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1 Chapter 4 Quantities of Reactants and Products 4.1 Chemical Equations (p. 124) A chemical equation is a shorthand description of a chemical reaction using symbols and formulas to represent the elements and compounds involved. It is a symbolic representation of a chemical reaction in terms of chemical formulas. Reactants -----> Products or Reactants = Products The symbols > or =, mean forms, yields, or produces. The reactants are the starting substances and the products are the substances formed. The following abbreviations indicate the phases of the reacting participants. gas =( ); liquid = ( ); solid = ( ); aqueous = ( ) Combustion of candle wax, C 50 H 102, with molecular oxygen, O 2, produces carbon dioxide, CO 2, and water, H 2 O, and is communicated as follows. Indicate the phases in the following equation. 2Na( ) + 2H 2 O( ) ----> 2NaOH( ) + H 2 ( ) 2 C 50 H 102 (g) + 51 O 2 (g) ---> 100 CO 2 (g) H 2 O(g) are the numbers placed in front of formulas in a chemical equation to balance the equation, thereby indicating the combining ratios of the reactants and the products. Balanced equations obey the law of conservation of mass. That is the total mass before a reaction takes place will equal the total mass after the reaction is complete. This also holds true for number of atoms involved in the reaction. For example, the reaction of solid sodium metal with gaseous chlorine is indicated as follows. 2 Na(s) + Cl 2 (g) ----> 2 NaCl(s) This balanced reaction can be interpreted on at least three scales or levels: 1. On the smallest scale possible (nanoscale level or atomic level), 2 atoms of sodium react with 1 molecule of chlorine to produce 2 formula units of sodium chloride. 2. On a molar scale, 2 moles of sodium atoms react with 1 mole of chlorine molecules to produce 2 moles of sodium chloride. 3. On a mass scale (practical scale) g of sodium react with g of chlorine to produce g o sodium chloride. Stoichiometry is the relationship that describes the relation between the masses of reactants and products. 4.2 Patterns of Chemical Reactions (p. 126) In a combination reaction, two substances combine to form a third substance. {A + B ----> AB} In a decomposition reaction, a single compound reacts to give two or more substances. {AB ----> A + B} In a displacement reaction (or single replacement reaction), an element reacts with a compound and takes the place of (displaces) one of the elements in the original compound. {A + BC -----> AB + C}

2 An exchange (metathesis reaction or double-replacement reaction), is a reaction between compounds that, when written as a molecular equation, appears to involve the exchange of parts between the two reactants. Such reactions often occur between ions in solution that can result in the formation of insoluble substances, weak electrolytes, or nonelectrolytes. {AX + BY -----> AY + BX} In combustion reactions a substance reacts with oxygen, usually with the rapid release of heat. 4.3 Balancing Chemical Equations (p. 133) 1. The equation will be understood to proceed from left to right. The reactants are on the left side of the arrow and the products are on the right. H 2 + Cl > 2 HCl 2. Equations are balanced by adjusting coefficients in front of formulas, never by changing subscripts within formulas. Remember that a 1 is understood when a coefficient is not present. 3. It is best to start with an element that appears in only one compound on each side of the arrow. 4. Next balance any element that appears in more than one compound on either the right or left. 5. Balance free elements last. That is, balance any element that appears in elemental form on the right or left. 6. When polyatomic species (NH 4 +, SO 4-2, OH -, etc.) appear on both sides of the arrow in compounds or as ions, balance them as units rather than individual elements. 7. Leave the coefficients in the lowest whole number ratio. Balance the following equations. #1 H 2 + O > H 2 O #2 N 2 + H > NH 3 #3 Na(s) + H 2 O(l) ---> NaOH(aq) + H 2 (g) #4 H 3 PO 4 + Ca(OH) > Ca 3 (PO 4 ) 2 + H 2 O #5 Ca 3 (PO 4 ) 2 + SiO 2 + C ---> P 4 + CaSiO 3 + CO #6 As 2 S 3 + O > As 2 O 3 + SO 2 Balance the following combustion of hydrocarbon rxns. #1 CH 4 + O 2 ---> CO 2 + H 2 O #2 C 2 H 4 + O 2 ---> CO 2 + H 2 O #3 C 2 H 6 + O 2 ---> CO 2 + H 2 O #4 C 4 H 8 + O 2 ---> CO 2 + H 2 O #5 C 5 H 10 + O 2 ---> #6 C 50 H O 2 ---> #7 Balance the equation for the combustion of C 6 H 6. #8 What is the sum of the coefficients in the balanced equation for the combustion of heptane, C 7 H 14?

3 4.4 The Mole and Chemical Reactions: The Macro-Nano Connection (p. 136) A balanced equation can be used to derive molar ratios of the participants in the reaction. 2 C 4 H 10 (g) + 13 O 2 (g) ---> 8 CO 2 (g) + 10 H 2 O(g) Using the equation above, derive the following relationships. mol O 2 / mol C 4 H 10 or mol CO 2 / mol C 4 H 10 or mol H 2 O/ mol O 2 Stoichiometry Problem #1 How many moles of oxygen are required to react with 0.50 mol of butane, C 4 H 10? [3.25 mol O 2 ] Stoichiometry Problem #2 How many moles of water can be formed from the complete reaction of 3 moles of oxygen with an excess of hydrogen? [6 moles] Stoichiometry Problem #3 How many moles of water can be formed from the reaction of moles of phosphoric acid with sufficient calcium hydroxide? [0.75 moles] 2 H 3 PO Ca(OH) > Ca 3 (PO 4 ) H 2 O Stoichiometry Problem #4 How many moles of phosphoric acid are required to react with g of calcium hydroxide? [Molar mass of calcium hydroxide is g/mol.] [ mol] 2 H 3 PO Ca(OH) > Ca 3 (PO 4 ) H 2 O Stoichiometry Problem #5 What mass of oxygen is needed to combust 1.00 g of C 6 H 6? [3.072g] Stoichiometry Problem #6 What mass of water could be formed from the complete combustion g of C 8 H 18? [49.68 g] 4.5 Reactions with One Reactant in Limited Supply (p. 141) A balanced equation shows the relative proportions between reactants and between products. This relation is fixed but the actual amounts of reactants present can vary. This means that sometimes there can be an excess of one or more of the reactants. In other words the quantity of one reactant will control the amount of products that will be formed. The following steps can be used to calculate the mass produced of a given substance when one or more of the reactants are present in a limiting mass. Step 1. Calculate the amount of a specific product (moles or grams, as needed) that can be formed from each reactant. Step 2. The reactant that gives the least amount of product is the limiting reactant. The limiting reactant will determine the amount of product formed in the reaction.)

4 Limiting Reagent Problem #1 How many moles of H 2 O will be formed when 4.0 moles of H 2 are allowed to react with 1.0 mole of O 2? [2.0 mol H 2 O] 2 H 2 + O > 2 H 2 O Limiting Reagent Problem #2 How many moles of BaSO 4 will be produced from a mixture of 3.5 moles H 2 SO 4 and 2.5 moles BaCl 2? (2.5 moles BaSO 4 ) H 2 SO 4 + BaCl > BaSO HCl Limiting Reagent Problem #3 A mixture of 35.0 g of hydrogen and 270 grams of oxygen react to form water. How many grams of water will form? (304 g H 2 O) Limiting Reagent Problem #4 How many grams of silicon will form when 15.0 grams of boron are reacted with 50.0 grams of silicon dioxide according to the following equation? (23.4 g Si) 4 B + 3 SiO > 3 Si + 2 B 2 O 3 Limiting Reactant Demonstration The reaction of Vinegar and Sodium Bicarbonate 7 bottles contain 0.25 mol of acetic acid. Balloons are attached containing 3g, 6g, 12g, 18g, 24g, 30g, and 36g of sodium bicarbonate (FW = 84.01). #1 What is the limiting reagent in each bottle? #2 How much carbon dioxide will be produced in each bottle? HC 2 H 3 O 2 (l) + NaHCO 3 (s) > NaC 2 H 3 O 2 (aq) + H 2 O(l) + CO 2 (g) Bottle Moles HC 2 H 3 O 2 Moles NaHCO 3 Limiting Reagent Moles CO 2 (g) mol 3.00 g = mol mol 6.00 g = mol mol 12.0 g = mol mol 18.0 g = mol mol 24.0 g = mol mol 30.0 g = mol mol 36.0 g = mol

5 4.6 Evaluating the Success of a Synthesis: Percent Yield (p. 146) Theoretical yield is the calculated maximum amount of product that can be obtained from a given amount of reactant, according to the chemical equation. The actual yield is the measured amount of product that is finally obtained. The actual yield is often less than the theoretical yield. (And certainly is never more!) The percent yield is the ratio of the actual yield to the theoretical yield multiplied by 100. Percent yield = actual yield x 100 theoretical yield Percent Yield Problem #1 If the theoretical yield calculated for a reaction is 14.8 g, and the amount of product obtained is 9.25 g, what is the percent yield? [62.5%] Percent Yield Problem #2 Carbon tetrachloride, CCl 4, was prepared by reacting excess carbon disulfide, CS 2, with 100. g of chlorine. Calculate the percent yield if 65.0 g of CCl 4 was obtained from the reaction. [89.9%] CS Cl 2 ---> CCl 4 + S 2 Cl 2 Percent Yield Problem #3 Silver bromide was prepared by reacting g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if g of silver bromide was obtained from the reaction. [91.9%] MgBr AgNO 3 ---> Mg(NO 3 ) AgBr Percent Yield Problem #4 [Hint: This involves a limiting reagent problem!] When 100 g of oxalic acid (H 2 C 2 O 4 ) and 100 g of methanol (CH 3 OH) were allowed to react, 38 g of methyl oxalate (C 4 H 6 O 4 ) were isolated. What is the percent yield? [29%] H 2 C 2 O CH 3 OH <---> C 4 H 6 O H 2 O oxalic acid methanol methyl oxalate 4.7 Percent Composition and Empirical Formulas (p. 150) In a combustion analysis of a compound containing carbon and hydrogen, the compound reacts with oxygen and all of the carbon in the compound is converted to carbon dioxide and the hydrogen in the compound is converted to water. 2 C 4 H 10 (g) + 13 O 2 (g) ---> 8 CO 2 (g) + 10 H 2 O(g) The law of conservation of mass and the concept of following the mass of elements from one single compound to several compounds provide a basis for determining percent composition, empirical formula and molecular formula. Combustion Analysis Problem #1 Benzene is a liquid compound composed of carbon and hydrogen. A sample of benzene weighing 342 mg is burned in excess oxygen and forms 1156 mg of carbon dioxide. What is the percent composition of benzene? [92.25%C & 7.75% H] Combustion Analysis Problem #2 n-butyl phthalate is used as an insect repellant and is composed of carbon, hydrogen, and oxygen. When a g sample was subjected to combustion analysis, g of water and g of carbon dioxide were produced. In another analysis, the molecular weight was determined to be g/mol. Calculate the empirical formula and the molecular formula. [C 8 H 11 O 2 & C 16 H 22 O 4 ] Combustion Analysis Problem #3 A g sample of a compound composed of carbon, hydrogen and oxygen is subjected to combustion analysis, producing g CO 2 and g H 2 O. What is the empirical formula for the compound? [C 3 H 6 O] If the molecular weight is g/mol, what is the molecular formula? [C 6 H 12 O 2 ]