CHAPTER 3: PART 2 8/9/2015. A chemical change (a chemical reaction) converts one substance into another.

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1 8/9/015 A chemical change (a chemical reaction) converts one substance into another. CHAPTER 3: PART Chemical Equations and Stoichiometry Chemical reactions involve: 1. Breaking bonds in the reactants. (reactants are the stuff you start with). Forming new bonds in the products. (products are the new stuff you make) CH 4 and O CO and H O A Chemical Equation Uses Specific Notation to Show a Chemical Reaction H (g) + O (g) H O(l) Symbols in parentheses to the right of each formula indicate the state or form in which the substance exists. H (g) + O (g) H O(l) reactants products gas gas liquid The reactants ( on the left) and products (on the right) are separated by an arrow that points to the products. A plus sign (+) is used to separate individual reactants and products. Symbols: (aqueous) (g) gas (l) liquid (s) solid (aq) for a substance dissolved in water Chemical Equations Are Balanced Law of Conservation of Matter: Atoms are neither created nor destroyed in chemical reactions. Coefficients (numbers) are written to the left of the molecular formula to make the equation obey the conservation Law. There must be the same number of each type of atom on the reactant side and on the product side. H (g) + O (g) H O(l) H (g) + O (g) H O(l) 4.0 g 3.0 g 36.0 g What this law means: Atoms are not lost or gained. The total # of hydrogen and oxygen is the same on the left and right sides of the equation. The mass of the reactants must equal the mass of the products. 4 H O 4 H & O The coefficient one is never written. When no coefficient appearance before the molecular formula, it is assumed to be one. Coefficients must be integers, not fractions or decimals are allowed. Without coefficients, the equation is said to be unbalanced. 1

2 8/9/015 Balancing Chemical Equations How to Balance a Chemical Equation Balance a chemical equation by writing coefficients to the left of the molecular formulas until the number of of each element are equal in the reactants and products. Example Write a balanced equation for the reaction of propane (C 3 H 8 ) with oxygen (O ) to form carbon dioxide (CO ) and water (H O). Unbalanced H (g) + Cl (g) HCl(g) Step [1] Write the equation with the correct formulas. H Cl 1 H & 1 Cl atom C 3 H 8 + O CO + H O Balanced H (g) + Cl (g) HCl(g) The subscripts in a formula can NEVER be changed to balance an equation, because changing a subscript changes the identity of a compound. How to Balance a Chemical Equation (CHO Method) Step [] Balance the equation with coefficients one element at a time. Balance the C s first: How to Balance a Chemical Equation (CHO Method) Step [3] Check to make sure that the smallest set of whole numbers is used. Balance the H s next: Balance the O s last: Balancing Chemical Equations By Inspection NaBr(aq) + Cl (aq) NaCl(aq) + Br (aq) NaBr(aq) + Cl (aq) NaCl(aq) + Br (aq) Balancing Chemical Equations: Leave the Simplest Molecules for Last Unbalanced H S(g) + O (g) SO (g) + H O(l) H O (aq) + H S(aq) H O(l) + S(s) Balanced H S(g) + 3 O (g) SO (g) + H O(l) H O (aq) + H S(aq) H O(l) + S(s)

3 8/9/015 Balancing Chemical Equations: Start with the Most Complicated Molecule Balancing Chemical Equations That Contain Polyatomic Ions: the MINOH Method Unbalanced Ag CO 3 (s) Ag(s) + CO (g) + O (g) Balanced Ag CO 3 4 Ag(s) + CO (g) + O (g) MINOH: Me know how to balance equations! Metals Ions(Polyatomic) Nonmetals Oxygen Hydrogen Unbalanced Na CO 3 (aq) + Ca(NO 3 ) NaNO 3 (aq) + CaCO 3 (s) Balanced Na CO 3 (aq) + Ca(NO 3 ) (aq) NaNO 3 (aq) + CaCO 3 (s) Balancing Chemical Equations: the CHO Method and Using Temporary Fractions Unbalanced Step 1: balance C and H C H 6 (g) + O (g) C H 6 (g) + O (g) CO (g) + H O(l) CO (g) + 3H O(l) Mole Relationships in Chemical Equations The coefficients of a balanced equation indicate how many molecules of reactants and products are needed and created, respectively, in the reaction. A balanced chemical equation also indicates the number of moles of reactants and products Step : add up O C H 6 (g) + O (g) CO (g) + 3H O(l) O = 7 O Step 3: temporarily use a C H 6 (g) + 7 O (g) CO (g) + 3H O(l) fraction to balance the O Step 4: clear the fraction C H 6 (g) + 7O (g) 4CO (g) + 6H O(l) moles of CO molecule CO 1 mole of O 1 molecule O moles of CO molecules CO Coefficients are used to form Mole Ratios, which can serve as Conversion Factors A molar ratio is a ratio of the moles (from the coefficients of a balanced equation) for any two substances in an equation. 4 Fe(s) + 3 O (g) Fe O 3 (s) Use the mole ratios from the coefficients in the balanced equation to convert mole of one compound (A) into moles of another compound (B) Fe and O : 4 moles Fe and 3 moles O 3 moles O 4 moles Fe Fe and Fe O 3 : 4 moles Fe and moles Fe O 3 moles Fe O 3 4 moles Fe O and Fe O 3 : 3 moles O and moles Fe O 3 moles Fe O 3 3 moles O 3

4 8/9/015 Moles to Moles Calculation Calculate the number of moles of CO produced when 3.8 moles of CO reacts with an excess of oxygen. Moles to Moles Calculation Using the balanced chemical equation, how many moles of CO are produced from 3.5 moles of C H 6? C H 6 (g) + 5 O (g) 4 CO(g) + 6 H O(g) moles CO produced = 3.8 mol CO mol CO mol CO = 3.8 mol CO 3.5 mol C H 6 x 4 mol CO mol C H 6 = Unwanted unit cancels. 7.0 mol CO Mole to Mole Conversions From the Balanced Equation Al O 3 4 Al + 3 O Every time we use moles of Al O 3 we make 3 moles of O moles Al O 3 or 3 mole O 3 mole O moles Al O 3 How many moles of O are produced when 3.34 moles of Al O 3 decompose? (5.01) Mass in Chemical Reactions How much do you make? How much do you need? Stoichiometry is the calculation of quantities in chemical reactions based on a mole ratio from the balanced equation. Remember that the coefficients in a balanced equation tell us how many moles of each kind of substance. We cannot measure moles. What can we do? We can convert grams to moles using the. Then do the math with the molar ratio using the coefficients from the balanced equation. Then use the to convert the moles back to grams. Steps to Follow to Solve a Stoichiometry Problem 1. Make sure that the equation is balanced.. Identify known (given) and unknown (what you are trying to find). 3. Convert grams of known to moles by dividing by the molar mass of the known. 4. Convert moles of known to moles of unknown by multiplying by the molar ratio obtained from the coefficients in the balanced equation. 5. Convert moles of unknown to grams of unknown by multiplying by the of the unknown. Substance Mass/Moles/Particles Conversions Moles of Substance Avogadro s number Avogadro s number # of Atoms or Molecules grams of known moles of known moles of unknown grams of unknown 4

5 8/9/015 How to Convert Moles of Reactant to Product How to Convert Reactant to Product Using the balanced equation, how many grams of O 3 (ozone) are formed from 9.0 mol of O? Moles of reactant sunlight 3 O (g) O 3 (g) product Moles of reactant [1] [] Moles of product [3] 9.0 mol O x mol O 3 3 mol O x 48.0 g O 3 1 mol O 3 = 90 g O 3 reactant product Mol O cancel. Mol O 3 cancel. How to Convert Reactant to Product Ethanol (C H 6 O, 46.1 g/mol) is synthesized by reacting ethylene (C H 4, 8.1 g/mol) with water. How many grams of ethanol are formed from 14 g of ethylene? C H 4 + H O C H 6 O reactant 14 g C H 4 x 1 mol C H g C H x 1 mol C H 6 O 4 1 mol C H x 46.1 g C H 6 O 4 1 mol C H 6 O Grams C H 4 cancel. Moles C H 4 cancel. = 3 g C H 6 O product Moles C H 6 O cancel. Stoichiometry Practice Problems To make silicon for computer chips, the following reaction is used: SiCl 4 + Mg MgCl + Si a. How many grams of Mg are needed to make 9.3 g of Si? (Mg = 4.3 g/mol and Si = 8.1 g/mol) (? = 16 g) b. How many grams of SiCl 4 are needed to make 9.3 g of Si? (SiCl 4 = g/mol and Si = 8.1 g/mol) (? = 56 g) c. How many grams of MgCl are produced along with 9.3 g Si? (MgCl = 95.3 g/mol and Si = 8.1 g/mol) (? = 63 g) grams of known moles of known moles of unknown grams of unknown Stoichiometry Practice Problems The US Space Shuttle boosters use this reaction: 3 Al(s) + 3 NH 4 ClO 4 Al O 3 + AlCl NO + 6 H O How much Al must be used to react with 65 g of NH 4 ClO 4? (Al = 7.0 g/mol and NH 4 ClO 4 = g/mol) (? = 15 g) Stoichiometry Practice Problems Sometimes, we do not need to use all of the steps when solving a stoichiometry problem. Suppose we want to determine the mass in grams of NH 3 that can form from.50 moles of N? N (g) + 3 H (g) NH 3 (g) grams of known moles of known moles of unknown grams of unknown (Note: in this problem, we do not need to convert grams of given into moles, because the given is already in moles.) (NH 3 = 17.0 g/mol) (? = 85 g) 5

6 8/9/015 The Limiting Reactant is the Reactant that is Completely Used Up in the Reaction 1. Determine how much of one reactant is needed to react with a second reactant. H (g) + O (g) H O(l) chosen to be Original Quantity chosen to be Unknown Quantity There are 4 molecules of H in the picture.. Write out the s that relate the numbers of moles (or molecules) of reactants. H (g) + O (g) H O(l) 3. Calculate the number of moles (molecules) of the second reactant needed for complete reaction. H (g) + O (g) H O(l) molecules H 1 molecule O 1 molecule O molecules H 4 molecules H 1 molecule O x = molecules O molecules H Choose this to cancel molecules H 4. Analyze the two possible outcomes: If the amount present of the second reactant is less than what is needed, the second reactant is the limiting reactant. If the amount present of the second reactant is greater than what is needed, the second reactant is in excess. 6

7 8/9/015 Using the balanced equation, determine the limiting reactant when 10.0 g of N (8.0 g/mol) reacts with 10.0 g of O (3.00 g/mol). N (g) + O (g) NO(g) 1. Convert the number of grams of each reactant into moles using the es. N (g) + O (g) NO(g). Determine the limiting reactant by choosing N as the original quantity and converting to mole O. Conversion mol N 1 mol O x = mol O 1 mol N The amount of O we started with (0.313 mol) is less than the amount we would need (0.357 mol) so O is the limiting reagent. Consider the reaction between 5 moles of CO and 8 moles of H to produce methanol. CO(g) + H (g) CH 3 OH(l) How many moles of H are necessary in order for all the CO to react? mol H moles of H = 5 mol CO = 10 mol H 1 mol CO How many moles of CO are necessary in order for all of the H to react? 1 mol CO moles of CO = 8 mol H = 4 mol CO mol H 10 moles of H required; 8 moles of H available; limiting reactant. 4 moles of CO required; 5 moles of CO available; excess reactant. Worked Example 8.7 Alka-Seltzer tablets contain aspirin, sodium bicarbonate, and citric acid. When they come into contact with water, the sodium bicarbonate (NaHCO 3 ) and citric acid (H 3 C 6 H 5 O 7 ) react to form carbon dioxide gas, among other products. 3NaHCO 3 (aq) + H 3 C 6 H 5 O 7 (aq) 3CO (g) + 3H O(l) + Na 3 C 6 H 5 O 7 (aq) The formation of CO causes the trademark fizzing when the tablets are dropped into a glass of water. An Alka-Seltzer tablet contains g of sodium bicarbonate and g citric acid. Determine, for a single tablet dissolved in water, (a) which ingredient is the limiting reactant, (b) what mass of the excess reactant is left over when the reaction is complete, and (c) what mass of CO forms. Strategy Convert each of the reactant masses to moles. Use the balanced equation to write the stoichiometric and determine which reactant is limiting. Next, determine the number of moles of excess reactant remaining and the number of moles of CO produced. Finally, use the appropriate es to convert moles of excess reactant and moles of CO to grams. Worked Example 8.7 (cont.) Solution 1 mol NaHCO g NaHCO 3 3 = mol NaHCO g NaHCO g H 3 C 6 H 5 O 7 (a) To determine which reactant is limiting, calculate the amount of citric acid necessary to react completely with mol sodium bicarbonate mol NaHCO 3 1 mol H 3 C 6 H 5 O g H 3 C 6 H 5 O 7 = mol H 3 C 6 H 5 O 7 1 mol H 3 C 6 H 5 O 7 3 mol NaHCO 3 = mol H 3 C 6 H 5 O 7 The amount of H 3 C 6 H 5 O 7 required to react with mol of NaHCO 3 is more than a tablet contains. Therefore, citric acid is the limiting reactant and sodium bicarbonate is the excess reactant. Worked Example 8.7 (cont.) Solution (b) To determine the mass of excess reactant (NaHCO 3 ) left over, first calculate the amount of NaHCO 3 that will react: mol H 3 C 6 H 5 O 7 3 mol NaHCO 3 1 mol H 3 C 6 H 5 O 7 = mol NaHCO 3 Thus, mole of NaHCO 3 will be consumed, leaving mole unreacted. Convert the unreacted amount to grams as follows: mol NaHCO g NaHCO 3 1 mol NaHCO 3 = g NaHCO 3 7

8 8/9/015 Worked Example 8.7 (cont.) Solution (c) To determine the mass of CO produced, first calculate the moles of CO produced from the number of moles of limiting reactant (H 3 C 6 H 5 O 7 ) consumed. 3 mol CO mol H 3 C 6 H 5 O 7 Think About It In a problem 1 such mol as H 3this, C 6 H 5it O is 7 a = good idea mol to check CO your work by calculating the amounts of the other products in the reaction. Convert According this amount to the to law grams of conservation as follows: of mass, the combined starting mass of the two reactants ( g g + CO mol CO g =.700 g) should equal the sum of the masses of products and = g CO 1 mol leftover CO excess reactant. In this case, the masses of H O and Na 3 C 6 H 5 O 7 produced are g and g, To summarize respectively. the The results: mass (a) of citric CO acid is the limiting reactant, (b) g produced is g [from part (c)] and sodium the amount bicarbonate of excess remains NaHCO unreacted, and (c) g carbon dioxide is 3 is g [from part (b)]. The total, produced g g g g, is.700 g, identical to the total mass of reactants. Theoretical, Actual, and Percent Yield Theoretical Yield is the maximum amount of product, which is CALCULATED using the balanced equation. Because of side reactions and loss of product in handling, you almost never get out 100% of product. The amount isolated from a reaction is called the actual yield. æ actual yield (g) ö % yield = ç 100 è theoretical yield (g) ø Calculating Percent Yield Suppose you have prepared cookie dough to make 5 dozen cookies. The phone rings and you answer. While you talk, a sheet of 1 cookies burns, and you have to throw them out. The rest of the cookies you make are okay. What is the percent yield of edible cookies? Theoretical yield: 60 cookies possible Actual yield: 48 cookies to eat Simple Percent Yield Problem If the reaction of ethylene with water to form ethanol has a calculated theoretical yield of 3 g of ethanol, what is the percent yield if only 15 g of ethanol are actually formed? Percent yield = actual yield (g) theoretical yield (g) x 100% Percent yield: 48 cookies x 100% = 80.% yield 60 cookies = 15 g 3 g x 100% = 65% More Difficult Percent Yield Problem A sample of HgO weighing 7. g was heated. The collected mercury weighed 5.95 g. What was the percentage yield of the reaction? HgO(s) Hg(l) + O (g) More Difficult Percent Yield Problem A sample of HgO weighing 7. g was heated. The collected mercury weighed 5.95 g. What was the percentage yield of the reaction? known HgO(s) unknown Hg(l) + O (g) grams of known moles of known moles of unknown grams of unknown æ actual yield of Hg (g) ö % yield = ç 100 = 5.95 è theoretical yield of Hg (g) ø? First, calculate the theoretical yield by following the steps to solving a stoichiometry problem. Known: Unknown: 7. g HgO ( HgO = 16.6 g/mol grams Hg ( Hg = 00.6 g/mol) (? = 6.69 g Hg) 8

9 8/9/015 More Difficult Percent Yield Problem A sample of HgO weighing 7. g was heated. The collected mercury weighed 5.95 g. What was the percentage yield of the reaction? Worked Example 8.8 Aspirin, acetylsalicylic acid (C 9 H 8 O 4 ), is the most commonly used pain reliever in the world. It is produced by the reaction of salicylic acid (C 7 H 6 O 3 ) and acetic anhydride (C 4 H 6 O 3 ) according to the following equation: HgO(s) Hg(l) + O (g) æ actual yield of Hg (g) ö % yield = ç 100 = 5.95 è theoretical yield of Hg (g) ø? æ = 5.95 ö ç 100 = 88.9% è 6.69 ø C 7 H 6 O 3 salicylic acid + C 4 H 6 O 3 C 9 H 8 O 4 + acetic anhydride acetylsalicylic acid HC H 3 O acetic acid In a certain aspirin synthesis, g of salicylic acid and g of acetic anhydride are combined. Calculate the percent yield if g of aspirin are produced. Strategy Convert reactant grams to moles, and determine which is the limiting reactant. Use the balanced equation to determine the moles of aspirin that can be produced and convert to grams for the theoretical yield. Use this and the actual yield given to calculate the percent yield. Worked Example 8.8 (cont.) Solution 1 mol C g C 7 H 6 O 3 7 H 6 O g C 7 H 6 O = mol C 7 H 6 O g C 4 H 6 O 3 1 mol C 4 H 6 O g C 4 H 6 O 3 = mol C 4 H 6 O 3 Because the two reactants combine in a 1:1 mole ratio, the reactant present in the smallest number of moles (in this case, salicylic acid) is the limiting reactant. According to the balanced equation, one mole of aspirin is produced for every mole of salicylic acid consumed. Worked Example 8.8 (cont.) Solution Thus, the theoretical yield is g. If the actual yield is 105.6, the percent yield is % yield = g g 100% = 77.5% Think About It Make sure you have used the proper es and remember that percent yield can never exceed 100 percent. Therefore, the theoretical yield of aspirin is mol. We convert this to grams using the of aspirin: mol C 9 H 8 O g C 9 H 8 O 4 1 mol C 9 H 8 O 4 = g C 9 H 8 O 4 9