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Composition Stoichiometry. Mass Percentage mass of element in compound molar mass of compound x 100%

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Presentation on theme: "Composition Stoichiometry. Mass Percentage mass of element in compound molar mass of compound x 100%"— Presentation transcript:

1 Composition Stoichiometry

2 Mass Percentage mass of element in compound molar mass of compound x 100%

3 Percent Composition The sum of the mass percents of all elements that make up the compound. % A + %B + … + … = 100%

4 Determining Chemical Formulas

5 Mass Percentage mass of element in compound molar mass of compound x 100%

6 Percent Composition The sum of the mass percents of all elements that make up the compound. % A + %B + … + … = 100%

7 Determining Chemical Formulas

8 Empircal (Simplest) Formula The smallest whole-number ratio of atoms (moles) in the compound Calculated from composition data Empirical formula is accepted formula for ionic compounds

9 Determining Empirical Formula Convert composition data to masses in grams –Assume 100g of substance ( if given percent composition) Convert masses in grams to amounts in moles Find the simplest whole-number ratio

10 CH % C, 14.4%H C2H4C2H4 C4H8C4H8 C 6 H % C, 14.4%H

11 Calculating Molecular Formula Empirical FormulaMolecular Formula: Empirical Formula MassMolecular Mass : The molecular formula is proportional to the empirical formula The molecular formula mass must be proportional to the empirical formula mass

12 Calculating Molecular Formula Determine the empirical formula Calculate empirical formula mass Determine ratio of empirical formula mass to molecular mass –molecular mass is experimentally determined Apply the ratio to the simplest formula

13 Reaction Stoichiometry

14 Balanced Formula Equations CH 4 (g) + O 2 (g)  CO 2 (g) + H 2 O (g) 1 atom C 4 atoms H 2 atoms O 1 atom C 2 atoms H 3 atoms O CH 4 (g) + O 3 (g)  CO 2 (g) + H 4 O (g) CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (g)

15 The Balanced Formula Equation CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (g) Establishes relationships among reactants and products Can compare two species at once Mole ratios Mass relationships derive from the mole relationships

16 Basic Stoichometric Relationships Mole-to-mole moles G  moles W Mole-to mass moles G  moles W  mass W

17 Basic Stoichometric Relationships Mass-to-mole mass G  moles G  moles W Mass-to-mass mass G  moles G  moles W  mass W

18 Basic Stoichometric Relationships Can convert from G  W only in moles Must use molar mass to convert G or W into moles

19 Limiting Reagent and Excess Reagent

20 Definitions Limiting Reagent –Reactant that runs out first –Causes reaction to stop –Limits the amount of product Excess Reagent –Reactant left over when reaction stops

21 How can I tell which one is the limiting reagent?

22 Finding the Limiting Reagent *How much do I have? *How much do I need?

23 Determining Limiting Reactant “The Easy Way” Step 1: Find quantities of all reactants in moles –convert from grams, if necessary

24 Determining Limiting Reactant “The Easy Way” Step 2: Divide each reactant by its coefficient from the balanced equation –Tells “how many” reactions can be fueled by that reactant Coefficient is moles per rxn

25 Determining Limiting Reactant “The Easy Way” Step 3: The smallest quotient from Step 2 is LR –can “fuel” the fewest number of reactions before it runs out

26 How can I tell how much of the excess reagent is left over?

27 Unreacted Excess Reactant Step 1: Determine LR –as shown previously Step 2: Use given quantity of LR as “starting point” Step 3: Calculate amount of ER needed to “use up” the LR Step 4: Subtract amount needed from amount available

28 Theoretical Yield Actual Yield Per Cent Yield

29 2H 2 + O 2  2H 2 O Theoretical Yield –Maximum amount of product you can possibly get –Based on the chemical equation –Determined by the limiting reactant

30 Determining the Theoretical Yield Find the limiting reactant Use the given amount of the limiting reactant to calculate product –Do a regular stoichiometry problem

31 Per Cent Yield Compares actual yield to theoretical yield Measures the efficiency of the reaction Will usually be less than 100%

32 Actual Yield The measured amount of product actually obtained from the experiment –Cannot be calculated from the chemical equation –Must be determined in the lab –Usually less than theoretical yield

33 Per Cent Yield % Yield = x 100% Part Whole Actual Theo.

34 Empircal (Simplest) Formula The smallest whole-number ratio of atoms (moles) in the compound Calculated from composition data Empirical formula is accepted formula for ionic compounds

35 Determining Empirical Formula Convert composition data to masses in grams –Assume 100g of substance ( if given percent composition) Convert masses in grams to amounts in moles Find the simplest whole-number ratio

36 CH % C, 14.4%H C2H4C2H4 C4H8C4H8 C 6 H % C, 14.4%H

37 Calculating Molecular Formula Empirical FormulaMolecular Formula: Empirical Formula MassMolecular Mass : The molecular formula is proportional to the empirical formula The molecular formula mass must be proportional to the empirical formula mass

38 Calculating Molecular Formula Determine the empirical formula Calculate empirical formula mass Determine ratio of empirical formula mass to molecular mass –molecular mass is experimentally determined Apply the ratio to the simplest formula

39 Reaction Stoichiometry

40 Balanced Formula Equations CH 4 (g) + O 2 (g)  CO 2 (g) + H 2 O (g) 1 atom C 4 atoms H 2 atoms O 1 atom C 2 atoms H 3 atoms O CH 4 (g) + O 3 (g)  CO 2 (g) + H 4 O (g) CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (g)

41 The Balanced Formula Equation CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (g) Establishes relationships among reactants and products Can compare two species at once Mole ratios Mass relationships derive from the mole relationships

42 Basic Stoichometric Relationships Mole-to-mole moles G  moles W Mole-to mass moles G  moles W  mass W

43 Basic Stoichometric Relationships Mass-to-mole mass G  moles G  moles W Mass-to-mass mass G  moles G  moles W  mass W

44 Basic Stoichometric Relationships Can convert from G  W only in moles Must use molar mass to convert G or W into moles

45 Limiting Reagent and Excess Reagent

46 Definitions Limiting Reagent –Reactant that runs out first –Causes reaction to stop –Limits the amount of product Excess Reagent –Reactant left over when reaction stops

47 How can I tell which one is the limiting reagent?

48 Finding the Limiting Reagent *How much do I have? *How much do I need?

49 Determining Limiting Reactant “The Easy Way” Step 1: Find quantities of all reactants in moles –convert from grams, if necessary

50 Determining Limiting Reactant “The Easy Way” Step 2: Divide each reactant by its coefficient from the balanced equation –Tells “how many” reactions can be fueled by that reactant Coefficient is moles per rxn

51 Determining Limiting Reactant “The Easy Way” Step 3: The smallest quotient from Step 2 is LR –can “fuel” the fewest number of reactions before it runs out

52 How can I tell how much of the excess reagent is left over?

53 Unreacted Excess Reactant Step 1: Determine LR –as shown previously Step 2: Use given quantity of LR as “starting point” Step 3: Calculate amount of ER needed to “use up” the LR Step 4: Subtract amount needed from amount available

54 Theoretical Yield Actual Yield Per Cent Yield

55 2H 2 + O 2  2H 2 O Theoretical Yield –Maximum amount of product you can possibly get –Based on the chemical equation –Determined by the limiting reactant

56 Determining the Theoretical Yield Find the limiting reactant Use the given amount of the limiting reactant to calculate product –Do a regular stoichiometry problem

57 Per Cent Yield Compares actual yield to theoretical yield Measures the efficiency of the reaction Will usually be less than 100%

58 Actual Yield The measured amount of product actually obtained from the experiment –Cannot be calculated from the chemical equation –Must be determined in the lab –Usually less than theoretical yield

59 Per Cent Yield % Yield = x 100% Part Whole Actual Theo.


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