= 8.27 g CO g CO 2 1 mol CO g O 2 x Limiting reagent xx= 4.50 g H 2 O 1 mol O g O 2 4 mol H 2 O 5 mol O g H 2 O 1 mol H 2 O

24 CS 2 (l) + 3 O 2 (g)  CO 2 (g) + 2 SO 2 (g) You have 1.55 g of CS 2 (l) and excess O 2 (g). What mass of each product forms? Calculations of this type are important! Practice on your own with ALEKS, or textbook p 209d, #24 + # g CS 2 xxx= g CO 2 1 mol CS g CS 2 1 mol CO 2 1 mol CS g CO 2 1 mol CO g CS 2 xxx= 2.61 g SO 2 1 mol CS g CS 2 2 mol SO 2 1 mol CS g SO 2 1 mol SO 2 Limiting Reagents Example:

25 Percent Yield actual yield: mass of product formed (exprimental determination) theoretical yield: mass of product that should form according to limiting reactant calculation. (according to stoichiometry) Percentage yield: a comparison of actual to theoretical yield. Note: percentage yields can be calculated using units of either moles or grams. Actual yield Theoretical yield x 100% Percent yield =

26 Analysis for sulfate (SO 4 2- ) uses barium cation (Ba 2+ ). SO 4 2- (aq) + BaCl 2 (aq)  BaSO 4 (s) + 2 Cl - (aq) If a sample containing 1.15 g of SO 4 2- is reacted with excess barium chloride, how much BaSO 4 should form? If a chemist actually collects 2.02 g BaSO 4, what is the percent yield? Molar masses: SO 4 2- = g/mol. BaSO 4 = g/mol. Percent Yield Example:

27 Percent Yield How much BaSO 4 should form? SO 4 2- (aq) + BaCl 2 (aq)  BaSO 4 (s) + 2 Cl - (aq) If a chemist actually collects 2.02 g BaSO 4, what is the percent yield? 1.15 g SO 4 2- xxx= 2.79 g BaSO 4 1 mol SO g SO mol BaSO 4 1 mol SO g BaSO 4 1 mol BaSO 4 = 72.4 % 2.02 g BaSO g BaSO 4 x 100%

28 At-Home Practice Practice problem: A chemist reacts 7.67 g of H 2 gas with g of O 2 gas. What is the theoretical yield of water in grams? (Limiting reactant calculation.) The actual experimentally measured amount is only 28.6 g of H 2 O. What is the percent yield? Next lecture, our clicker questions will be:  What is the theoretical yield?  What is the percent yield?


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