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Chapter 3

Chapter 3. Stoichiometry. Preview. the contents of this chapter will introduce you to the following topics: Atomic mass, Mole concept, and Molar mass (average atomic mass). Number of atoms per amount of element. Percent composition and Empirical formula of molecules.

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Chapter 3

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  1. Chapter 3 Stoichiometry

  2. Preview the contents of this chapter will introduce you to the following topics: • Atomic mass, Mole concept, and Molar mass (average atomic mass). • Number of atoms per amount of element. • Percent composition and Empirical formula of molecules. • Chemical equations, Balancing this equate and Stoichiometric calculations including limiting reactants.

  3. Mass spectrometer will be used to determine: 1. Accurate mass value for individual atoms. 2. The isotopic composition of a natural element Heavy Heavy Light Light KE = 1/2 x m x v2 v = (2 x KE/m)1/2 F = q x v x B 3.1 3.1 Atomic Masses • in 1961, 12C-isotope has been instituted as the standard for determining the atomic masses "12C is assigned a mass of exactly 12 atomic mass unit (amu)" • The most accurate method currently available for comparing the masses of atoms involves the use of Mass Spectrometer (figure 3.1)

  4. 3.1 Atomic Masses • All the mass in the periodic table are related to the 12C-mass. Sometimes, they are called the atomic weight for each element. • For carbon element the mass in the table is "12.01" and not exact 12.00?? • Natural carbon found on earth is a mixture of the isotopes C-12, C-13 and C-14 isotopes, and the mass in table is an average value reflecting the average of the isotopes composing it. The average masses are calculating using the following equation: A.A.M = ∑ (atomic mass X fractions of its abundance) Fraction = z%⁄ 100 • Average atomic mass is traditionally has been called atomic weight of element. • the average atomic mass will enable us to count atoms by weighing a sample of any element.

  5. 3.1 Atomic Masses Example 3.1: When a sample of natural copper is vaporized and injected into a mass spectrometer we got two isotopes: 63.29Cu (69.09%) and 65.29Cu (30.91%). The mass value for 63Cu & 65Cu are 62.93 and 64.93 amu, respectively. Use these data to compute the average mass of natural copper.

  6. eggs shoes Molar mass is the mass of 1 mole of in grams marbles atoms 3.2 The Mole The mole is “the number equal to number of carbon atoms in exactly 12 grams of pure 12C” Avogadro has determined this number to be NA = 6.02214 X 1023Particle/mole 1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g 1 12C atom = 12.00 amu For any element atomic mass (amu) = molar mass (grams) 6.022 X 1023 amu = 1 g exact

  7. 3.2 The Mole Example 3.3: Aluminum (Al) is high resistance to corrosion. Compute both the number of moles of atoms and the number of atoms in a 10.0-g sample of aluminum.

  8. 1S 32.07 amu 2O + 2 x 16.00 amu SO2 SO2 64.07 amu 3.3 Molar Mass of Compounds Compounds are a collection of atoms, and Molar Mass is the mass in grams of 1 Mole of the compound (traditionally is called Molecular weight) Molar mass for molecule or compound = ∑atomic masses in g For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2

  9. 3.3 Molar Mass of Compounds Example 3.7: Calcium carbonate (Ca CO3), also called calcite, is the principle found in limestone, marble, pearls… • Calculate the molar mass of calcium carbonate. • A certain sample of calcium carbonate contains 4.86 moles. What is the mass in grams of this sample? • What is the mass of the CO32- ion present? Example 3.8: Isopentylacetate (C7H14O2), the compound responsible for the scent of bananas interestingly, bees release about 1 mg of this compound when they sting. the resulting scent attracts other bees to join the attack. • How many molecules of isopentyl acetate are releases in a typical bee sting? • How many atoms of carbon are present?

  10. n x molar mass of element x 100% molar mass of compound 1 x (16.00 g) 2 x (12.01 g) 6 x (1.008 g) %C = %H = %O = x 100% = 13.13% x 100% = 34.73% x 100% = 52.14% 46.07 g 46.07 g 46.07 g C2H6O 3.4 percent Composition of Compounds The percent composition of elements in any compound con be determined using the comparing ratio equation: n is the number of moles of the element in 1 mole of the compound 52.14% + 13.13% + 34.73% = 100.0%

  11. 3.4 percent Composition of Compounds Example 3.10: Penicillin, the first of a now large number of antibiotic was discovered accidentally by the Scottish bacteriologist Alexander Aeming in 1928, but he was never able to isolate it as a pure component. This and similar antibiotics have saved millions of lives that might have been lost due to infections. Penicillin-F has the formula C14 H20 N2 SO4. Compute the mass percent of each element.

  12. figure 3.5: is a schematic diagram for combustion device: 3.5 Determining the Formula of a Compound For any new compound the 1st item of interest is the determination of its chemical formula. This performed by decomposing know sample mass into its component or reacting it with Oxygen to produce its oxides e.g. O2 Organic Compound CO2, H2O, N3 complete reaction

  13. g CO2 g H2O mol CO2 mol H2O mol C mol H g C g H 3.5 Determining the Formula of a Compound Masses of CO2, H2O and other gases will be used to determine: • Empirical Formula and then Molecular Formula if the Molar Mass is known Exercise: (Determining Empirical Formula) Combust 11.5 g ethanol Produce 22.0 g CO2 and 13.5 g H2O 6.0 g C = 0.5 mol C 1.5 g H = 1.5 mol H 4.0 g O = 0.25 mol O g of O = g of sample – (g of C + g of H) Empirical formula C0.5H1.5O0.25 Divide by smallest subscript (0.25) Empirical formula C2H6O Empirical formula is the simplest chemical formula for any compound

  14. 3.5 Determining the Formula of a Compound molecular formula = (empirical formula)n [n = integer] molecular formula = C6H6 = (CH)6 empirical formula = CH If the Molar mass for any compound is known one can get the chemical or molecular formula by the following: a. calculate empirical mass = ∑ atomic mass b. find the multiplication factor = MM/EM c. molecular formula = multiplicity x EF Note: if MM = EM molecular formula = empirical formula

  15. Solution: Assume 100g of caffeine • mC = 49.48 g mc = 49.48/12 = 4 • mH = 5.15 g mH =5.15/1 = 5 • mN = 28.87 g mN = 28.87/14 = 2 • mO = 16.49 g mO = 16.49/16 = 1 3.5 Determining the Formula of a Compound Example 3.13: Caffeine a stimulant found in coffee, tea, and chocolate, contain 49.48 % C, 5.159% H, 28.87 % N, and 16.49 % O by mass has a molar mass 194.2 g/mol. Determine the molecular formula of caffeine. empirical formula ---(C4 H5 N2 O)n E.M. = (12x4) + (5x1) + (14x2) + (16x1) = 48 + 5 + 28 + 16 = 97 E.F x n = 194.2 ----- n = 194.2/ 97 ~ 2 Molecular Formula ----> C8 H10 N4 O2

  16. 3.6 Chemical Equations: A process in which one or more substances is changed into one or more new substances is a chemical reaction Note: In chemical reaction atoms are neither created nor destroyed, i.e., all atoms present in reactants must be accounted for among the products. this is the called balancing process. For example: The combustion (oxidation) of methane (CH4) produces (yields) carbon dioxide (CO2) and dihydrogen oxide (H2O) [common name, water] CH4 + O2  CO2 + H2O Reactants Products (this equation is not balanced)

  17. 3.6 Chemical Equations: How to “Read” Chemical Equations Chemical Equation for a reaction gives two important types of information: • The nature of reactants and products (including the physical state: "s, l, g, aq“) • the relative number of each of the components (including coefficient moles, and number of species) A representation of a chemical reaction 2 Mg + O2  2 MgO 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO 2 moles Mg + 1 mole O2 makes 2 moles MgO IS NOT:2 grams Mg + 1 gram O2 makes 2 g MgO 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO

  18. C2H6 + O2 CO2 + H2O C2H6 + O2 C2H6 + O2 CO2 + H2O 2CO2 + H2O 2 hydrogen on right multiply H2O by 3 6 hydrogen on left 3.7 Balancing Chemical Equation: Most chemical equations can be balanced by inspection, that is, by trial and error, but you have to follow some procedure: • Write the correct formula(s) for the reactants and the product(s) • e.g. Ethane reacts with oxygen to form carbon dioxide and water • Start by balancing those elements that appear in only one reactant and one product or the most complicated one. start with C or H but not O multiply CO2 by 2 2 carbon on left 1 carbon on right

  19. 4 oxygen (2x2) C2H6 + O2 2CO2 + 3H2O 2 oxygen on left C2H6 + O2 2CO2 + 3H2O + 3 oxygen (3x1) C2H6 + O2 2CO2 + 3H2O 2C2H6 + 7O2 4CO2 + 6H2O Reactants Products 7 4 C 4 C 2 12 H 12 H 14 O 14 O 3.7 Balancing Chemical Equation: • Balance those elements that appear in two or more reactants or products. = 7 oxygen on right remove fraction multiply both sides by 2 • Check to make sure that you have the same number of each type of atom on both sides of the equation.

  20. 3.8 Stoichiometric Calcalations: Amount of reactants and Products • Write balanced chemical equation • Convert quantities of known substances into moles • Use coefficients in balanced equation to calculate the number of moles of the sought quantity • Convert moles of sought quantity into desired units

  21. 2CH3OH + 3O2 2CO2 + 4H2O grams CH3OH moles CH3OH moles H2O grams H2O molar mass CH3OH molar mass H2O coefficients chemical equation 4 mol H2O 18.0 g H2O 1 mol CH3OH x = x x 2 mol CH3OH 32.0 g CH3OH 1 mol H2O 209 g CH3OH 235 g H2O 3.8 Stoichiometric Calcalations: Amount of reactants and Products Methanol burns in air according to the equation If 209 g of methanol are used up in the combustion, what mass of water is produced?

  22. 3.9 Calculations Involving a Limiting Reactant When chemical reaction undergo, the reactants are often Mixed in Stoichiometric quantities, but if not what happen: The limiting reactantis the reactant that is consumed first,limiting the amounts of products formed.

  23. Actual Yield % Yield = x 100 Theoretical Yield 3.9 Calculations Involving a Limiting Reactant 1. Balance the equation. 2. Convert masses to moles. 3. Determine which reactant is limiting. 4. Use moles of limiting reactant and mole ratios to find moles of desired product. 5. Convert from moles to grams. Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction.

  24. Do You Understand Limiting Reagents? 2Al + Fe2O3 Al2O3 + 2Fe g Al mol Al mol Fe2O3 needed g Fe2O3 needed OR g Fe2O3 mol Fe2O3 mol Al needed g Al needed 1 mol Fe2O3 160. g Fe2O3 1 mol Al = x x x 27.0 g Al 2 mol Al 1 mol Fe2O3 Start with 124 g Al need 367 g Fe2O3 In one process, 124 g of Al are reacted with 601 g of Fe2O3 Calculate the mass of Al2O3 formed. 367 g Fe2O3 124 g Al Have more Fe2O3 (601 g) so Al is limiting reagent

  25. 2Al + Fe2O3 Al2O3 + 2Fe g Al mol Al mol Al2O3 g Al2O3 1 mol Al x 27.0 g Al 1 mol Al2O3 102. g Al2O3 = x x 2 mol Al 1 mol Al2O3 Use limiting reagent (Al) to calculate amount of product that can be formed. 234 g Al2O3 124 g Al

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