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Stoichiometry Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations John D. Bookstaver St. Charles Community College St. Peters, MO.

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Presentation on theme: "Stoichiometry Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations John D. Bookstaver St. Charles Community College St. Peters, MO."— Presentation transcript:

1 Stoichiometry Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice-Hall Chemistry, The Central Science, 10th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten

2 Stoichiometry Law of Conservation of Mass We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends. --Antoine Lavoisier, 1789

3 Stoichiometry Chemical Equations Concise representations of chemical reactions

4 Stoichiometry Anatomy of a Chemical Equation Coefficients are inserted to balance the equation. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

5 Stoichiometry Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in a molecule

6 Stoichiometry Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in a molecule Coefficients tell the number of molecules

7 Stoichiometry Reaction Types

8 Stoichiometry Combination Reactions Examples: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) C 3 H 6 (g) + Br 2 (l) C 3 H 6 Br 2 (l) 2 Mg (s) + O 2 (g) 2 MgO (s) Two or more substances react to form one product

9 Stoichiometry 2 Mg (s) + O 2 (g) 2 MgO (s) Play movie

10 Stoichiometry Decomposition Reactions Examples: CaCO 3 (s) CaO (s) + CO 2 (g) 2 KClO 3 (s) 2 KCl (s) + O 2 (g) 2 NaN 3 (s) 2 Na (s) + 3 N 2 (g) One substance breaks down into two or more substances Play air bag animation

11 Stoichiometry Combustion Reactions Examples: CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) Rapid reactions that produce a flame Most often involve hydrocarbons reacting with oxygen in the air

12 Stoichiometry Formula Weights

13 Stoichiometry Formula Weight (FW) Sum of the atomic weights for the atoms in a chemical formula So, the formula weight of calcium chloride, CaCl 2, would be Ca: 1(40.1 amu) + Cl: 2(35.5 amu) amu These are generally reported for ionic compounds

14 Stoichiometry Molecular Weight (MW) Sum of the atomic weights of the atoms in a molecule For the molecule ethane, C 2 H 6, the molecular weight would be C: 2(12.0 amu) + H: 6(1.0 amu) 30.0 amu

15 Stoichiometry Percent Composition One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation: % element = (number of atoms)(atomic weight) (FW of the compound) x 100

16 Stoichiometry Percent Composition So the percentage of carbon in ethane is… %C = (2)(12.0 amu) (30.0 amu) 24.0 amu 30.0 amu = x 100 = 80.0%

17 Stoichiometry PRACTICE EXERCISE Calculate the percentage of nitrogen, by mass, in Ca(NO 3 ) 2. Answer: 17.1 % SAMPLE EXERCISE 3.6 continued

18 Stoichiometry Moles

19 Stoichiometry Avogadros Number 6.02 x mole of 12 C has a mass of 12 g

20 Stoichiometry Molar Mass By definition, these are the mass of 1 mol of a substance (i.e., g/mol) –The molar mass of an element is the mass number for the element that we find on the periodic table –The formula weight (in amus) will be the same number as the molar mass (in g/mol)

21 Stoichiometry Using Moles Moles provide a bridge from the molecular scale to the real-world scale

22 Stoichiometry Mole Relationships One mole of atoms, ions, or molecules contains Avogadros number of those particles One mole of molecules or formula units contains Avogadros number times the number of atoms or ions of each element in the compound

23 Stoichiometry PRACTICE EXERCISE How many oxygen atoms are in (a) 0.25 mol Ca(NO 3 ) 2 and (b) 1.50 mol of sodium carbonate? Answer: (a) , (b) SAMPLE EXERCISE 3.8 continued

24 Stoichiometry Finding Empirical Formulas

25 Stoichiometry Calculating Empirical Formulas One can calculate the empirical formula from the percent composition

26 Stoichiometry Calculating Empirical Formulas The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

27 Stoichiometry Calculating Empirical Formulas Assuming g of para-aminobenzoic acid, C:61.31 g x = mol C H: 5.14 g x= 5.09 mol H N:10.21 g x= mol N O:23.33 g x = mol O 1 mol g 1 mol g 1 mol 1.01 g 1 mol g

28 Stoichiometry Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: C:= H:= N:= O:= mol mol 5.09 mol mol mol mol

29 Stoichiometry Calculating Empirical Formulas These are the subscripts for the empirical formula: C 7 H 7 NO 2

30 Stoichiometry PRACTICE EXERCISE Ethylene glycol, the substance used in automobile antifreeze, is composed of 38.7% C, 9.7% H, and 51.6% O by mass. Its molar mass is 62.1 g/mol. (a) What is the empirical formula of ethylene glycol? (b) What is its molecular formula? Answers: (a) CH 3 O, (b) C 2 H 6 O 2 SAMPLE EXERCISE 3.14 continued

31 Stoichiometry Combustion Analysis Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this –C is determined from the mass of CO 2 produced –H is determined from the mass of H 2 O produced –O is determined by difference after the C and H have been determined

32 Stoichiometry Elemental Analyses Compounds containing other elements are analyzed using methods analogous to those used for C, H and O

33 Stoichiometry SAMPLE EXERCISE 3.15 Determing Empirical Formula by Combustion Analysis Isopropyl alcohol, a substance sold as rubbing alcohol, is composed of C, H, and O. Combustion of g of isopropyl alcohol produces g of CO 2 and g of H 2 O. Determine the empirical formula of isopropyl alcohol. Solution Analyze: We are told that isopropyl alcohol contains C, H, and O atoms and given the quantities of CO 2 and H 2 O produced when a given quantity of the alcohol is combusted. We must use this information to determine the empirical formula for isopropyl alcohol, a task that requires us to calculate the number of moles of C, H, and O in the sample. Plan: We can use the mole concept to calculate the number of grams of C present in the CO 2 and the number of grams of H present in the H 2 O. These are the quantities of C and H present in the isopropyl alcohol before combustion. The number of grams of O in the compound equals the mass of the isopropyl alcohol minus the sum of the C and H masses. Once we have the number of grams of C, H, and O in the sample, we can then proceed as in Sample Exercise 3.13: Calculate the number of moles of each element, and determine the mole ratio, which gives the subscripts in the empirical formula. Solve: To calculate the number of grams of C, we first use the molar mass of CO 2, 1 mol CO 2 = 44.0 g CO 2, to convert grams of CO 2 to moles of CO 2. Because there is only 1 C atom in each CO 2 molecule, there is 1 mol of C atoms per mole of CO 2 molecules. This fact allows us to convert the moles of CO 2 to moles of C. Finally, we use the molar mass of C, 1 mol C = 12.0 g C, to convert moles of C to grams of C. Combining the three conversion factors, we have:

34 Stoichiometry SAMPLE EXERCISE 3.15 continued The calculation of the number of grams of H from the grams of H 2 O is similar, although we must remember that there are 2 mol of H atoms per 1 mol of H 2 O molecules: The total mass of the sample, g, is the sum of the masses of the C, H, and O. Thus, we can calculate the mass of O as follows: We then calculate the number of moles of C, H, and O in the sample: To find the empirical formula, we must compare the relative number of moles of each element in the sample. The relative number of moles of each element is found by dividing each number by the smallest number, The mole ratio of C : H : O so obtained is 2.98 : 7.91 : The first two numbers are very close to the whole numbers 3 and 8, giving the empirical formula C 3 H 8 O. Check: The subscripts work out to be moderately sized whole numbers, as expected.

35 Stoichiometry PRACTICE EXERCISE (a) Caproic acid, which is responsible for the foul odor of dirty socks, is composed of C, H, and O atoms. Combustion of a g sample of this compound produces g CO 2 and g H 2 O. What is the empirical formula of caproic acid? (b) Caproic acid has a molar mass of 116 g/mol. What is its molecular formula? Answers: (a) C 3 H 6 O, (b) C 6 H 12 O 2 SAMPLE EXERCISE 3.15 continued

36 Stoichiometry Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products

37 Stoichiometry Stoichiometric Calculations From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if its a product) or used (if its a reactant)

38 Stoichiometry Limiting Reactants

39 Stoichiometry How Many Cookies Can I Make? You can make cookies until you run out of one of the ingredients Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat)

40 Stoichiometry How Many Cookies Can I Make? In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make

41 Stoichiometry Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount Play animation

42 Stoichiometry Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount –In other words, its the reactant youll run out of first (in this case, the H 2 )

43 Stoichiometry Limiting Reactants In the example below, the O 2 would be the excess reagent

44 Stoichiometry SAMPLE EXERCISE 3.19 Calculating the Amount of Product Formed from a Limiting Reactant Consider the following reaction: Solution Analyze: We are asked to calculate the amount of a product, given the amounts of two reactants, so this is a limiting reactant problem. Suppose a solution containing 3.50 g of Na 3 PO 4 is mixed with a solution containing 6.40 g of Ba(NO 3 ) 2. How many grams of Ba 3 (PO 4 ) 2 can be formed? Plan: We must first identify the limiting reagent. To do so, we can calculate the number of moles of each reactant and compare their ratio with that required by the balanced equation. We then use the quantity of the limiting reagent to calculate the mass of Ba(PO 4 ) 2 that forms. Using the molar mass of each substance, we can calculate the number of moles of each reactant: Solve: From the balanced equation, we have the following stoichiometric relations:

45 Stoichiometry SAMPLE EXERCISE 3.19 continued Thus, there are slightly more moles of Ba(NO 3 ) 2 than moles of Na 3 PO 4 The coefficients in the balanced equation indicate, however, that the reaction requires 3 mol Ba(NO 3 ) 2 for each 2 mol Na 3 PO 4 [That is, 1.5 times more moles of Ba(NO 3 ) 2 are needed than moles of Na 3 PO 4.] Thus, there is insufficient Ba(NO 3 ) 2 to completely consume the Na 3 PO 4 That means that Ba(NO 3 ) 2 is the limiting reagent. We therefore use the quantity of Ba(NO 3 ) 2 to calculate the quantity of product formed. We can begin this calculation with the grams of Ba(NO 3 ) 2 but we can save a step by starting with the moles of Ba(NO 3 ) 2 that were calculated previously in the exercise: Check: The magnitude of the answer seems reasonable: Starting with the numbers in the two conversion factors on the right, we have 600/3 = 200; = 5. The units are correct, and the number of significant figures (three) corresponds to the number in the quantity of Ba(NO 3 ) 2. Comment: The quantity of the limiting reagent, Ba(NO 3 ) 2 can also be used to determine the quantity of NaNO 3 formed (4.16 g) and the quantity of Na 3 PO 4 used (2.67 g). The number of grams of the excess reagent, Na 3 PO 4 remaining at the end of the reaction equals the starting amount minus the amount consumed in the reaction, 3.50 g 2.67 g = 0.82 g.

46 Stoichiometry PRACTICE EXERCISE A strip of zinc metal having a mass of 2.00 g is placed in an aqueous solution containing 2.50 g of silver nitrate, causing the following reaction to occur: Answers:(a) AgNO 3, (b) 1.59 g, (c) 1.39 g, (d) 1.52 g Zn (a) Which reactant is limiting? (b) How many grams of Ag will form? (c) How many grams of Zn(NO 3 ) 2 will form? (d) How many grams of the excess reactant will be left at the end of the reaction? SAMPLE EXERCISE 3.19 continued

47 Stoichiometry Theoretical Yield The theoretical yield is the amount of product that can be made –In other words its the amount of product possible as calculated through the stoichiometry problem This is different from the actual yield, the amount one actually produces and measures

48 Stoichiometry SAMPLE EXERCISE 3.20 Calculating the Theoretical Yield and Percent Yield for a Reaction Adipic acid, H 2 C 6 H 8 O 4, is used to produce nylon. The acid is made commercially by a controlled reaction between cyclohexane (C 6 H 12 ) and O 2 : Solution Analyze: We are given a chemical equation and the quantity of the limiting reactant (25.0 g of C 6 H 12 ). We are asked first to calculate the theoretical yield of a product (H 2 C 6 H 8 O 4 ) and then to calculate its percent yield if only 33.5 g of the substance is actually obtained. (a) Assume that you carry out this reaction starting with 25.0 g of cyclohexane and that cyclohexane is the limiting reactant. What is the theoretical yield of adipic acid? (b) If you obtain 33.5 g of adipic acid from your reaction, what is the percent yield of adipic acid? (a) The theoretical yield, which is the calculated quantity of adipic acid formed in the reaction, can be calculated using the following sequence of conversions: Plan: (b) The percent yield is calculated by comparing the actual yield (33.5 g) to the theoretical yield using Equation 3.14.

49 Stoichiometry SAMPLE EXERCISE 3.20 continued Solve: Check: Our answer in (a) has the appropriate magnitude, units, and significant figures. In (b) the answer is less than 100% as necessary.

50 Stoichiometry Answers: (a) 105 g Fe, (b) 83.7% (a) If you start with 150 g of Fe 2 O 3 as the limiting reagent, what is the theoretical yield of Fe? (b) If the actual yield of Fe in your test was 87.9 g, what was the percent yield? PRACTICE EXERCISE Imagine that you are working on ways to improve the process by which iron ore containing Fe 2 O 3 is converted into iron. In your tests you carry out the following reaction on a small scale: SAMPLE EXERCISE 3.20 continued

51 Stoichiometry Percent Yield A comparison of the amount actually obtained to the amount it was possible to make Actual Yield Theoretical Yield Percent Yield =x 100


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