Chapter 6 Chemical Calculations

Chapter 6 Chemical Calculations 1 Submicroscopic Macroscopic

2 Chapter Outline 1. Formula Masses (Ch 6.1) 2. Percent Composition (supplemental material) 3. The Mole & Avogadro s Number (Ch 6.2) 4. Molar Mass (Ch 6.3) 5. Chemical Formula & Mole Calculations (Ch 6.4 & 6.5) 6. Empirical & Molecular Formula (supplemental material & Lab Exp 8) 7. Chemical Equations & Stoichiometric Calculations (Ch 6.6, 6.7, 6.8) 8. Heat of Reaction (supplemental material) 9. Percent Yield (supplemental material) 10. Limiting Reagent (supplemental material)

3 1. Formula Masses (Ch 6.1) Molecular mass/weight sum of atomic masses of the atoms in a molecule. Example: M.W. = 46.0 amu C: 2 x 12.0 amu = 24.0 amu H: 6 x 1.0 amu = 6.0 amu O: 1 x 16.0 amu = 16.0 amu 46.0 amu

4 Formula mass/weight sum of atomic masses of atoms in an ionic substance (formula unit, NOT a molecule). Example: Ammonium sulfide F.W. = 68.1 amu N: 2 x 14.0 amu = 28.0 amu H: 8 x 1.0 amu = 8.0 amu S: 1 x 32.1 amu = 32.1 amu 68.1 amu Due to its offensive smell, ammonium sulfide it is the active ingredient in a variety of foul pranks, including the common stink bomb.

5 2. Percent Composition (supplemental material) % Composition = # of g of each element in 100 g of a compound = mass of element x 100 total mass Example: Ammonium sulfide (NH 4 ) 2 S F.W. = 68.1 amu % N = 2 x 14.0 amu N x 100 = 41.111 % N 68.1 amu (NH 4 ) 2 S % H = 8 x 1.0 amu H x 100 = 11.747 % H 68.1 amu (NH 4 ) 2 S % S = 32.1 amu S x 100 = 47.136 % S 68.1 amu (NH 4 ) 2 S

6 3. The Mole & Avogadro s Number (Ch 6.2) 1 dozen = 12 objects 1 ream = 500 sheets 1 mole = objects 6.0221415 x 10 23 Avogadro s Number

7 Sample problem: How many He atoms are in 2.55 moles of He? information x = information given factor sought 2.55 mole He x 6.02 x 10 23 He atoms = He atoms 1 mole He

8 4. Molar Mass (Ch 6.3) 1 mole 12 C = exactly 12 g 12 C demo sample Molar mass of 12 C = We now have three conversion factors to relate moles, number of atoms, and mass of Carbon: 6.02 x 10 23 C atoms 12 g C 1 mole 1 mole 6.02 x 10 23 C atoms 12 g C

9 Figure 6.5 1 mole of S, Zn, C, Mg, Pb, Si, Cu, Hg (start counterclockwise from yellow S and Hg in center) 12 C is our standard and when we compare it to other elements, we find that there are Avogadro s number of atoms of any element in a sample whose mass in grams is numerically equal to its atomic weight. Mg = 24.31 amu 1 mole Mg = 24.31 g Mg = 6.02 x 10 23 atoms Mg Pb = amu 1 mol Pb = g Pb = 6.02 x 10 23 atoms Pb

10 Now we can do the same for ionic compounds as well as for molecules because the molar mass is the mass (in grams) of a substance that is numerically equal to the substance s formula mass. Ammonium sulfide (NH 4 ) 2 S formula mass = 68.1 amu 68.1 g (NH 4 ) 2 S = 1 mole (molar mass or formula weight, F.W.: ) 68.1 g (NH 4 ) 2 S = 6.02 x 10 23 formula units of (NH 4 ) 2 S Carbon dioxide CO 2 formula mass = 44.01 amu 44.01 g CO 2 = 1 mole (molar mass or molecular weight, M.W.: ) 44.01 g CO 2 = 6.02 x 10 23 molecules of CO 2

11 Sample Problem: If 7.50 moles of ammonia, NH 3, are required for a certain experiment, what mass of ammonia is needed? Formula mass = 3 x 1.0 (H) + 14.0 (N) = 17.0 amu 1 mole NH 3 (molar mass) = 17.0 g NH 3 7.50 moles NH 3 x 17.0 g NH 3 = g NH 3 1 mole NH 3

12 5. Chemical Formula & Mole Calculations (Ch 6.4 & 6.5) The subscripts in a chemical formula give the number of moles of atoms present in 1 mole of the substance: Example: Ammonium sulfide (NH 4 ) 2 S For N: 2 moles of N atoms or 1 mole (NH 4 ) 2 S 1 mole (NH 4 ) 2 S formula units moles N For H: 8 moles of H atoms or 1 mole (NH 4 ) 2 S 1 mole (NH 4 ) 2 S formula units moles H For S: 1 mole of S atoms or 1 mole (NH 4 ) 2 S 1 mole (NH 4 ) 2 S formula units mole S

13 Sample calculation: How many H atoms are in 35.6 g of (NH 4 ) 2 S? (NH 4 ) 2 S formula mass = 68.1 amu 1 mole (NH 4 ) 2 S = 68.1 g 1 mole (NH 4 ) 2 S = moles H atoms 1 mole H atoms = 6.02 x 10 23 H atoms Strategy: mass (NH 4 ) 2 S moles (NH 4 ) 2 S moles H atoms H 35.6 g (NH 4 ) 2 S x 1 mol (NH 4 ) 2 S x 8 mol H x 6.02 x 10 23 H 68.1 g (NH 4 ) 2 S 1 mol (NH 4 ) 2 S 1 mol H = x 10 H atoms

14 Fig 6.7 Transitions allowed in solving chemical-formula bases problems: Drill problem: How many g of (NH 4 ) 2 S are required to obtain moles of NH 4 +? (NH 4 ) 2 S = 68.1 g/mol (molar mass) moles NH 4 + mole (NH 4 ) 2 S g (NH 4 ) 2 S 0.50 moles NH 4 + x 1 mole (NH 4 ) 2 S + 2 moles NH 4 x 68.1 g (NH 4 ) 2 S 1 mol (NH 4 ) 2 S = (NH 4 ) 2 S

15 6. Empirical & Molecular Formula (supplemental material & Lab Exp 8) C 6 H 12 O 6 Molecular Formula CH 2 O Empirical Formula The Empirical Formula (E.F.) is the simplest ratio of atoms in a compound. acid C 2 H 4 O 2 CH 2 O Molecular Formula Empirical Formula Both compounds have the same composition: 40.0 % C, 6.7 % H, 53.3 % O

16 If we are given the experimentally determined composition of a compound, we can calculate the E.F. Sample calculation. Composition: 40.0 % C, 6.7 % H, 53.3 % O Step 1: assume a 100 g sample and convert to For C: 40.0 g C x 1 mole C = 3.33 moles C 12.0 g C For H: 6.7 g H x 1 mole H = 6.7 moles H 1.0 g H For O: 53.3 g O x 1 mole O = 3.33 moles O 16.0 g O Step 2: divide each number of moles by the smallest of the numbers to obtain mole ratios = E.F. For C: 3.33/3.33 = 1.0 For H: 6.7/3.33 = 2.0 For O: 3.33/3.33 = 1.0 Empirical Formula =

17 Drill problem: Composition of Borazole = 40.28%B, 52.20%N, 7.52%H Molar mass = 80.5 amu Calculated the molecular formula B: 40.28g x 1 mole = 3.726 moles = 1.0 10.81 g 3.726 N: 52.20g x 1 mole = 3.726 moles = 1.0 14.01 3.726 H: 7.52g x 1 mole = 7.45 moles = 2.0 1.01 g 3.726 E.F.= E.F. mass: 10.81 + 14.01 + (2x1.01) = 26.84 Molar mass = amu = 3 E.F. mass amu Molecular Formula = 3 x (BNH 2 ) = B 3 N 3 H 6 Borazole

18 Mole ratios must be within 0.1 of a whole number. If they are not, each result must be multiplied by the same multiplication factor until every value is + 0.1 of a whole number. Example for a hypothetical set of mole ratios obtained from % composition: 2.247 for C 1.98 for H 1.000 for O The result for C is not + 0.1 of a whole number; therefore, each result must be multiplied by an integer until all of the values are. For the above example, the multiplication factor that works is 4: 2.247 x 4 = 8.988 for C 1.98 x 4 = 7.92 for H 1.000 x 4 = 4.000 for O Empirical Formula =

7. Chemical Equations & Stoichiometric Calculations (Ch 6.6, 6.7, 6.8) Summary of submicroscopic and macroscopic levels of a chemical equation: 2 Na(s) + 2 H 2 O 2 NaOH(aq) + H 2 (g) 2 atoms + 2 molecules 2 formula units + 1 molecule 2 moles + 2 moles 2 moles + 1 mole 19 2x23=46g + 2x18=36g 2x40=80g + 1x2=2g reactants products Law of Conservation of Mass

20 The coefficients in a balanced equation give the numerical relationships among formula units consumed or produced in a chemical reaction. Keys to the calculations = N 2 + 3 H 2 2 NH 3 1 mole N 2 3 moles H 2 2 moles NH 3 3 moles H 2 1 mole N 2 1 mole N 2 1 mole N 2 3 moles H 2 2 moles NH 3 2 moles NH 3 2 moles NH 3 3 mole H 2

21 Figure 6.9 In solving stoichiometric calculations, only the following transitions are allowed: Sample calculation: How many g O 2 are needed to convert 45 g glucose (C 6 H 12 O 6 ) into CO 2 and H 2 O? First write a balanced equation and calculated the molecular masses of glucose and oxygen: M.W. C 6 H 12 O 6 + 6 O 2 6 CO 2 180 amu + 32 amu Then set up the calculations according to Figure 6.9 + 6 H 2 O 45 g glu x 1 mol glu x mol O 2 x 32 g O 2 180 g glu 1 mol glu 1 mol O 2 = O 2

22 Drill Problem: Figure 6.10 The chemical equation for the deployment of airbags is 2 NaN 3 (s) 2 Na(s) + 3 N 2 (g) How many g NaN 3 would have to decompose on order to generate 253 million molecules of N 2? F.W. NaN 3 = 65.0 amu Avogadro s # = 6.02 x 10 23 molecules N 2 moles N 2 moles NaN 3 g NaN 3 2.53x10 8 molecules N 2 x 1 mole N 2 = 4.203 x 10-16 moles N 2 6.02 x 10 23 molecules N 2 4.203 x 10-16 moles N 2 x moles NaN 3 x 65.0 g NaN 3 moles N 2 mol NaN 3 = x 10-14 g NaN 3

23 8. Heat of Reaction (Ch 9.5 & supplemental material) When the chemical energy stored in reactants is greater than that stored in the products, energy is released by the reaction, and it is termed exothermic. The change in energy is called the enthalpy change and is represented by H; the value is negative for an exothermic reaction. Example: combustion of propane is exothermic. The heat released can be represented with energy on the product side: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(l) + 530 kcal However, the reaction is also represented by placing the enthalpy change to the right of the equation: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(l) H =

24 An endothermic reaction is a chemical reaction in which a continuous input of energy is needed for the reaction to occur. Energy is a reactant. Example: photosynthesis is endothermic H = positive 6 CO 2 (g) + 6 H 2 O(g) + 678 kcal C 6 H 12 O 6 (aq) + 6 O 2 (g) 6 CO 2 (g) + 6 H 2 O(g) C 6 H 12 O 6 (aq) + 6 O 2 (g) H =

25 Sample Problem: How much energy is produced when 0.50 g of butane, C 4 H 10, is burned in a butane lighter? C 4 H 10 = 58.1 g/mole 2 C 4 H 10 (g) + 13 O 2 (g) 8 CO 2 (g) + 10 H 2 O(l) + 1365 kcal The equation shows that 1365 kcal of heat are produced when 2 moles of butane undergo combustion. 0.50g C 4 H 10 x 1 mol C 4 H 10 x 1365 kcal = kcal 58.1 g C 4 H 10 moles C 4 H 10 Is this reaction exothermic or endothermic? What is the sign for H?

26 9. Percent Yield (supplemental material) The theoretical yield in a reaction is the amount of product that could be obtained if a given reactant reacted completely. In most cases the actual yield is smaller that the theoretical yield because of side reactions, incomplete reaction, or other experimental limitations. The discrepancy between the theoretical yield and the actual yield is reported as the percent yield, which is calculated as shown: % yield = actual yield x 100 theoretical yield The theoretical yield is calculated from the given amount of the specified reactant. The actual yield is identified in the problem.

27 Fe 2 O 3 + 3 C 2 Fe + 3 CO 1. Calculate the theoretical yield: g Fe 2 O 3 mol Fe 2 O 3 mol Fe g Fe Pure iron can be produced from iron oxide in a blast furnace. Sample Problem. When 884 g of Fe 2 O 3 was reduced with excess carbon, 507 g of Fe were obtained. What was the percent yield? 884g Fe 2 O 3 x 1 mol Fe 2 O 3 x 2 moles Fe x 55.85g Fe = g Fe 159.7g Fe 2 O 3 1 mol Fe 2 O 3 1 mol Fe 2. Calculate the % yield: % yield = actual yield x 100 = 507 g x 100 = % theoretical yield g

28 10. Limiting Reagent (supplemental material) Two batteries are required for these flashlights to work. So, if you have 10 flashlights and 17 batteries, how many working flashlights do you have? The batteries are the LIMITING REAGENT. The flashlights are IN EXCESS.

29 Reaction A: Stoichiometric amounts of reactants are used. NaOH + HCl NaCl + H 2 O 1 mol + 1 mol 1 mol + 1 mol Reaction B: One reactant is limiting. NaOH + HCl NaCl + H 2 O 0.6 mol + 0.3 mol mol + mol Which is the limiting reagent? Which reagent is in excess? What are the amounts of each product? How much of each compound is present at the end of the reaction? NaOH = HCl = NaCl = H 2 O = Please take note that you must calculate the theoretical yield of any reaction only from the limiting reagent!

30 Sample Problem: If you have 25 moles of N 2 reacting with 45 moles of H 2 according to the following reaction, which is the limiting reagent? N 2 (g) + 3 H 2 (g) 2 NH 3 (g) moles of A x mole ratio = moles of B available needed 25 moles N 2 x 3 moles H 2 = moles H 2 1 mole N 2 For 25 moles N 2 we would need moles H 2 but only 45 moles H 2 are available. H 2 = limiting reagent You can do the same analysis starting with H 2 45 moles H 2 x 1 moles N 2 = moles N 2 moles H 2 We have more N 2 than we need. N 2 =

31 Here is another way to finding the limiting reagent. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 25 moles 45 moles The limiting reagent has the lowest mole-to-coefficient ratio: 25 moles N 2 = 25 45 moles H 2 = 15 1 mole N 2 3 moles H 2

32 Demo: Identify the limiting reagent HC 2 H 3 O 2 (aq) + NaHCO 3 (s) H 2 O(l) + CO 2 (g) + NaC 2 H 3 O 2 (aq) A B 0.19 mol + 0.048 mol in Rxn I 0.19 mol + 0.095 mol in Rxn II 0.19 mol + 0.19 mol in Rxn III 0.19 mol + 0.38 mol in Rxn IV Reagent A and B are mixed and CO 2 evolution is measured. Which is the limiting reagent? Rxn I Rxn II Rxn III Rxn IV B is limiting B is limiting stoichiometric amounts (balanced) A is limiting

33 The following drill problem summarizes the important chemical calculations you have learned in this chapter: When aqueous solutions of CaCl 2 and AgNO 3 are mixed, a white precipitate of AgCl forms. If 3.33 g CaCl 2 are reacted with 8.50 g AgNO 3 and 5.63 g AgCl are obtained, what is the % yield? CaCl 2 (aq) + 2 AgNO 3 (aq) 2 AgCl(s) + Ca(NO 3 ) 2 (aq) Calculated formula masses Convert g to moles Check if you have a limiting reagent Calculate theoretical yield Calculate % yield

34 CaCl 2 (aq) + 2 AgNO 3 (aq) 2 AgCl(s) + Ca(NO 3 ) 2 (aq) 3.33 g 8.50 g 5.63 g 111.0 amu 169.9 amu 143.3 amu 3.33 g CaCl 2 x 1 mol = 0.0300 mol CaCl 2 111.0 g 8.50 g AgNO 3 x 1 mol = 0.0500 mol AgNO 3 169.9 g Mole-to-coefficient ratios: 0.0300 CaCl 2 = 0.0300 0.0500 AgNO 3 = 0.0250 Theoretical yield: 0.0500 mol AgNO 3 x 2 mol AgCl x 143.3 g AgCl = g AgCl 2 mol AgNO 3 1 mol AgCl % Yield: actual yield x 100 = 5.63 g x 100 = % theoretical yield g