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Stoichiometry of Formulas and Equations

Chapter 3. Stoichiometry of Formulas and Equations. The Mole. mole(mol) - the amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12. This amount is 6.0221 x 10 23 . The number is called Avogadro’s number and is abbreviated as N A.

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Stoichiometry of Formulas and Equations

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  1. Chapter 3 Stoichiometry of Formulas and Equations

  2. The Mole mole(mol) - the amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12 This amount is 6.0221 x 1023. The number is called Avogadro’s number and is abbreviated asNA One mole (1 mol) contains 6.0221 x 1023 entities (to five significant figures)

  3. Figure 3.2 Oxygen 32.00 g One mole of common substances. Water 18.02 g CaCO3 100.09 g Copper 63.55 g

  4. Table: Summary of Mass Terminology (from 4th ed.) Term Definition Unit Isotopic mass Mass of an isotope of an element amu Atomic mass Average of the masses of the naturally occurring isotopes of an element weighted according to their abundance amu (also called atomic weight) Molecular (or formula) mass (also called molecular weight) Sum of the atomic masses of the atoms (or ions) in a molecule (or formula unit) amu Molar mass (M) Mass of 1 mole of chemical entities (atoms, ions, molecules, formula units) g/mol

  5. Information Contained in the Chemical Formula of Glucose C6H12O6 ( M = 180.16 g/mol) Table 3.1 Oxygen (O) Carbon (C) Hydrogen (H) Atoms/molecule of compound 6 atoms 12 atoms 6 atoms Moles of atoms/ mole of compound 6 moles of atoms 12 moles of atoms 6 moles of atoms Atoms/mole of compound 6(6.0221 x 1023) atoms 12(6.0221 x 1023) atoms 6(6.0221 x 1023) atoms Mass/moleculeof compound 6(12.01 amu) =72.06 amu 12(1.008 amu) =12.10 amu 6(16.00 amu) =96.00 amu Mass/mole of compound 72.06 g 12.10 g 96.00 g

  6. Find molar mass of caffeine, which we all need, C8H10N4O2, and the hydrate BaCl2.2H2O. Which mole weighs more? Also do percent composition of caffeine if you remember how. Caffeine: BaCl2.2H2O: 8*C = 96.08 49.48% 1*Ba = 137.327 10*H=10.079 5.1903% 2*Cl = 70.9054 4*N =56.026 28.851% 2H2O = 36.0304 2*O =31.9988 16.478% 244.263 g/mol 194.19 g/mol NOTE: PERCENTAGES MUST OBEY SIG FIG RULES! PRACTICE:

  7. MOVING BETWEEN ATOMS, MOLES AND MASS START OF A MOLE MAP – SEE FIG 3.3 IN TEXT, NEXT SLIDE AND LECTURE NOTES NA STANDS FOR AVOGADRO’S NUMBER M STANDS FOR MOLAR MASS

  8. Atoms, ions, molecules, formula units… /NA X NA MOLES X M / M MASS, grams

  9. How many atoms in 0.222 g of He? 0.222g * NA = 3.34 x 1022 atoms 4.003 g/mol How many grams in 1.0 x 109 atoms of lead? 1.0 x 109 atoms * 207.2 g/mol = 3.4 x 10-13 g NA EXAMPLES:

  10. How many moles of caffeine in 0.485g? How many molecules? How many carbon atoms? (You previously determined molar mass to be 194.19 g/mol.) (2.50 x 10-3 MOLES, 1.50 x 1021 molecules, *8 = 1.20 x 1022 atoms C) PRACTICE:

  11. DEFINITIONS: Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis and gives rise to the smallest set of whole numbers of atoms. Molecular Formula - The formula of the compound as it exists, it may be a multiple of the empirical formula.

  12. From molecular formula to empirical formula Molecular Divider Empirical FormulaFormula C2H2 2 CH C6H6 6 CH C5H10 5 CH2 C6H12O6 6 CH2O N2H4 2 NH2

  13. If we only know percent composition and want the molecular formula, there are two possible ways: Way 1: Assume 100 g of it, and plug in mass for percent. Convert mass to moles, then find lowest whole number mole ratio to get empirical formula. Use experimental molar mass to get molecular formula. (This is very common method because we don’t always know the molar mass.) Way 2: If we already know molar mass and % comp, then multiply molar mass by % to get actual mass of each element, then convert to moles to obtain molecular formula Determining empirical formula from percent composition

  14. Make grams with percent composition from previous slide 8 about caffeine: 49.48 g of C/12.011 g/mol =4.120 mol C 5.19 g of H/1.0079 g/mol = 5.149 mol H 28.85 g of N/14.0067 g/mol =2.059 mol N 16.49 g of O/15.9994 g/mol = 1.031 mol O Now divide all by one with lowest # of moles: 4.120 mol C/1.031 mol O = 3.996/1 C/O, 2.059 mol N/1.031 mol O = 1.998/1 N/O, 5.149 mol H/1.031 mol O = 4.99/1 H/O 4 C, 2 N, 5 H, 1 O: C4H5N2O is the empirical formula Have to find molar mass to get molecular formula: we know caffeine has molar mass of 194.193 g/mol Divide molar mass by empirical mass to see how many EF units in molecule: 194.193/82.1 = ~2 Double everything in EF for molecular formula of C8H10N4O2 Example with caffeine using Way 1:

  15. Convert percents back to fractions and multiply by molar mass: 194.193*.4948 = 96.087 g /12.011g/mol =8.000 mol C 194.193*.0519 = 10.079 g /1.0079g/mol= 10.0 mol H 194.193*.2885 = 56.0247g/14.0067g/mol=4.000 mol N 194.193*.1649 = 32.022g/ 15.9994g/mol=2.001 mol O Therefore molecular formula is C8H10N4O2 Example with caffeine using Way 2:

  16. Magnetite is a magnetic form of an iron and oxygen compound that is 72.4% Fe and 27.6% O. Find its empirical formula, which is same as its chemical formula. USE METHOD #1: 100 g of compound 72.4 g Fe/55.846 g/mol=1.30 mol Fe 27.6 g O/15.9994 g/mol = 1.73 mol O 1.73 mol O/1.30 mol Fe = 1.33 O TO 1 Fe Have to be whole numbers, so multiply to 3: 1.33 * 3 = 4 O, 1 * 3 = 3 Fe TO MAKE Fe3O4 ****Note I did not round down 1.33 to 1 to make it FeO!!! This is important; the ratio should be within 0.05 of a whole number to round off PRACTICE ON AN IONIC COMPOUND:

  17. Video: nice review of method 1 • ‪Finding Empirical Formulas‬‏ - YouTube

  18. m 2 m 2 CnHm + (n+ ) O2 = n CO2(g) + H2O(g) Figure 3.5 Combustion train for the determination of the chemical composition of organic compounds. A sample of compound that contains C and H (and perhaps other elements) is burned in a stream of O2 gas. The CO2 and H2O formed are absorbed separately, while any other element oxides are carried through by the O2 gas stream. The increases in mass of the absorbers are used to calculate the amounts (mol) of C and H in the sample.

  19. If 1.505 g of THIOPHENE (C,H,S) yields 3.149 g CO2, 0.645 g H2O & 1.146 g SO2 then find empirical formula. CxHySz + O2 x CO2 + y/2 H2O + z SO2 1.505 3.1490.6451.146 Do lots of practice problems to learn how to do this!! PRACTICE: a little harder than the AP Chem examples

  20. PRACTICE WITH PERCENT COMPOSITION • Work on problems 30c, 33. • 30c: Determine empirical formula for a compound that is 79.9%-mass carbon and 20.1%-mass hydrogen. • 33: Find EF & MF for cmpd that is 69.6%-mass C, 8.32%-mass H, 22.1%-mass O, with M of ~362 g/mol. • (Practice combustion analysis: prob 34)

  21. Figure 3.6 The formation of HF gas on the macroscopic and molecular levels. How to view a chemical equation: as moles, as atoms, or even as mass.

  22. 1. Predict the products if not given (will learn how to do this) 2. Formulas: determine that each has the correct chemical formula 3. Conservation of matter: balance the # of atoms of each kind on each side of arrow, usually by inspection. (Never change subscripts on formulas to make it balance.) (If an element occurs in only one compound on each side of equation, balance it first; if an element occurs as free element on either side, balance it last.) LEARN STEPS TO BALANCING EQ'NS IF YOU HAVE NOT ALREADY LEARNED HOW TO DO SO!

  23. C8H18 + O2 CO2 + H2O C8H18 + O2 CO2 + H2O 2C8H18 + 25O216CO2 + 18H2O 2C8H18 + 25O216CO2 + 18H2O 2C8H18(l) + 25O2 (g) 16CO2 (g) + 18H2O (g) Sample Problem 3.7 Balancing Chemical Equations PROBLEM: Within the cylinders of a car’s engine, the hydrocarbon octane (C8H18), one of many components of gasoline, mixes with oxygen from the air and burns to form carbon dioxide and water vapor. Write a balanced equation for this reaction. 25/2 8 9 Hint: memorize products for combustion of hydrocarbons!

  24. 1. COMBINATION OR SYNTHESIS: A + B  C A. formation of compounds from element METALS + HALOGENS : Zn + I2 ZnI2 METALS + OXYGEN (oxidation - rusting of iron, etc) Fe + O2 (CAN GO TO FeO OR Fe3O4) NONMETALS + OXYGEN : C OR N OR S + O2 --> gases that make acid rain 2. DECOMPOSITION: compound decomposes to simpler compounds or elements CA+B CaCO3 CaO + CO2 TYPES OF CHEMICAL EQUATIONS:(NOT IN CHP 3 BUT YOU NEED TO MEMORIZE ALL FIVE OF THESE)

  25. 3. COMBUSTION: A rapid reaction with oxygen. Organic compounds: C3H8+ O2CO2 + H2O Practice balancing it: balance C, then H, leave O til last SINGLE REPLACEMENT: A + BC  AC + B (Use of Activity Series) __Cu + __AgNO3 __Ag + __Cu(NO3)2 5. DOUBLE REPLACEMENT: AB + CD  AD + CB. Notice how written: C is positive ion and gets written first, goes with B, the negative ion. (ppt, neut, or gas-forming reactions) TYPES OF CHEMICAL EQUATIONS:

  26. PRACTICE WITH CHEMICAL EQUATIONS • Work on problems 37bd and 38a. Then ID each type of reaction they are. • 37b: ___P4O10(s) + ___H2O(l) ___H3PO4(l) • 37d: ___CH3NH2(l) + ___O2(g)  ___CO2(g) + ___H2O(l) + ___N2(g) • 38a: ___Cu(NO3)2(aq) + ___KOH(aq)  ___Cu(OH)2(s) + ___KNO3(aq)

  27. 1. Start by writing the balanced chemical equation. 2. Convert mass (or whatever) to moles (divided by molar mass). 3. Use stoichiometric conversion factors: mole/mole ratio from balanced equation. 4. Convert moles back to mass (or whatever). Stoichiometry Steps – A Review

  28. Suppose we want to make 10,000. kg of CaO to use in the chemical industry. Theoretically, how much CaCO3 is needed as starting material? 1. Start by writing the equation, predict products, balance: CaCO3 CaO + CO2 2. Convert mass to moles (dividing by molar mass) – we want CaO: 10,000. kg * 1000 g/kg * mol CaO/56.08 g = 1.783 x 105 mol CaO 3. Use stoichimetric conversion factors: 1.783 x 105 mol CaO * 1 CaCO3/1 CaO = 1.783 X 105 mol CaCO3 starting 4. Convert moles back to mass (multiply by molar mass) 1.783 x 105 mol CaCO3 * 100.09 g/mol = 1.784 x 107 g Practice Stoichiometry:

  29. Find the mass of CO2 produced by the combustion of one 10.0 gallon tank of octane (C8H18), DENSITY = 0.75 g/mL. 2 C8H18 + 25 O2 16 CO2 + 18 H2O (Answer 8.63 x 104 g) Then do problem 43 in chp 3. (See note below) Practice Stoichiometry:

  30. Write the balanced chemical equation Go to moles of each of the reactants of interest Use stoichiometric coefficients to determine moles of product each reactant would form, assuming other reactants present in excess Choose the least moles of product case. That reactant limits the reaction, and others are present in excess May have to determine excess by subtracting what was used with the limiting reagent STEPS TO DETERMINING THE LIMITING REAGENT

  31. EXAMPLE: given exactly 1 mol iron (III)sulfide, 2 mol water, 3 mol oxygen gas, find moles of iron (III) hydroxide produced and moles of excess reagents. Fe2S3(s) + H2O(l) + O2(g)Fe(OH)3(s) + S(s) 1. Balance: 2, 6, 3, 4, 6 2. Already in moles! 3. 1.0 mol Fe2S3 * 4 Fe(OH)3/2 Fe2S3 = 2.0 mol Fe(OH)3 2.0 mol H2O * 4 Fe(OH)3/6 H2O = 1.33 mol Fe(OH)3**limiting 3.0 mol O2 * 4 Fe(OH)3/3 O2 = 4.0 mol Fe(OH)3 4. see ** marked above. (Could convert to mass of product: 1.33 moles Fe(OH)3 * 106.069 g/mol = 142 g) 5. 2.0 mol H2O * 2 Fe2S3/6 H2O = 0.67 mol Fe2S3 consumed. 1.00 mol - 0.67 mol = 0.33 mol remaining 2.0 mol H2O * 3 O2/6 H2O = 1.0 mol O2 consumed. 3.00 mol - 1.0 mol = 2.0 mol remaining LIMITING REAGENT:

  32. If 1.50 g Al reacts with 342 mL of 0.350 Molar HCl, what mass of AlCl3 could be recovered? Which reactant is present in excess and how much is left over at the end of the reaction? (make 5.32 g AlCl3, 0.4236g of Al leftover) If 10.00 g of Cu is put in 1.00 L of 0.100 M AgNO3, what mass of Ag will be produced? (Cu(NO3)2 is the other product.) How many moles of excess reagent is left? (Is it Cu?) (10.79 g Ag, 0.1074 moles Cu leftover) When a reactant runs out, the reaction stops! LIMITING REAGENT (assuming you know M and molar mean mol/L)

  33. THEORETICAL YIELD: amount possible based on stoichiometric calculations ACTUAL YIELD: mass of product obtained (always < theoretical) PERCENT YIELD = ACTUAL/THEOR * 100 (ALWAYS < 100%) WHY DON'T WE ALWAYS GET 100% YIELD: 1. Reaction may not be driven to completion (equilibrium, heat evolved, whatever) 2. Side reactions use up reactants to make unwanted products 3. Purification/recovery process causes loss. THEORETICAL AND ACTUAL YIELDS, PERCENT YIELD:

  34. Given a reaction with 0.473 grams of phosphorous with excess chlorine gas that makes only 2.12 g of phosphorous pentachloride. What is percent yield? 1. Balance reaction: 2 P + 5 Cl2 2 PCl5 2. Convert mass to moles: 0.473 g/30.97g/mol = 1.527x10-2 mol P 3. STOICH RATIO: 1.527x 10-2 mol P * 2 PCl5/2 P = SAME 4. Moles to mass: 1.527x 10-2 mol PCl5 * 208.2 g/mol = 3.18 g 5. % YIELD = 2.12 g/3.18 g * 100 = 66.7% (very typical) PERCENT YIELD EXAMPLE:

  35. PRACTICE WITH PERCENT YIELD Work on problem 59. Given: 200. grams of PCl3 react with excess H2O to form 128 g of HCl and some aqueous phosphorous acid. Find the percent yield. PCl3 + 3 H2O  3 HCl + H3PO3 200.g(1 mol/137.33g)(3HCl/1PCl3)(36.46g/mol) = 159.3 g HCl theoretical yield % yield = (128/159.3) * 100 = 80.4% yield

  36. Solutions have concentrations – we will use Molarity Molarity = moles solute/liter solution M = n/V Go between moles, volume and molarity: n = M*V; V = n/M Example: How many moles of NaOH are in 25.0 mL of 0.555 M NaOH? 25.0 mL * 1L/103 mL * 0.555 mol/L = 0.0138 mol NaOH AQUEOUS REACTIONS/COMPOSITION

  37. PLAN: Molarity is the number of moles of solute per liter of solution. 0.715 mol glycine 1000mL 495 mL soln 1 L Sample Problem 3.13 Calculating the Molarity of a Solution PROBLEM: Glycine (H2NCH2COOH) is the simplest amino acid. What is the molarity of an aqueous solution that contains 0.715 mol of glycine in 495 mL of solution ? SOLUTION: mol of glycine divide by volume = 1.44 M glycine concentration(mol/mL) glycine 103mL = 1L molarity(mol/L) glycine

  38. SOLUTION INVENTORY: WHAT IONS AND WHAT IS CONCENTRATION? Example 1: if NaOH dissolved and is 0.100 M, then the concentration of each ion is 0.100 M Example 2: if Na2CO3 is dissolved and is 0.100 M, then [Na+] = 2 x 0.100 M = 0.200 M, and [CO32-] is 0.100 M Practice 1: Do the solution inventory for 0.022 M iron (III) chloride. Practice 2: Do the solution inventory for 0.0015 M nitric acid. IONS IN SOLUTIONS:

  39. Method 1: determine concentration and volume needed 1) convert to moles needed, then to grams 2) put some water in a volumetric flask, add solid and let it completely dissolve, then dilute with water to mark Example: Need 0.500 L of 0.500 M NaOH (Add about 10 g to a 500-mL volumetric flask) MAKING SOLUTIONS BY TWO METHODS

  40. A • Weigh the solid needed. • Transfer the solid to a volumetric flask that contains about half the final volume of solvent. CAdd solvent until the solution reaches its final volume, then mix thoroughly again. BDissolve the solid thoroughly by swirling. Figure 3.12 from 4th ed. Laboratory preparation of molar solutions. Note there is already some water in the flask!

  41. Method 2: C1V1 = C2V2 Dilution Equation C is for concentration in any units V is for volume in any units Example: Make 1.0 L of 0.100 M NaOH from 6.0M NaOH 6.00 M x ? L = 0.100 M x 1.0 L Use 0.0167 L or 16.7 mL MAKING SOLUTIONS BY TWO METHODS

  42. Figure 3.11 Converting a concentrated solution to a dilute solution. When a solution is diluted, only solvent is added. The solution volume increases while the total number of moles of solute remains the same. Therefore, as shown in the blow-up views, a unit volume of concentrated solution contains more solute particles than the same unit volume of dilute solution. (

  43. How would you make 250.0 ml of a 0.555 M standard NaOH solution? You have some 6.00 M NaOH available and some solid NaOH available. Show both methods. (Be prepared to show your work to class on document camera.) (Method 1: 5.52 g solid dry NaOH; Method 2: 23.1 mL of 6.0 M NaOH) MAKING SOLUTIONS BY TWO METHODS

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