Stoichiometry Chapter 3 Chemical Stoichiometry The study of

Stoichiometry Chapter 3

Chemical Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.

Schematic diagram of a mass spectrometer.

Average Atomic Mass Elements occur in nature as mixtures of isotopes Carbon = 98. 89% 12 C 1. 11% 13 C <0. 01% 14 C (0. 9889)(12. 0000 amu) = 11. 87 amu (0. 0111)(13. 0034 amu) = 0. 144 amu 12. 01 amu

The Mole The number equal to the number of carbon atoms in exactly 12 grams of pure 12 C. 1 mole of anything = 6. 022 1023 units of that thing Equal moles of substances have equal numbers of atoms, molecules, ions, formula units, etc.

Counting Atoms are too small to be seen or counted individually. Atoms can only be counted by weighing them. • all jelly beans are not identical. • jelly beans have an average mass of 5 g. • How could 1000 jelly beans be counted?

Jelly Beans & Mints have an average mass of 15 g. How would you count out 1000 mints? Why do 1000 mints have a mass greater than 1000 jelly beans?

Atomic Mass Unit Atoms are so tiny that the gram is much too large to be practical. The mass of a single carbon atom is 1. 99 x 10 -23 g. The atomic mass unit (amu) is used for atoms and molecules.

AMU’s and Grams 1 amu = 1. 661 x 10 -24 g Conversion Factors 1. 661 x 10 -24 g/amu 6. 022 x 1023 amu/g

The Mole • One mole of rice grains is more than the number of grains of all rice grown since the beginning of time! • A mole of marshmallows would cover the U. S. to a depth of 600 miles! • A mole of hockey pucks would be equal in mass to the moon.

The Mole Substance Na Cu S Al Average Atomic Mass (g) 22. 99 63. 55 32. 06 26. 98 # Moles 1 1 # Atoms 6. 022 x 1023 6 6. 022 x 1023

Measurements dozen = 12 gross = 12 dozen = 144 ream = 500 mole = 6. 022 x 1023

Avogadro’s number equals 23 6. 022 10 units

Unit Cancellation How many dozen eggs would 36 eggs be? (36 eggs)(1 dozen eggs/12 eggs) = 3 dozen eggs How many eggs in 5 dozen? (5 dozen eggs)(12 eggs/1 dozen eggs) = 60 eggs

Calculating Mass Using AMU’S 1 N atom = 14. 01 amu (23 N atoms)(14. 01 amu/1 N atom) = 322. 2 amu

Calculating Number of Atoms from Mass 1 O atom = 16. 00 amu (288 amu)(1 O atom/16. 00 amu) = 18 atoms O

AMU’s & Grams 1 atom C = 12. 011 amu = 1. 99 x 10 -23 g 1 mol C = 12. 011 g Use TI-83 or TI-83 Plus to store 6. 022 x 1023 to A.

Calculating Moles & Number of Atoms 1 mol Co = 58. 93 g (5. 00 x 1020 atoms Co)(1 mol/6. 022 x 1023 atoms) = 8. 30 x 10 -4 mol Co (8. 30 x 10 -4 mol)(58. 93 g/1 mol) = 0. 0489 g Co Moles are the doorway grams <---> moles <---> atoms

Molar Mass A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound. CO 2 = 44. 01 grams per mole

Calculating Mass from Moles Ca. CO 3 1 Ca = 1 (40. 08 g) = 40. 08 g 1 C = 1 (12. 01 g) = 12. 01 g 3 O = 3 (16. 00 g) = 48. 00 g 100. 09 g (4. 86 mol. Ca. CO 3)(100. 09 g/1 mol) = 486 g Ca. CO 3

Calculating Moles from Mass Juglone 10 C = 10(12. 01 g) 6 H = 6(1. 008 g) 3 O = 3(16. 00 g) = 120. 1 g = 6. 048 g = 48. 00 g 174. 1 g (1. 56 g juglone)(1 mol/174. 1 g) = 0. 00896 mol juglone

Percent Composition Mass percent of an element: For iron in iron (III) oxide, (Fe 2 O 3)

% Composition Cu. SO 4. 5 H 2 O 1 Cu = 1 (63. 55 g) = 63. 55 g 1 S = 1 (32. 06 g) = 32. 06 g 4 O = 4 (16. 00 g) = 64. 00 g 5 H 2 O = 5 (18. 02 g) = 90. 10 g 249. 71 g

% Composition (Continued) % Cu = 63. 55 g/249. 71 g (100 %) = 25. 45 % Cu % S = 32. 06 g/249. 71 g (100 %) = 12. 84 % S % O = 64. 00 g/249. 71 g (100 %) = 25. 63 % O % H 2 O = 90. 10 g/249. 71 g (100 %) = 36. 08 % H 2 O Check: Total percentages. Should be equal to 100 % plus or minus 0. 01 %.

Formulas Ionic compounds -- empirical formula Na. Cl Ca. Cl 2 Covalent compounds -- molecular formula C 6 H 12 O 6 C 2 H 6

Formula of a Compound Calculations Example for Problems 75 -78 on page 125. A compound contains only carbon, hydrogen, and nitrogen. When 0. 1156 g of it is reacted with oxygen, 0. 1638 g of CO 2 and 0. 1676 g of water are collected. Molar mass of CO 2 is 44. 01 g Molar mass of HOH is 18. 02 g

Formula of a Compound Calculations (0. 1638 g CO 2)(12. 01 g C/44. 01 g CO 2) = 0. 04470 g C %C = (0. 04470 g/0. 1156 g)(100%) = 38. 67% C (0. 1676 g HOH)(2. 016 g H/18. 02 g HOH) = 0. 01875 g H %H = (0. 01875 g/0. 1156 g)(100%) = 16. 22% H %H + %C + %N = 100% -38. 67% -16. 22% = 45. 11% N

Formula of a Compound Calculations (38. 67 g C)(1 mol/12. 01 g) = 3. 220 mol C = 1 C 3. 219 mol (16. 22 g H)(1 mol/1. 008 g) = 16. 09 mol H = 5 H 3. 219 mol (45. 11 g N)(1 mol/14. 01 g) = 3. 219 mol N = 1 N 3. 219 mol Empirical Formula is CH 5 N.

Formulas molecular formula = (empirical formula)n [n = integer] molecular formula = C 6 H 6 = (CH)6 empirical formula = CH

Empirical Formula Determination 1. Base calculation on 100 grams of compound. 2. Determine moles of each element in 100 grams of compound. 3. Divide each value of moles by the smallest of the values. 4. Multiply each number by an integer to obtain all whole numbers.

Calculating Empirical Formulas 4. 151 g Al & 3. 692 g O (4. 151 g Al)(1 mol/26. 98 g) = 0. 1539 mol Al/0. 1539 mol = 1. 000 (3. 692 g O)(1 mol/16. 00 g) = 0. 2308 mol O/0. 1539 mol = 1. 500 1 Al (2) = 2 Al 1. 5 O (2) = 3 O Al 2 O 3

Molecular Formulas 71. 65 % Cl, 24. 27 % C, & 4. 07 % H (71. 65 g Cl)(1 mol/35. 45 g) = 2. 021 mol/2. 021 mol = 1 (24. 27 g C)(1 mol/12. 01 g) = 2. 021 mol/2. 021 mol = 1 (4. 07 g H )(1 mol/1. 01 g) = 4. 03 mol/2. 021 mol = 2 (EM)x = (MM) (49. 46)x = (98. 96) x = 2 (EF)x = (MF) (Cl. CH 2)2 = Cl 2 C 2 H 4

Chemical Equations Chemical change involves a reorganization of the atoms in one or more substances.

Chemical Equation A representation of a chemical reaction: C 2 H 5 OH + 3 O 2 2 CO 2 + 3 H 2 O reactants products

Physical States • solid (s) • liquid (l) • gas (g) • aqueous (aq)

Important Equation Symbols • yields ------> • heat -------> cat. • catalyst -------> • light ----> H 2 SO 4 • catalyst ------> elect. • electricity ------>

Chemical Equations Quantitative Significance 4 Al(s) + 3 O 2(g) ---> 2 Al 2 O 3(s) This equation means 4 Al atoms + 3 O 2 molecules ---give---> 2 molecules of Al 2 O 3 4 moles of Al + 3 moles of O 2 ---give---> 2 moles of Al 2 O 3

Balancing Equation Prerequisites Student must have memorized: • 44 chemical symbols • Table 2. 3 on page 61 in text • Table 2. 4 on page 62 • Table 2. 5 on page 66 • Table 2. 6 on page 67 • Type I, III, and acid nomenclature • Count HOFBr. INCl

Four Steps in Balancing Equations 1. Get the facts down. 2. Check for diatomic molecules (subscripts). 3. Balance charges on compounds containing a metal, ammonium compounds, and acids (subscripts). 4. Balance the number of atoms (coefficients). a. Balance most complicated molecule first. b. Balance other elements. c. Balance hydrogen next to last. d. Balance oxygen last.

Balancing Equations Caution The identities (formulas) of the compounds must never be changed in balancing a chemical equation! Only coefficients can be used to balance the equationsubscripts will not change!

Chemical Equation C 2 H 5 OH + 3 O 2 2 CO 2 + 3 H 2 O The equation is balanced. 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

Balancing Equations When solid ammonium dichromate decomposes, it produce solid chromium(III) oxide, nitrogen gas, and water vapor. (NH 4)2 Cr 2 O 7(s) ----> Cr 2 O 3(s) + N 2(g) + HOH(g) (NH 4)2 Cr 2 O 7(s) ----> Cr 2 O 3(s) + N 2(g) + 4 HOH(g)

Calculating Masses of Reactants and Products 1. 2. 3. 4. Balance the equation. Convert mass to moles. Set up mole ratios. Use mole ratios to calculate moles of desired substituent. 5. Convert moles to grams, if necessary.

Gram to Mole & Gram to Gram __Al(s) + __I 2(s) ---> __Al. I 3(s) 2 Al(s) + 3 I 2(s) ---> 2 Al. I 3(s) How many moles and how many grams of aluminum iodide can be produce from 35. 0 g of aluminum?

Gram to Mole & Gram to Gram 2 Al(s) + 3 I 2(s) ---> 2 Al. I 3(s) (35. 0 g Al) (1 mol/26. 98 g)(2 mol Al. I 3/2 mol Al) = 1. 30 mol Al. I 3 (35. 0 g Al) (1 mol/26. 98 g)(2 mol Al. I 3/2 mol Al)(407. 68 g/1 mol) = 529 g Al. I 3

Limiting Reactant The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed.

Solving a Stoichiometry Problem 1. 2. 3. 4. Balance the equation. Convert masses to moles. Determine which reactant is limiting. Use moles of limiting reactant and mole ratios to find moles of desired product. 5. Convert from moles to grams.

Limiting Reactant Problem If 56. 0 g of Li reacts with 56. 0 g of N 2, how many grams of Li 3 N can be produced? __Li(s) + __N 2(g) ---> __Li 3 N(s) 6 Li(s) + N 2(g) ---> 2 Li 3 N(s) (56. 0 g Li) (1 mol/6. 94 g)(1 mol N 2/6 mol Li) (28. 0 g/1 mol) = 37. 7 g N 2 Since there were 56. 0 g of N 2 and only 37. 7 g used, N 2 is the excess and Li is the Limiting Reactant.

Limiting Reactant Problem 6 Li(s) + N 2(g) ---> 2 Li 3 N(s) (56. 0 g Li)(1 mol/6. 94 g)(2 mol Li. N 3/6 mol Li) (34. 8 g/1 mol) = 93. 6 g Li 3 N How many grams of nitrogen are left? 56. 0 g N 2 given - 37. 7 g used = 18. 3 g excess. N 2

% Yield Values calculated using stoichiometry are always theoretical yields! Values determined experimentally in the laboratory are actual yields!

Limiting Reactant & % Yield If 68. 5 kg of CO(g) is reacted with 8. 60 kg of H 2(g), what is theoretical yield of methanol that can be produced? __H 2(g) + __CO(g) ---> __CH 3 OH(l) 2 H 2(g) + CO(g) ---> CH 3 OH(l) (68. 5 kg CO)(1 mol/28. 0 g)(2 mol H 2/1 mol CO) (2. 02 g/1 mol) = 9. 88 kg H 2

Limiting Reactant & % Yield 2 H 2(g) + CO(g) ---> CH 3 OH(l) Since only 8. 60 kg of H 2 were provided, the H 2 is the limiting reactant, and the CO is in excess. (8. 60 kg H 2)(1000 g/1 kg)(1 mol/2. 02 g)(1 mol CH 3 OH/2 mol H 2)(32. 0 g/1 mol) = 6. 85 x 104 g CH 3 OH

Limiting Reactant & % Yield 2 H 2(g) + CO(g) ---> CH 3 OH(l) If in the laboratory only 3. 57 x 104 g of CH 3 OH is produced, what is the % yield? % Yield = 52. 1 %