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Chapter 6 Chemical Calculations Submicroscopic

Macroscopic

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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)

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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

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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.

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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 (NH4)2S F.W. = 68.1 amu % N = 2 x 14.0 amu N x 100 = 41.111 % N 68.1 amu (NH4)2S % H = 8 x 1.0 amu H x 100 = 11.747 % H 68.1 amu (NH4)2S % S = 32.1 amu S x 100 = 47.136 % S 68.1 amu (NH4)2S

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3. The Mole & Avogadro’s Number (Ch 6.2)

1 dozen = 12 objects 1 ream = 500 sheets 1 mole = _________ objects 6.0221415 x 1023 Avogadro’s Number

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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 1023 He atoms = __________ He atoms 1 mole He

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4. Molar Mass (Ch 6.3) 1 mole 12C = exactly 12 g 12C demo sample

Molar mass of 12C = __________ We now have three conversion factors to relate moles, number of atoms, and mass of Carbon: 6.02 x 1023 C atoms 12 g C 1 mole 1 mole 6.02 x 1023 C atoms 12 g C

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Figure 6.5 1 mole of S, Zn, C, Mg, Pb, Si, Cu, Hg (start counterclockwise from yellow S and Hg in center)

12C 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 1023 atoms Mg Pb = ________ amu 1 mol Pb = _______ g Pb = 6.02 x 1023 atoms Pb

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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 (NH4)2S formula mass = 68.1 amu

68.1 g (NH4)2S = 1 mole (molar mass or formula weight, F.W.: ____________)

68.1 g (NH4)2S = 6.02 x 1023 formula units of (NH4)2S Carbon dioxide CO2 formula mass = 44.01 amu

44.01 g CO2 = 1 mole (molar mass or molecular weight, M.W.: _____________)

44.01 g CO2 = 6.02 x 1023 molecules of CO2

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Sample Problem: If 7.50 moles of ammonia, NH3, 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 NH3 (molar mass) = 17.0 g NH3 7.50 moles NH3 x 17.0 g NH3 = _______ g NH3 1 mole NH3

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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 (NH4)2S For N: 2 moles of N atoms or 1 mole (NH4)2S 1 mole (NH4)2S formula units ___ moles N

For H: 8 moles of H atoms or 1 mole (NH4)2S 1 mole (NH4)2S formula units ___ moles H For S: 1 mole of S atoms or 1 mole (NH4)2S 1 mole (NH4)2S formula units ___ mole S

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Sample calculation:

How many H atoms are in 35.6 g of (NH4)2S? (NH4)2S formula mass = 68.1 amu

1 mole (NH4)2S = 68.1 g

1 mole (NH4)2S = ___ moles H atoms

1 mole H atoms = 6.02 x 1023 H atoms Strategy: mass (NH4)2S → moles (NH4)2S → moles H → atoms H 35.6 g (NH4)2S x 1 mol (NH4)2S x 8 mol H x 6.02 x 1023 H 68.1 g (NH4)2S 1 mol (NH4)2S 1 mol H = ______ x 10___ H atoms

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Fig 6.7 “Transitions” allowed in solving chemical-formula bases problems:

Drill problem:

How many g of (NH4)2S are required to obtain ___ moles of NH4+?

(NH4)2S = 68.1 g/mol (molar mass) moles NH4

+ → mole (NH4)2S → g (NH4)2S 0.50 moles NH4

+ x 1 mole (NH4)2S x 68.1 g (NH4)2S 2 moles NH4

+ 1 mol (NH4)2S = ______ (NH4)2S

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6. Empirical & Molecular Formula (supplemental material & Lab Exp 8) _______ C6H12O6 Molecular Formula CH2O Empirical Formula The Empirical Formula (E.F.) is the simplest ratio of atoms in a compound. _____ acid C2H4O2 Molecular Formula CH2O Empirical Formula Both compounds have the same composition: 40.0 % C, 6.7 % H, 53.3 % O

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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 = _______

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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 (BNH2) = B3N3H6 Borazole

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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 = __________

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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 H2O → 2 NaOH(aq) + H2(g) 2 atoms + 2 molecules → 2 formula units + 1 molecule 2 moles + 2 moles → 2 moles + 1 mole 2x23=46g + 2x18=36g → 2x40=80g + 1x2=2g ______ reactants → ______ products

Law of Conservation of Mass

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The coefficients in a balanced equation give the numerical relationships among formula units consumed or produced in a chemical reaction. Keys to the calculations = _______________ N2 + 3 H2 → 2 NH3 1 mole N2 3 moles H2 2 moles NH3 3 moles H2 1 mole N2 1 mole N2 1 mole N2 3 moles H2 2 moles NH3 2 moles NH3 2 moles NH3 3 mole H2

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Figure 6.9 In solving stoichiometric calculations, only the following “transitions” are allowed:

Sample calculation:

How many g O2 are needed to convert 45 g glucose (C6H12O6) into CO2 and H2O?

First write a balanced equation and calculated the molecular masses of glucose and oxygen:

C6H12O6 + 6 O2 → 6 CO2 + 6 H2O M.W. 180 amu + 32 amu Then set up the calculations according to Figure 6.9

45 g glu x 1 mol glu x ___mol O2 x 32 g O2 180 g glu 1 mol glu 1 mol O2 = ________ O2

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Drill Problem: Figure 6.10 The chemical equation for the deployment of airbags is 2 NaN3(s) → 2 Na(s) + 3 N2(g) How many g NaN3 would have to decompose on order to generate 253 million molecules of N2? F.W. NaN3 = 65.0 amu Avogadro’s # = 6.02 x 1023

molecules N2 → moles N2 → moles NaN3 → g NaN3 2.53x108 molecules N2 x 1 mole N2 = 4.203 x 10-16 moles N2 6.02 x 1023molecules N2 4.203 x 10-16 moles N2 x ___ moles NaN3 x 65.0 g NaN3 ___ moles N2 ___mol NaN3 = _______ x 10-14 g NaN3

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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:

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) + 530 kcal However, the reaction is also represented by placing the enthalpy change to the right of the equation: C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ∆H = ________

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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 CO2(g) + 6 H2O(g) + 678 kcal → C6H12O6(aq) + 6 O2(g) 6 CO2(g) + 6 H2O(g) → C6H12O6(aq) + 6 O2(g)

∆H = _________

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Sample Problem: How much energy is produced when 0.50 g of butane, C4H10, is burned in a butane lighter? C4H10 = 58.1 g/mole

2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(l) + 1365 kcal The equation shows that 1365 kcal of heat are produced when 2 moles of butane undergo combustion. 0.50g C4H10 x 1 mol C4H10 x 1365 kcal = _____ kcal 58.1 g C4H10 ___ moles C4H10 Is this reaction exothermic or endothermic? What is the sign for ∆H?

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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.

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Pure iron can be produced from iron oxide in a blast furnace. Sample Problem. When 884 g of Fe2O3 was reduced with excess carbon, 507 g of Fe were obtained. What was the percent yield?

Fe2O3 + 3 C → 2 Fe + 3 CO 1. Calculate the theoretical yield: g Fe2O3 → mol Fe2O3 → mol Fe → g Fe 884g Fe2O3 x 1 mol Fe2O3 x 2 moles Fe x 55.85g Fe = ___g Fe 159.7g Fe2O3 1 mol Fe2O3 1 mol Fe 2. Calculate the % yield: % yield = actual yield x 100 = 507 g x 100 = _______ % theoretical yield ____ g

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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.

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Reaction A: Stoichiometric amounts of reactants are used.

NaOH + HCl → NaCl + H2O 1 mol + 1 mol → 1 mol + 1 mol

Reaction B: One reactant is limiting.

NaOH + HCl → NaCl + H2O 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 = H2O = Please take note that you must calculate the theoretical yield of any reaction only from the limiting reagent!

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Sample Problem: If you have 25 moles of N2 reacting with 45 moles of H2 according to the following reaction, which is the limiting reagent?

N2(g) + 3 H2(g) → 2 NH3(g) moles of A x mole ratio = moles of B available needed 25 moles N2 x 3 moles H2 = ____ moles H2 1 mole N2

For 25 moles N2 we would need ____ moles H2 but only 45 moles H2 are available.

H2 = limiting reagent You can do the same analysis starting with H2

45 moles H2 x 1 moles N2 = ____ moles N2 ___ moles H2

We have more N2 than we need. N2 = _________

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Here is another way to finding the limiting reagent.

N2(g) + 3 H2(g) → 2 NH3(g) 25 moles 45 moles The limiting reagent has the lowest mole-to-coefficient ratio: 25 moles N2 = 25 45 moles H2 = 15 1 mole N2 3 moles H2

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Demo: Identify the limiting reagent HC2H3O2(aq) + NaHCO3(s) → H2O(l) + CO2(g) + NaC2H3O2(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 CO2 evolution is measured. Which is the limiting reagent? Rxn I B is limiting Rxn II B is limiting Rxn III stoichiometric amounts (balanced) Rxn IV A is limiting

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The following drill problem summarizes the important chemical calculations you have learned in this chapter:

When aqueous solutions of CaCl2 and AgNO3 are mixed, a white precipitate of AgCl forms. If 3.33 g CaCl2 are reacted with 8.50 g AgNO3 and 5.63 g AgCl are obtained, what is the % yield?

CaCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ca(NO3)2(aq)

• Calculated formula masses • Convert g to moles • Check if you have a limiting reagent • Calculate theoretical yield • Calculate % yield

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CaCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ca(NO3)2(aq)

3.33 g 8.50 g 5.63 g

111.0 amu 169.9 amu 143.3 amu 3.33 g CaCl2 x 1 mol = 0.0300 mol CaCl2 111.0 g 8.50 g AgNO3 x 1 mol = 0.0500 mol AgNO3 169.9 g Mole-to-coefficient ratios:

0.0300 CaCl2 = 0.0300 0.0500 AgNO3 = 0.0250 ______________ Theoretical yield: 0.0500 mol AgNO3 x 2 mol AgCl x 143.3 g AgCl = ___ g AgCl 2 mol AgNO3 1 mol AgCl % Yield: actual yield x 100 = 5.63 g x 100 = ______ % theoretical yield ____ g