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Stoichiometry Objectives

Stoichiometry Objectives. Identify the quantitative relationships in a balanced chemical chemical equation. Determine the mole ratios from from a balanced chemical equation. Explain the sequence of steps used in solving stoichiometric problems. Use the steps to solve stoichiometric problems.

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Stoichiometry Objectives

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  1. Stoichiometry Objectives • Identify the quantitative relationships in a balanced chemical chemical equation. • Determine the mole ratios from from a balanced chemical equation. • Explain the sequence of steps used in solving stoichiometric problems. • Use the steps to solve stoichiometric problems. • Identify the limiting reactant in a chemical equation. • Identify the excess reactant and calculate the amount remaining after the reaction is complete. • Calculate the mass of a product when the amounts of more than one reactant are given. • Calculate the theoretical yield of a chemical reaction from data. • Determine the percent yield for a chemical reaction.

  2. The Mole (Ch 11) Chemists need a convenient method for counting accurately the number of atoms, molecules, or formula units in a sample of a substance. -atoms and molecules are extremely small -even in the smallest sample it’s impossible to actually count each individual atom. To fix this problem chemists created their own counting unit called the mole. -mole: commonly abbreviated mol, is the SI base unit used to measure the amount of a substance

  3. The Mole Through experimentation, it has been established 1 mole = 6.022 136 7 x 1023 representative particles. -representative particle: any particle such as atoms, molecules, formula units, electrons, or ions. -called Avogadro’s number in honor of the Italian physicist and lawyer Amedeo Avogadro who, in 1811, determined the volume of one mole of a gas. -we round Avogadro’s number to three significant figures— 6.02 x 1023. - If you write out Avogadro’s number, it looks like this: - 602 000 000 000 000 000 000 000

  4. One Mole Quantities

  5. Other Mole & Stoichiometry Vocabulary Study of quantitative relationships among amounts of reactants and products. 1. What is stoichiometry? • What is a mole ratio? Ratio between the moles of any two substances in a balanced chemical equation 3. What is molar mass? Mass, in grams, of one mole of pure substance. -numerically equal to its atomic mass -has the units g/mol

  6. Converting Moles to Particles (11.1) Determine how many particles of sucrose are in 3.50mol of sucrose. - write a conversion factor using Avogadro’s number that relates representative particles to moles of a substance.

  7. Converting Particles to Moles (11.1) Now, suppose you want to find out how many moles are represented by a certain number of representative particles, such as 4.50 x 1024 atoms of zinc. -You can use the inverse of Avogadro’s number as a conversion factor.

  8. Mole Practice 1 Identify and calculate the number of representative particles in each of the following quantities. 1. 2.15 moles of gold 2. 0.151 mole of nitrogen oxide 3. 11.5 moles of potassium bromide Calculate the number of moles of the substance that contains the following number of representative particles. 4. 8.92 x 1023 atoms of barium 5. 5.50 x 1023 molecules of carbon monoxide 6. 2.66 x 1023 formula units of potassium iodide

  9. Practice-Homework p 311 # 1-3; p 312 # 4a-c • Determine the number of atoms in 2.50 mol Zn • Determine the number of formula units in 3.25 mol AgNO3 • Determine the number of molecules in 11.5 mol H2O • Determine the number of moles in a. 5.75 x 1024 atoms Al b. 3.75 x 1024 molecules CO2 c. 3.58 x 1023 formula units ZnCl2

  10. Moles to Mass (11.2) To convert between moles and mass, you need to use the atomic mass found on the periodic table. Calculate the mass of 0.625 moles of calcium. -According to the periodic table, the atomic mass of calcium is 40.078 amu, so the molar mass of calcium is 40.078 g/mol.

  11. Mass to Moles (11.2) How many moles of copper are in a roll of copper that has a mass of 848g?

  12. Practice-Homework p 316 # 11ab-12ab • Determine the mass in grams of a. 3.57 mol Al b. 42.6 mol Si • Determine the number of moles of a. 25.5 g Ag b. 300.0 g S

  13. Mass to Atoms (11.2) To find the number of atoms in a sample, you must first determine the number of moles. Calculate the number of atoms in 4.77 g lead. • Determine moles

  14. Mass to Atoms (cont.) 2. Determine atoms You can also convert from number of particles to mass by reversing the procedure above and dividing the number of particles by Avogadro’s number to determine the number of moles present.

  15. Atoms to Mass Example problem 11-5, p 318 A party balloon has 5.50x1022 atoms of helium. What is the mass in grams of the helium? 5.50x1022 atoms He x 1 mol He = 0.0914 mol He 6.02 x1023 atoms He 0.0914 mol He x 4.00 g He = 0.366 g He 1 mol He

  16. Mole Practice 2 How many atoms are in the following samples? 1. 1.24 g cobalt 2. 0.575 g cesium How many grams are in the following samples? 3. 4.16 x1023 atoms of radium 4. 1.50 x 1020 atoms of cadmium

  17. Practice-Homwork p 316 # 11ab-12ab, p 318 # 13ab-14ab

  18. Moles of Compounds (11.3) A mole of a compound contains as many moles of each element as are indicated by the subscripts in the formula. -For example, a mole of ammonia (NH3) consists of one mole of nitrogen atoms and three moles of hydrogen atoms. -the molar mass of the compound is found by adding the molar masses of all of the atoms in the representative particle. Molar mass of NH3 = 1(molar mass of N) + 3(molar mass of H) Molar mass of NH3 = 1(14.007 g) + 3(1.008 g) = 17.031 g/mol

  19. Practice p 322 # 25 P 318 # 13ab-14ab p322 25. Determine the molar mass of each of the following: NaOH, CaCl2, Sr(NO3)2 p318 • How many atoms are in a. 55.2 g Li b. 0.230 g Pb • What is the mass of a. 6.02x1023 atoms Bi b. 1.00x1024 atoms Mn

  20. -Mole relationships from a formula (p 321) Determine the number of moles of aluminum ions in 1.25 moles of aluminum oxide (Al2O3). First we need the ratio of Al ions to Al2O3. 2 mol Al ions 1 mole Al2O3 1.25 mol Al2O3 x 2 mol Al ions = 2.50 mol Al ions 1 mole Al2O3

  21. -Mole relationships from a formula (p 321) P321# 20-21 Determine the number of moles of chloride ions in 2.50 mol ZnCl2. Calculate the number of moles of each element in 1.25 mole glucose (C6H12O6).

  22. -Mole to Mass for compounds ( p 323) What is the mass of 2.50 moles of allyl sulfide, (C3H5)2S? Calculate the mass of allyl sulfide. 6(12.01 g/mol) = 72.06 g/mol 10(1.01 g/mol) = 10.10 g/mol 1(32.07 g/mol) = 32.07 g/mol 114.23 g/mol

  23. -Mole to Mass for compounds ( p 323) What is the mass of 2.50 moles of allyl sulfide, (C3H5)2S? 2. Convert the moles to mass. 2.50mol (C3H5)2S x 114.23g (C3H5)2S 1 mol (C3H5)2S = 286g (C3H5)2S

  24. Mass of Compound to Moles Calculate the number of moles of water that are in 1.000 kg of water? 1. Before you can calculate moles, you must determine the molar mass of water (H2O). molar mass H2O = 2(molar mass H) + molar mass O

  25. 2. Now you can use the molar mass of water as a conversion factor to determine moles of water. -Notice 1.000 kg is converted to 1.000 x 103 g

  26. Practice p 323 # 27-28, p 324 # 30a&b 27. What is the mass of 3.25 moles H2SO4? • What is the mass of 4.35x10-2 moles of ZnCl2? • Determine the number of moles present in each of the following: a. 22.6 g AgNO3 b. 6.50 g ZnSO4

  27. Mole Practice 3 Calculate the molar mass of the following: • C2H5OH • HCN What is the mass of the following: • 2.25 moles of KMnO4 • 1.56 moles of H2O Determine the number of moles in the following: • 35.0 g HCl • 254 g PbCl4 What is the mass in grams of one molecule of the following: 7. H2SO4

  28. Percent Composition, Molecular & Empirical Formulas Recall that every chemical compound has a definite composition—a composition that is always the same wherever that compound is found. The composition of a compound is usually stated as the percent by mass of each element in the compound, using the following process.

  29. Percent Composition Example: Determine the percent composition of calcium chloride (CaCl2). 1. Determine mass of each ion in CaCl2. -1mol CaCl2 consists of 1mol Ca+2 ions and 2mol Cl- ions. 1mol Ca+2 ions x 40.08g Ca+2 ions = 40.08g Ca+2 ions 1mol Ca+2 ions 2mol Cl- ions x 35.45g Cl- ions = 70.90g Cl- ions 1mol Cl- ions

  30. Percent Composition Example: Determine the percent composition of calcium chloride (CaCl2). 2. Calculate molar mass of CaCl2. - 40.08g Ca+2 ions + 70.90g Cl- ions = 110.98 g CaCl2 1 mole CaCl2 1 mole CaCl2 3. Determine percent by mass of each element.

  31. Percent Composition Example: Determine the percent composition of calcium chloride (CaCl2). 3. Determine percent by mass of each element. % Ca = 40.08 g Ca+2 x 100 = 36.11 % Ca+2 110.98 g CaCl2 % Cl = 70.90 g Cl- x 100 = 63.89 % Cl- 110.98 g CaCl2 4. Make sure your percent compositions equal 100%. 36.11% Ca+2 + 63.98% Cl- = 100%

  32. Practice p 331 # 43, 45 • Calculate the percent composition of sodium sulfate (Na2SO4). 45. What is the percent composition of phosphoric acid (H3PO4).

  33. Empirical Formula You can use percent composition data to help identify an unknown compound by determining its empirical formula. -empirical formula-simplest whole-number ratio of atoms of elements in the compound. ~In many cases, the empirical formula is the actual formula for the compound. the empirical formula of sodium chloride is Na1Cl1, or NaCl, which is the true formula ~sometimes, the empirical formula is not the actual formula of the compound. the empirical formula for N2O4 (the actual) is NO2.

  34. Empirical Formula Example: The percent composition of an unknown compound is found to be 38.43% Mn, 16.80% C, and 44.77% O. Determine the compound’s empirical formula. - Because percent means “parts per hundred parts,” assume that you have 100 g of the compound. 1. Calculate the number of moles of each element in the 100 g of compound.

  35. Empirical Formula Example: The percent composition of an unknown compound is found to be 38.43% Mn, 16.80% C, and 44.77% O. Determine the compound’s empirical formula. 1. Calculate the number of moles of each element in the 100 g of compound.

  36. Empirical Formula Example: The percent composition of an unknown compound is found to be 38.43% Mn, 16.80% C, and 44.77% O. Determine the compound’s empirical formula. 2. The results show the following relationship 3. Obtain the simplest whole-number ratio of moles: -divide each number of moles by the smallest number of moles. 0.6995 mol Mn : 1.339 mol C : 2.798 mol O 0.6995 mol 0.6995 mol 0.6995 mol 1 : 2 : 4

  37. Empirical Formula Example: The percent composition of an unknown compound is found to be 38.43% Mn, 16.80% C, and 44.77% O. Determine the compound’s empirical formula. 3. Obtain the simplest whole-number ratio of moles: Mn : C : O 1 : 2 : 4 4. Determine the empirical formula. MnC2O4

  38. Practice p 333 # 46-47 • A blue solid is found to contain 36.84% N and 63.16% O. What is the empirical formula? • Determine the empirical formula for a compound that contains 35.98% Al and 64.02% S.

  39. Molecular Formula For many compounds, the empirical formula is not the true formula. -Chemists have learned, though, that acetic acid is a molecule with the formula C2H4O2, which is the molecular formula for acetic acid. -molecular formula tells the exact number of atoms of each element in a molecule or formula unit of a compound. Notice the molecular formula for acetic acid (C2H4O2) has exactly twice as many atoms of each element as the empirical formula (CH2O). -The molecular formula is always a whole-number multiple of the empirical formula.

  40. Molecular Formula Example: Determine the molecular formula for maleic acid, which has a molar mass of 116.1g/mol. 1. empirical formula of the compound -composition of maleic acid is 41.39% C, 3.47% H, and 55.14% O (change the % to g)

  41. Molecular Formula Example: Determine the molecular formula for maleic acid. 1. empirical formula of the compound -the ratio of C:H:O is 1:1:1, making the empirical formula CHO 2. calculate the molar mass of CHO (empirical formula). -29.01g/mol 3. Determine the molecular formula for maleic acid,

  42. Molecular Formula Example: Determine the molecular formula for maleic acid. 3. Determine the molecular formula for maleic acid, -shows the molar mass of maleic acid is 4x that of CHO. 4. Multiply CHO by 4 to get C4H4O4

  43. Practice p 335 # 51, 53 51. A substance has a chemical composition of 65.45% C, 5.45% H and 29.09% O. The molar mass of the molecular formula is 110.0 g/mol. Determine the molecular formula. 53. A compound contains 46.68 % N and 53.32 % O. It has a molar mass of 60.01 g/mol. What is the molecular formula?

  44. Empirical Formula from Mass You can also calculate the empirical formula of a compound from mass of individual elements. Example: Determine the empirical formula for ilmenite, which contains 5.41g Fe, 4.64g Ti and 4.65g O. 1. Multiply the mass by molar mass to get moles 5.41g Fe x 1 mol Fe = 0.0969 mol Fe 55.85 g Fe 4.64g Ti x 1 mol Ti = 0.0969 mol Ti 47.88g Ti 4.65g O x 1 mol O = 0.291 mol O 16.00g O

  45. Empirical Formula from Mass Example: Determine the empirical formula for ilmenite, which contains 5.41g Fe, 4.64g Ti and 4.65g O. 2. Multiply by the smallest number to get the mole ratio. 0.0969 mol Fe : 0.0969 mol Ti : 0.291 mol O 0.0969 mol 0.0969 mol 0.0969 mol 1 : 1 : 3 3. Calculate the empirical formula. FeTiO3

  46. Practice p 337 # 54-55

  47. TEST!!!!

  48. Stoichiometry Practice – Conservation of Mass For the following balanced chemical equations, determine all possible mole ratios. • HCl(aq) + KOH(aq) KCl(aq) + H2O(l) • 2. 2Mg(s) + O2(g) 2MgO(s) • 2HgO(s) 2Hg(l) + O2(g)

  49. Stoichiometric Calculations Many times we need to determine a certain amount of product from a reaction or want to know how much product will form from a given amount of reactant. To do this you need: 1. Balanced chemical equations 2. Mole ratios Moles of known xmoles of unknown = moles of known

  50. Stoichiometric Calculations: mole-mole Example: If you put 0.0400 mol of K into water, how much hydrogen gas will be produced? 2K(s) + 2H2O(l) →2KOH(aq) +H2(g) Mole ratio between K and KOH is 2 mol K or 1 mol H2 1 mol H2 2 mol K moles of known xmoles of unknown moles of known 0.0400 mol K x 1 mol H2 = 2 mol K 0.0200 mol H2

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