1. 1 Chapter 3 Stoichiometry

2. 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. 3 3.1 Atomic Masses in 1961, 12 C-isotope has been instituted as the standard for determining the atomic masses " 12 C 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) KE = 1/2 x m x v 2 v = (2 x KE/m) 1/2 F = q x v x B 3.1 Light Heavy Mass spectrometer will be used to determine: 1. Accurate mass value for individual atoms. 2. The isotopic composition of a natural element

4. 4  All the mass in the periodic table are related to the 12 C-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. 3.1 Atomic Masses

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

6. 6 3.2 The Mole The mole is “the number equal to number of carbon atoms in exactly 12 grams of pure 12 C” Avogadro has determined this number to be N A = 6.02214 X 10 23 Particle/mole Molar mass is the mass of 1 mole of in grams eggs shoes marbles atoms 1 mole 12 C atoms = 6.022 x 10 23 atoms = 12.00 g 1 12 C atom = 12.00 amu For any element atomic mass (amu) = molar mass (grams) 6.022 X 10 23 amu = 1 g exact

7. 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. 8 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 SO 2 1S32.07 amu 2O+ 2 x 16.00 amu SO 2 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO 2 = 64.07 amu 1 mole SO 2 = 64.07 g SO 2

9. 9 Example 3.7: Calcium carbonate (Ca CO 3 ), also called calcite, is the principle found in limestone, marble, pearls… 1.Calculate the molar mass of calcium carbonate. 2.A certain sample of calcium carbonate contains 4.86 moles. What is the mass in grams of this sample? 3.What is the mass of the CO 3 2- ion present? 3.3 Molar Mass of Compounds Example 3.8: Isopentylacetate (C 7 H 14 O 2 ), the compound responsible for the scent of bananas interestingly, bees release about 1  g of this compound when they sting. the resulting scent attracts other bees to join the attack. 1.How many molecules of isopentyl acetate are releases in a typical bee sting? 2.How many atoms of carbon are present?

10. 10 3.4 percent Composition of Compounds The percent composition of elements in any compound con be determined using the comparing ratio equation: n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound C2H6OC2H6O %C = 2 x (12.01 g) 46.07 g x 100% = 52.14%H = 6 x (1.008 g) 46.07 g x 100% = 13.13%O = 1 x (16.00 g) 46.07 g x 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.0%

11. 11 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 C 14 H 20 N 2 SO 4. Compute the mass percent of each element. 3.4 percent Composition of Compounds

12. 12 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. O 2 Organic Compound CO 2, H 2 O, N 3 complete reaction figure 3.5: is a schematic diagram for combustion device:

13. 13 Masses of CO 2, H 2 O and other gases will be used to determine:  Empirical Formula and then Molecular Formula if the Molar Mass is known 3.5 Determining the Formula of a Compound g CO 2 mol CO 2 mol Cg C g H 2 O mol H 2 Omol Hg H g of O = g of sample – (g of C + g of H) Combust 11.5 g ethanol Produce 22.0 g CO 2 and 13.5 g H 2 O 6.0 g C = 0.5 mol C 1.5 g H = 1.5 mol H 4.0 g O = 0.25 mol O Empirical formula C 0.5 H 1.5 O 0.25 Divide by smallest subscript (0.25) Empirical formula C 2 H 6 O Exercise: (Determining Empirical Formula) Empirical formula is the simplest chemical formula for any compound

14. 14 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 3.5 Determining the Formula of a Compound molecular formula = (empirical formula) n [n = integer] molecular formula = C 6 H 6 = (CH) 6 empirical formula = CH

15. 15 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. 3.5 Determining the Formula of a Compound Assume 100g of caffeine 1.m C = 49.48 g m c = 49.48/12= 4 2.m H = 5.15 g m H =5.15/1 = 5 3.m N = 28.87 g m N = 28.87/14 = 2 4.m O = 16.49 g m O = 16.49/16= 1 Solution: empirical formula ---(C 4 H 5 N 2 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 ----> C 8 H 10 N 4 O 2

16. 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 (CH 4 ) produces (yields) carbon dioxide (CO 2 ) and dihydrogen oxide (H 2 O) [ common name, water] CH 4 + O 2  CO 2 + H 2 O ReactantsProducts (this equation is not balanced)

17. 17 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) How to “Read” Chemical Equations 2 Mg + O 2  2 MgO 2 atoms Mg + 1 molecule O 2 makes 2 formula units MgO 2 moles Mg + 1 mole O 2 makes 2 moles MgO IS NOT: 2 grams Mg + 1 gram O2 makes 2 g MgO 48.6 grams Mg + 32.0 grams O 2 makes 80.6 g MgO 3.6 Chemical Equations: A representation of a chemical reaction

18. 18 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: C 2 H 6 + O 2 CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O start with C or H but not O 2 carbon on left 1 carbon on right multiply CO 2 by 2 C 2 H 6 + O 2 2CO 2 + H 2 O 6 hydrogen on left 2 hydrogen on right multiply H 2 O by 3  Start by balancing those elements that appear in only one reactant and one product or the most complicated one.  Write the correct formula(s) for the reactants and the product(s) e.g. Ethane reacts with oxygen to form carbon dioxide and water

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

20. 20 3.8 Stoichiometric Calcalations: Amount of reactants and Products 1.Write balanced chemical equation 2.Convert quantities of known substances into moles 3.Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4.Convert moles of sought quantity into desired units

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

22. 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 reactant is the reactant that is consumed first, limiting the amounts of products formed.

23. 23 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. % Yield = Actual Yield Theoretical Yield x 100

24. 24 Do You Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe Calculate the mass of Al 2 O 3 formed. g Almol Almol Fe 2 O 3 neededg Fe 2 O 3 needed OR g Fe 2 O 3 mol Fe 2 O 3 mol Al neededg Al needed 124 g Al 1 mol Al 27.0 g Al x 1 mol Fe 2 O 3 2 mol Al x 160. g Fe 2 O 3 1 mol Fe 2 O 3 x = 367 g Fe 2 O 3 Start with 124 g Alneed 367 g Fe 2 O 3 Have more Fe 2 O 3 (601 g) so Al is limiting reagent

25. 25 Use limiting reagent (Al) to calculate amount of product that can be formed. g Almol Almol Al 2 O 3 g Al 2 O 3 124 g Al 1 mol Al 27.0 g Al x 1 mol Al 2 O 3 2 mol Al x 102. g Al 2 O 3 1 mol Al 2 O 3 x = 234 g Al 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe