Section2_FundamentalsII

Section2_FundamentalsII - Fundamentals II Mass Relations,...

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Unformatted text preview: Fundamentals II Mass Relations, Yields and Limiting Reagents By the end of this section, you should be able to: (1) Determine the percentage composition by mass of elements in compounds (2) Balance equations for chemical reactions (3) Calculate the yield of a reaction, and identify the limiting reagent Page 1 1.10 Mass Relations in Chemical Formulae (MH5, 3.3) Recall: Molar Mass The molar mass of pure substance (element or compound) is the mass in grams of 1 mole of that substance. The molar mass (or, molecular weight) of a compound is the sum of the average atomic masses of its constituent elements. e.g., glucose OH "CH2" HO HO O "CH" OH OH Molecular formula: C6H12O6 Molar mass (molecular weight): 6 12.011 + 12 1.008 + 6 15.999 = 180.156 g mol-1 Page 2 The number of moles, n, of a pure substance is given by: n = m / MM where m is the mass of the substance in grams, and MM is its molecular weight in grams per mole. e.g., How many moles of glucose are there in 1.15 g? n = 1.15 g / 180.156 g mol-1 = 6.38 10-3 mol Percentage Composition by Mass The composition of a compound is often specified by citing the mass percents of the elements present. Let's take a 100 g sample of glucose. What is the mass in grams of each element? glucose Let's first figure out how many moles of water we are dealing with: m = MM n 100g = 180.156 g/mol * n n = 100g / (180.156g/mol) = 0.555mol Page 3 Therefore there are: 6 * 0.555 mol = 3.33 ___ moles of C atoms ___ moles of H atoms ___ moles of O atoms in 100 g of glucose 12 * 0.555 mol = 6.66 3.33 3.33 ___ moles of C atoms weigh 3.33 mol * 12.01 g/mol = 40.0 g 6.66 ___ moles of H atoms weigh 6.66 * 1.008 = 6.71 g 3.33 ___ moles of O atoms weigh 3.33 * 16.00 = 53.29 g lucose Therefore the mass composition of water is C : 40% ; H : 6.71 % ; O : 53.29% NOTE: These values are in mass %. Page 4 C6H12O6 The composition of gluocse in mole % is: %C = %O = 6 / (6+12+6) = 25% %H = 12/(6+12+6) = 50% The (weight !) percent composition of a compound is often obtained as the result of a chemical analysis. This can be used to obtain the simplest (= empirical) formula of the compound. Example: (MH5 3.38) A chemical analysis reveals that the percent composition of saccharin (an artificial sweetener) is 45.90 % C, 2.75 % H, 26.2 % O, 17.50 % S, and 7.65 % N. Determine the simplest formula. Imagine : 100g sachharin, ie., 45.9g C, 2.75g H, 26.2g O, 17.5g S, 7.65g N C: 45.9g/12.01g/mol = 3.82 mol; H = 2.75/ 1.008 = 2.73 mol O: 26.2/16.00 = 1.64 mol; S: 17.50/32.07 = 0.546 mol N: 7.65/14.07 = 0.544 mol lowest common denominator C:H:O:S:N = 7:5:3:1:1 C7H5O3SN Page 5 1.11 Mass Relations in Reactions (MH5, 3.4) Chemical reactions are written as equations, which have the form: Reactants Products It would be better to write: Reactants Products to emphasize that reactions are actually equilibria, but we won't worry about this now. In order to write an equation, you must know the chemical identities and physical states of the reactants and the products. In a balanced equation, there are the same number of atoms of a given element on both sides, and the total charge on both sides is also the same. e.g., H2 + O2 H2O 2 H2 + O2 2 H2O is not balanced is balanced Page 6 Use the symbols g, l, s to designate the physical state of reactants and products; aq is used for species dissolved in water (= aqueous solution), especially ions. You also have to "know" what happens during the reaction, e.g., when hydrogen is burned (combusted), a lot of heat is generated, and thus product H2O is g, not l. 2 H2(g) + O2(g) 2 H2O(g) Table salt is dissolved in water NaCl(s) Na+(aq) + Cl-(aq) aqueous or aquated Combustion of glucose Combustion means "burning in excess oxygen." All carbon is transformed to carbon dioxide, all hydrogen to water. C6H12O6 + O2 CO2 + H2O There are 6 reactant C atoms and 12 reactant H atoms, so there must 6 and 12, respectively, in the products, too: C6H12O6 + O2 6 CO2 + 6 H2O Page 7 Now, there are 18 product O atoms, so there must be 18 in the reactants: C6H12O6 + 6 O2 6 CO2 + 6 H2O Now add the states: g C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(l) And we're done! All chemical equations can be balanced using this iterative procedure. You should be very comfortable balancing chemical equations. Example The "rocket fuel" used by NASA is a mixture of two liquids: hydrazine (N2H4) and dinitrogen tetraoxide (N2O4) The products of the reaction are gaseous nitrogen and water vapor. (H2O) (N2) Page 8 A lot of thermal energy is generated during the reaction. The gas molecules are expelled from the rocket engine with an extremely high velocity; that is what makes the rocket "fly". Write a balanced equation for the rocket "engine". N2H4 + N2O4 ---> N2 + H2O N2H4 + N2O4 ----> N2 + 4H2O 2N2H4 + N2O4 ----> N2 + 4H2O 2N2H4(l) + N2O4(l) ---> 3N2(g) + 4H2O(g) + heat! Page 9 Example (MH5 3.56) Phosphine gas reacts with oxygen according to the following (unbalanced) equation: 4 PH3(g) + 8 O2(g) 1 P4O10(s) + 6 H2O(g) a) How many grams of P4O10 are produced from 12.43 mol of phosphine? many grams of oxygen gas are required for producing 1.000 g of water? b) How A) PH3 : P4O10 = 411 => 12.43 mol PH3 == 12.43/4 = 3.11 mol P4O10 mm(P4O10) = 4 * 30.97 + 10 * 16.00 = 283.88 g/mol => mass of P4O10 = 3.11 mol * 283.88 g/ mol = 882.16g b) 1g H2O = 1g / 18.016g/mol = .555 mol H2O : O2 = 6:8 = 1:1.3333 0.555 mol * 1.33333 = .74 mol O2 required 0.74mol * (2* 16.00)g/mol = 2.73 g O2 are needed Page 10 1.13 Limiting Reactant and Yield of a Reaction (MH5, 3.4) A cheese sandwich consists of 1 bun and 1 slice of cheese. You have 100 buns, and 32 slices of cheese. How many cheese sandwiches can you make? 32 Cheese is the limiting reactant, buns are the excess reactant. 1 mol of oxygen is mixed with 1 mol of hydrogen, ignition of this mixture leads to the formation of water. What is the limiting reactant? What is the excess reactant? 2H2(g) + O2(g) ---> 2H2O(g) Require 2 : 1 Available 1 : 1 (stoichiometric) H2 O2 Page 11 Reactants are not always mixed exactly in their stoichiometric proportions. Often you will have to decide which is the limiting reactant. The final concentration or pressure of the limiting reactant is always ZERO. Page 12 Example (MH5 3.64) Aluminum reacts with sulfur gas to form aluminum sulfide, Al2S3. Initially, 1.18 mol of aluminum and 2.25 mol of sulfur are combined. What is the limiting reactant? Al What is the maximum amount (MH5 calls this the "theoretical yield") of aluminum sulfide that can be obtained (mol)? a) 16AL(s) + 3S8(s) ----> 8 Al2S3(s) 1.18 mol Al requies 3/16 * 1.18 = .221 mol S8 for complete reaction therefore there is around 10 times excess of S8 b) Al: Al2S3 = 16:8 (2:1) All Al will be consumed : 1.18 therefore 1/2 * 1.18 = 0.59 mol Al2S3 wil form Page 13 Yield Suppose you carry out a chemical reaction. Based on the reactant amounts used, and on the equation of the reaction you expect 10 g of product. However, due to experimental imperfections only 9 g are obtained. (Some material was "lost" during handling ...) This means you only obtained 9/10 = 0.9 (= 90%) of the expected product. We say that the yield of the reaction is 0.9 = 90% (MH5 calls this the experimental yield). In general: Yield = mass obtained maximum mass expected "practical yield" "chemical yield" which is the same as Yield = moles obtained maximum number of moles expected (usually expressed in %) Chemical engineers optimize chemical procecces to maximize the yield (maximum yield = maximum $$$). Page 14 Example: 4 mol of H2 and 2 mol of O2 are mixed and ignited. 1.3 mol H2O are found as reaction product. What is the yield? 2H2(g) + O2(g) ----> 2H2O(g) Required 2:1 Available 4:2 both will be consumed H2 :H2O 2:2 4 mol H2 should produce 4 mol H2O But only 1.3 mol are recovered therefore yield = 1.3 / 4.0 *100% = 32.5% Page 15 Example: (MH5, 3.66) When iron and steam react at high temperatures the following reaction takes place: 3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4 H2(g) How much iron (in grams) must react with excess steam to form 897 g of Fe3O4 if the yield is 69% ? 879 879g Fe3O4 = 69% yield => 100% yield = 100/69% * 879 = 1274g Fe3O4 1274g Fe3O4 = 1273g/ (3*55.85 + *16.00) g/mol = 5.502 mol Fe3O4 Fe3O4 : Fe = 1:3 => 3 * 5.502 = 16.506 mol Fe are required 16.506 mol = 16.506 mol * 55.85 g/mol = 921.8g Fe Page 16 ...
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This note was uploaded on 06/20/2008 for the course CHEM 024b taught by Professor Jones during the Winter '07 term at UWO.

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