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Chapter 9 Chemical Quantities Chemistry B2A 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 B2A Formula and Molecule Ionic & covalent compounds  Formulaformula of NaCl Covalent compounds  Molecule molecule."— Presentation transcript:

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2 Chapter 9 Chemical Quantities Chemistry B2A

3 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

4 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

5 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

6 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

7 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

8 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

9 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

10 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

11 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

12 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

13 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

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:

15 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

16 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

17 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.

18 Limiting Reagents Example: C(s) + O 2 (g)  CO 2 (g) 12g of C 64g of O 2 ? Limiting reagent ? g of CO 2 will be formed Make sure that chemical equation is balanced. 12g C ( 1 mole C 12g C ) = 1.0 mole C 64g O 2 ( 1 mole O 2 32g O 2 ) = 2.0 mole O 2 C is the limiting reagent. C(s) + O 2 (g)  CO 2 (g) 1 mol

19 Limiting Reagents We should use the mass of the limiting reagent. (because it controls our reaction). 12g C ( 1 mole C 12g C )( 1 mole CO 2 1 mole C )( 44g CO 2 1 mole CO 2 ) = 44g CO 2 ? g of CO 2 will be formed: C(s) + O 2 (g)  CO 2 (g) A B 12g 64g ? g

20 Percent Yield Percent yield = actual yield theoretical yield × 100 actual yield: mass of product formed theoretical yield: mass of product that should form (according to stoichiometry) N 2 (g) + O 2 (g)  2NO(g) theoretical yield = 40.g NO actual yield = 37g NO Percent yield = 37g NO 40.g NO × 100 = 93% 7% error or lost


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