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A Study of Chemical Reactions Equations, Mole Conversions, & Stoichiometry Types of Reactions Many chemical reactions have defining characteristics which allow them to be classified as to type. Types of Chemical Reactions The five types of chemical reactions in this unit are: Combination/Synthesis Decomposition/Analysis Single Replacement/Displacement Double Replacement/Metathesis Combustion Combination Reactions Two or more substances combine to form one substance. The general form is A + X AX Example: Magnesium + oxygen magnesium oxide 2Mg + O2 2MgO Magnesium + Oxygen Combination Reactions Combination reactions may also be called composition or synthesis reactions. Some types of combination reactions: Combination of elements K + Cl2 One product will be formed Combination Reactions K + Cl2 Write the ions: K+ Cl- Balance the charges: KCl Balance the equation: 2K + Cl2 2KCl Combination Reactions Some types of combination reactions: Oxide + water Nonmetal oxide + water acid SO2 + H2O H2SO3 Metal oxide + water base BaO + H2O Ba(OH)2 Combination Reactions Some types of combination reactions: Metal oxides + nonmetal oxides Na2O + CO2 Na2CO3 CaO + SO2 CaSO3 Decomposition Reactions One substance reacts to form two or more substances. The general form is AX A + X Example: Water can be decomposed by electrolysis. 2H2O 2H2 + O2 Electrolysis of Water Decomposition Reactions Types of Decomposition Reactions: Decomposition of carbonates When heated, some carbonates break down to form an oxide and carbon dioxide. CaCO3 CaO + CO2 H2CO3 H2O + CO2 Decomposition Reactions Types of decomposition reactions: Some metal hydroxides decompose into oxides and water when heated. Ca(OH)2 CaO + H2O Note that this is the reverse of a similar combination reaction. Decomposition Reactions Types of decomposition reactions: Metal chlorates decompose into chlorides and oxygen when heated. 2KClO3 2KCl + 3O2 Zn(ClO3)2 ZnCl2 + 3O2 Some of these reactions are used in explosives. Decomposition Reactions Some substances can easily decompose: Ammonium hydroxide is actually ammonia gas dissolved in water. NH4OH NH3 + H2O Some acids decompose into water and an oxide. H2SO3 H2O + SO2 Decomposition Reactions Some decomposition reactions are difficult to predict. The decomposition of nitrogen triiodide, NI3, is an example of an interesting decomposition reaction. Nitrogen triiodide Single Replacement Reactions Cationic: A metal will replace a metal ion in a compound. The general form is A + BX AX + B Anionic: A nonmetal will replace a nonmetal ion in a compound. The general form is Y + BX BY + X Single Replacement Reactions Examples: Ni + AgNO3 Nickel replaces the metallic ion Ag+. The silver becomes free silver and the nickel becomes the nickel(II) ion. Ni + AgNO3 Ag + Ni(NO3)2 Balance the equation: Ni + 2AgNO3 2Ag + Ni(NO3) Activity Series Single Replacement Reactions Not all single replacement reactions that can be written actually happen. The metal must be more active than the metal ion. Aluminum is more active than iron in Al + Fe2O3 in the following reaction: Thermite Reaction Thermite Reaction Al + Fe2O3 Aluminum will replace iron(III) Iron(III) becomes Fe and aluminum metal becomes Al3+. 2Al + Fe2O3 2Fe + Al2O3 Single Replacement Reactions An active nonmetal can replace a less active nonmetal. The halogen (F2, Cl2, Br2, I2) reactions are good examples. F2 is the most active and I2 is the least. Cl2 +2 NaI 2 NaCl + I2 Double Replacement Reactions Ions of two compounds exchange places with each other. The general form is AX + BY AY + BX Metathesis is an alternate name for double replacement reactions. NaOH + CuSO4 Metathesis (sink or float?) NaOH + CuSO4 The Na+ and Cu2+ switch places. Na+ combines with SO42- to form Na2SO4. Cu2+ combines with OH- to form Cu(OH)2 NaOH + CuSO4 Na2SO4 + Cu(OH)2 2NaOH + CuSO4 Na2SO4 + Cu(OH)2 CuSO4 + Na2CO3 Double Replacement CuSO4 + Na2CO3 Cu2+ combines with CO32- to form CuCO3. Na+ combines with SO42- to form Na2SO4. CuSO4 + Na2CO3 CuCO3 + Na2SO4 Na2CO3 + HCl Double Replacement Na2CO3 + HCl Notice that gas bubbles were produced rather than a precipitate. What was the gas? Write the double replacement reaction first. Double Replacement Na2CO3 + HCl Na+ combines with Cl- to form NaCl. H+ combines with CO32- to form H2CO3. Na2CO3 + 2HCl 2NaCl + H2CO3 H2CO3 breaks up into H2O and CO2. Double Replacement The gas formed was carbon dioxide. The final balanced reaction is: Na2CO3 + HCl NaCl + H2O + CO2. Balance the equation. Na2CO3 + 2HCl 2NaCl + H2O + CO2 Combustion Reaction When a substance combines with oxygen, a combustion reaction results. The combustion reaction may also be an example of an earlier type such as 2Mg + O2 2MgO. The combustion reaction may be burning of a fuel. Combustion Reaction Methane, CH4, is natural gas. When hydrocarbon compounds are burned in oxygen, the products are water and carbon dioxide. CH4 + O2 CO2 + H2O CH4 + 2O2 CO2 + 2H2O Combustion Reactions Combustion reactions involve light and heat energy released. Natural gas, propane, gasoline, etc. are burned to produce heat energy. Most of these organic reactions produce water and carbon dioxide. Practice Classify each of the following as to type: H2 + Cl2 2HCl Combination Ca + 2H2O Ca(OH)2 + H2 Single replacement Practice 2CO + O2 2CO2 Combination and combustion 2KClO3 2KCl + 3O2 Decomposition Practice FeS + 2HCl FeCl2 + H2S Double replacement Zn + HCl ? Single replacement Zn + 2HCl ZnCl2 + H2 How molecules are symbolized Cl2 2Cl 2Cl2 • Molecules may also have brackets to indicate numbers of atoms. E.g. Ca(OH)2 • Notice that the OH is a group O Ca O H • The 2 refers to both H and O H • How many of each atom are in the following? a) NaOH Na = 1, O = 1, H = 1 b) Ca(OH)2 Ca = 1, O = 2, H = 2 c) 3Ca(OH)2 Ca = 3, O = 6, H = 6 Balancing equations: MgO The law of conservation of mass states that matter can neither be created or destroyed Thus, atoms are neither created or destroyed, only rearranged in a chemical reaction Thus, the number of a particular atom is the same on both sides of a chemical equation Example: Magnesium + Oxygen (from lab) Mg + O2 MgO Mg + O O Mg O • However, this is not balanced • Left: Mg = 1, O = 2 • Right: Mg = 1, O = 1 Balance equations by “inspection” Mg + O2 MgO 2Mg + O2 2MgO is correct Mg + ½O2 MgO is incorrect Mg2 + O2 2MgO is incorrect 4Mgwith + 2elements O2 4MgO is incorrect Hints: start that occur in one compound on each side. Treat polyatomic ions that repeat as if they were a single entity. a) P4 + 5 O2 P4O10 b) 2 Li + 2 H2O H2 + 2 LiOH c) 2 Bi(NO3)3 + 3 K2S Bi2S3 + 6 KNO3 d) C2H6 +3.5 O2 2 CO2 + 3 H2O 2 C2H6 + 7 O2 4 CO2 + 6 H2O From Balance these skeleton equations: a) b) c) d) e) f) g) h) i) Mg + 2HCl MgCl2 + H2 3Ca + N2 Ca3N2 NH4NO3 N2O + 2H2O 2BiCl3 + 3H2S Bi2S3 + 6HCl 2C4H10 + 13O2 8CO2 + 10H2O 6O2 + C6H12O6 6CO2 + 6H2O 3NO2 + H2O 2HNO3 + NO Cr2(SO4)3+ 6NaOH 2Cr(OH)3+ 3Na2SO4 Al4C3 + 12H2O 3CH4 + 4Al(OH)3 The Mole Q: how long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second? Background: atomic masses Look at the “atomic masses” on the periodic table. What do these represent? E.g. the atomic mass of C is 12 (atomic # is 6) We know there are 6 protons and 6 neutrons Protons and neutrons have roughly the same mass. So, C weighs 12 u (atomic mass units). What is the actual mass of a C atom? Answer: approx. 2 x 10-23 grams (protons and neutrons each weigh about 1.7 x10-24 grams) Two problems 1. Atomic masses do not convert easily to grams 2. They can’t be weighed (they are too small) The Mole With these problems, why use atomic mass at all? 1. Masses give information about # of p+, n0, e– 2. It is useful to know relative mass E.g. Q - What ratio is needed to make H2O? A - 2:1 by atoms, but 2:16 by mass It is useful to associate atomic mass with a mass in grams. It has been found that 1g H, 12g C, or 23g Na have 6.02x1023 atoms 6.02 x 1023 is a “mole” or “Avogadro’s number” “mol” is used in equations, “mole” is used in writing; one gram = 1 g, one mole = 1 mol. Mollionaire Q: how long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second? A: $ 6.02 x 1023 / $1 000 000 000 = 6.02 x 1014 payments = 6.02 x 1014 seconds 6.02 x 1014 seconds / 60 = 1.003 x 1013 minutes 1.003 x 1013 minutes / 60 = 1.672 x 1011 hours 1.672 x 1011 hours / 24 = 6.968 x 109 days 6.968 x 109 days / 365.25 = 1.908 x 107 years A: It would take 19 million years Comparing sugar (C12H22O11) & H2O Same 1 gram each 1 mol each No, they have dif. No, molecules volume? densities. have dif. sizes. Yes, that’s what No, molecules mass? grams are. have dif. masses No, they have dif. Yes. # of moles? molar masses No, they have dif. Yes (6.02x1023 # of in each) molecules? molar masses No, sugar has # of atoms? No more (45:3 ratio) Molar mass The mass of one mole is called “molar mass” E.g. 1 mol Li = 6.94 g Li This is expressed as 6.94 g/mol What are the following molar masses? S 32.06 g/mol SO2 64.06 g/mol Cu3(BO3)2 308.27 g/mol Calculate molar masses (to 2 decimal places) CaCl2 Cu x 3 = 63.55 x 3 = 190.65 (NH4)2CO3 B x 2 = 10.81 x 2 = 21.62 O2 O x 6 = 16.00 x 6 = 96.00 308.27 Pb3(PO4)2 C6H12O6 Molar mass The mass of one mole is called “molar mass” E.g. 1 mol Li = 6.94 g Li This is expressed as 6.94 g/mol What are the following molar masses? S 32.06 g/mol SO2 64.06 g/mol Cu3(BO3)2 308.27 g/mol Calculate molar masses (to 2 decimal places) CaCl2 110.98 g/mol (Cax1, Clx2) (NH4)2CO3 96.11 g/mol (Nx2, Hx8, Cx1, Ox3) O2 32.00 g/mol (Ox2) Pb3(PO4)2 811.54 g/mol (Pbx3, Px2, Ox8) C6H12O6 180.18 g/mol (Cx6, Hx12, Ox6) Converting between grams and moles If we are given the # of grams of a compound we can determine the # of moles, & vise-versa In order to convert from one to the other you must first calculate molar mass Formula HCl H2SO4 NaCl Cu g/mol 36.46 98.08 58.44 63.55 g mol (n) 9.1 0.25 53.15 0.5419 207 3.55 1.27 0.0200 The Mole Convert 36.0 grams of carbon into atoms. Convert 30 molecules of methane into Liters of gas. Simplest and molecular formulae Consider NaCl (ionic) vs. H2O2 (covalent) Na Cl Na Cl Cl Na Cl Na • Chemical formulas are either “simplest” (a.k.a. “empirical”) or “molecular”. Ionic compounds are always expressed as simplest formulas. • Covalent compounds can either be molecular formulas (I.e. H2O2) or simplest (e.g. HO) Q - Write simplest formulas for propene (C3H6), C2H2, glucose (C6H12O6), octane (C8H14) Q - Identify these as simplest formula, molecular formula, or both H2O, C4H10, CH, NaCl Answers Q - Write simplest formulas for propene (C3H6), C2H2, glucose (C6H12O6), octane (C8H14) Q - Identify these as simplest formula, molecular formula, or both H2O, C4H10, CH, NaCl A - CH2 CH CH2O C 4H 7 A - H2O is both simplest and molecular C4H10 is molecular (C2H5 would be simplest) CH is simplest (not molecular since CH can’t form a molecule - recall Lewis diagrams) NaCl is simplest (it’s ionic, thus it doesn’t form molecules; it has no molecular formula) Calculating percentage mass If you can work out Mr then this bit is easy… Percentage mass (%) = Mass of element Ar Relative formula mass Mr x100% Calculate the percentage mass of magnesium in magnesium oxide, MgO: Ar for magnesium = 24 Ar for oxygen = 16 Mr for magnesium oxide = 24 + 16 = 40 Therefore percentage mass = 24/40 x 100% = 60% Calculate the percentage mass of the following: 1) Hydrogen in hydrochloric acid, HCl 2) Potassium in potassium chloride, KCl 3) Calcium in calcium chloride, CaCl2 4) Oxygen in water, H2O Empirical formulae Empirical formulae is simply a way of showing how many atoms are in a molecule (like a chemical formula). For example, CaO, CaCO3, H20 and KMnO4 are all empirical formulae. Here’s how to work them out: A classic exam question: Find the simplest formula of 2.24g of iron reacting with 0.96g of oxygen. Step 1: Divide both masses by the relative atomic mass: For iron 2.24/56 = 0.04 For oxygen 0.96/16 = 0.06 Step 2: Write this as a ratio and simplify: 0.04:0.06 is equivalent to 2:3 Step 3: Write the formula: 2 iron atoms for 3 oxygen atoms means the formula is Fe2O3 25/05/2017 Example questions 1) Find the empirical formula of magnesium oxide which contains 48g of magnesium and 32g of oxygen. 2) Find the empirical formula of a compound that contains 42g of nitrogen and 9g of hydrogen. 3) Find the empirical formula of a compound containing 20g of calcium, 6g of carbon and 24g of oxygen. 25/05/2017 Stoichiometry Molar Mass of Compounds The molar mass (MM) of a compound is determined the same way, except now you add up all the atomic masses for the molecule (or compound) Ex. Molar mass of CaCl2 Avg. Atomic mass of Calcium = 40.08g Avg. Atomic mass of Chlorine = 35.45g Molar Mass of calcium chloride = 40.08 g/mol Ca + (2 X 35.45) g/mol Cl 110.98 g/mol CaCl2 20 Ca 40.08 17 Cl 35.45 Liters of Gas Mole Roadmap 22.4 Atoms or Molecules 6.02 X 1023 Moles molar mass from periodic table Mass (grams) Practice Calculate the Molar Mass of calcium phosphate Formula = Ca3(PO4)2 Masses elements: Molar Mass = Calculations molar mass Grams Avogadro’s number Moles particles Everything must go through Moles!!! Chocolate Chip Cookies!! 1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen How many eggs are needed to make 3 dozen cookies? How much butter is needed for the amount of chocolate chips used? How many eggs would we need to make 9 dozen cookies? How much brown sugar would I need if I had 1 ½ cups white Cookies and Chemistry…Huh!?!? Just like chocolate chip cookies have recipes, chemists have recipes as well Instead of calling them recipes, we call them reaction equations Furthermore, instead of using cups and teaspoons, we use moles Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients Chemistry Recipes Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe) Be sure you have a balanced reaction before you start! Example: 2 Na + Cl2 2 NaCl This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride What if we wanted 4 moles of NaCl? 10 moles? 50 moles? Practice Write the balanced reaction for hydrogen gas reacting with oxygen gas. 2 H2 + O 2 2 H2 O How many moles of reactants are needed? What if we wanted 4 moles of water? What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get? What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced? Mole Ratios These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)? 2 Na + Cl2 2 NaCl 5 moles Na 1 mol Cl2 2 mol Na = 2.5 moles Cl2 Mole-Mole Conversions How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal? Mole-Mass Conversions Most of the time in chemistry, the amounts are given in grams instead of moles We still go through moles and use the mole ratio, but now we also use molar mass to get to grams Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl2 2 NaCl 5.00 moles Na 1 mol Cl2 70.90g Cl2 2 mol Na 1 mol Cl2 = 177g Cl2 Practice Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum. Mass-Mole We can also start with mass and convert to moles of product or another reactant We use molar mass and the mole ratio to get to moles of the compound of interest Calculate the number of moles of ethane (C2H6) needed to produce 10.0 g of water 2 C2H6 + 7 O2 4 CO2 + 6 H20 10.0 g H2O 1 mol H2O 2 mol C2H6 = 0.185 18.0 g H2O 6 mol H20 mol C2H6 Practice Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide Mass-Mass Conversions Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!) Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in Mass-Mass Conversion Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. N2 + 3 H2 2 NH3 2.00g N2 1 mol N2 2 mol NH3 17.06g NH3 28.02g N2 1 mol N2 = 2.4 g NH3 1 mol NH3 Practice How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen? Limiting Reactant: Cookies 1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen If we had the specified amount of all ingredients listed, could we make 4 dozen cookies? What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies? What if we only had one egg, could we make 3 dozen cookies? Limiting Reactant Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant. That reactant is said to be in excess (there is too much). The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant. Limiting Reactant To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one. The lower amount of a product is the correct answer. The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it! Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same! Limiting Reactant Limiting 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl2 2 AlCl3 Start with Al: 10.0 g Al Reactant: Example 1 mol Al 2 mol AlCl3 27.0 g Al 2 mol Al 133.5 g AlCl3 1 mol AlCl3 = 49.4g AlCl3 Now Cl2: 35.0g Cl2 1 mol Cl2 71.0 g Cl2 2 mol AlCl3 133.5 g AlCl3 3 mol Cl2 1 mol AlCl3 = 43.9g AlCl3 LR Example Continued We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete . Limiting Reactant Practice 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made. Finding the Amount of Excess By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess. Can we find the amount of excess potassium in the previous problem? Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I2 2 KI We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced. 15.0 g I2 K 1 mol I2 2 mol K 39.1 g = 4.62 g K USED! 254 g I2 1 mol I2 1 mol K 15.0 g K – 4.62 g K = 10.38 g K EXCESS Given amount of excess reactant Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should only start with it! Limiting Reactant: Recap 1. 2. 3. 4. 5. 6. 7. You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. Convert ALL of the reactants to the SAME product (pick any product you choose.) The lowest answer is the correct answer. The reactant that gave you the lowest answer is the LIMITING REACTANT. The other reactant(s) are in EXCESS. To find the amount of excess, subtract the amount used from the given amount. If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!