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To understand how to use stoichiometry in scientific terms, you will have to understand how to use it in… EVERYDAY EQUATIONS!! • For example, when baking food, a recipe is given for you to follow. If you need to make a larger number of the item, you could double or even triple the quantity of ingredients you use. • “A balanced chemical equation provides the same kind of quantitative information that a recipe does.” • In a recipe you can imagine the ingredients as reactants and the cookies as the products. Stoichiometry • Balanced chemical equations tell you what amounts of reactants to mix and what amount of products to expect from the reaction – Chemist use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction – The calculation of quantities in chemical reactions is the subject of chemistry called stoichiometry • Stoichiometric calculations allow chemist to keep track of reactants and products in a chemical reaction Interpreting Chemical Equations • Consider the following equation N2(g) + 3H2(g) 2NH3(g) • The balanced chemical equation can be interpreted in terms of different quantities such as: numbers of atoms, molecules, moles, mass, and volume – Examples: • Atoms – In formation of ammonia two atoms of nitrogen react with 6 atoms of hydrogen. These eight atoms are recombined in the product • Molecules – One molecule of nitrogen reacts with 3 molecules of hydrogen to form two molecules of ammonia • Moles – One mole of nitrogen gas reacts with 3 moles of hydrogen gas to form 2 moles of ammonia gas • Mass – The mass of1mole of N2 (28 grams) plus the mass of 3 moles of H2 (6.0 grams) equals the mass of 2 moles of NH3 (34 grams) • Volume –22.4L of N2 gas reacts with 67.2L H2 (3 x 22.4L)gas to form 44.8L of NH3 (2 x 22.4L) gas (Remember that the volume of one mole of gas = 22.4L) Mass Conservation in Chemical Reactions • Mass and atoms are conserved in every chemical reaction Writing and using mole ratios N2(g) + 3H2(g) 2NH3(g) • A mole ratio is a conversion factor derived from the coefficients of a balanced chemical equation interpreted in terms of moles • Example: 1 mol N2 3 mol H2 2 mol NH3 1 mol N2 3 mol H2 2 mol NH3 • In chemical calculations, mole ratios are used to convert between moles of reactant and moles of product, between moles of reactants, or between moles of products Mole-Mole Calculations • How many moles of ammonia are produced when 0.6 moles of nitrogen reacts with hydrogen? 1 mol N2 N2(g) + 3H2(g) 2NH3(g) 3 mol H 2 x mole G Given 0.6 moles N2 b mole W =xb/a mol W a mole G Mole Ratio 2 mol NH3 1 mol N2 Calculated 2 mol NH3 = 1.2 moles NH3 1 mol N2 Entered into calculator 0.6 X 2 = 1.2 3 mol H2 2 mol NH3 Mole-Mole Calculations • How many moles of excess hydrogen are consumed when 0.6 moles of nitrogen reacts with hydrogen to form ammonia? 1 mol N2 3 mol H2 N2(g) + 3H2(g) 2NH3(g) x mole GX Given 0.6 moles N2 X b mole W =xb/a mol W a mole G Mole Ratio 3 mol H2 1 mol N2 Calculated = 1.8 moles H2 Entered into calculator 0.6 X 3 = 1.8 Flip mole ratio so N2 cancel each other 2 mol NH3 1 mol N2 3 mol H2 2 mol NH3 Chemical Calculations: • In chemical calculations, mole ratios are used to convert between moles of reactants and moles of product, between moles of reactants, or between moles of products. • A mole ratio is a conversion factor derived from the coefficients of a balanced chemical equation interpreted in terms of moles. • To change one units to another, you will need a map to ensure that the units as well as their labels are correct. Map on next page Stoichiometry Calculator: Coefficient Volume A 22.4 L mol Mass A mw mol Mole A Volume B 22.4 L mol Coefficient 6.02×1023 molec mol Molecule A Mole B mw mol 6.02×1023 molec mol Coefficient Molecule B Mass B How To Use The Map: • In order to use this map to convert units of a substance, you need to start by drawing a t-like outline and put the given information in the first empty space and nothing in the space beneath it. For example, if you have 20 mol H, and you want to know howAmany L of O there are in the equation 2H + O 20 mol H2O , this is how you would start : • Then, you need to follow the map and convert 20 mol A to liters by adding 22.4 L/mol into your outline like so: 20 mol A 22.4 L A mol A • Last, you will need to convert your liters of H to liters of O by following the map once again and adding the coefficient to your outline. (make sure that units match up with the information diagonal to it.) (coefficient is the number in front Continued on next page of the element. It will look like this now: 20 mol A 22.4 L A 1 L B mol A 2 L A • Now that you are done creating your outline, you will need to multiply the top and bottom numbers by each other and divide them. This is what you will end up with: 448 224 answer!! 2 • When you are done, make sure to add your units of your substa 224 Liters of O final answer Another example using “the map”: • How many molecules of oxygen are produced when 29.2 g of water is decomposed by electrolysis according to this electricity balanced 2H equation: 2O (L) 2H2 (g) + O 2 (g) 1. List the known: mass of water= 29.2 g 2. Calculate: 1 mol H2O 29.2 g H2O 18.0 g H2O 23 1 mol O2 6.02 10 molec. 1 mol O2 2 mol H2O 4.88 1023 molecules O2 3. Check answer: Yes, the answer is correct Mass-Mass Calculations • Calculate the number of grams of NH3 produced by the reaction of 5.40 grams of hydrogen with excess nitrogen N2(g) + 3H2(g) 2NH3(g) 1. Convert 5.40 grams H2 into moles 2. Change the moles of hydrogen to moles of ammonia 3. Change the moles of ammonia to grams of ammonia 5.40 g H2 X 1 mol H2 2.0 g H2 Step 1 X 2 mol NH3 17.0 g NH3 X = 31g NH3 3 mol H2 1 mol NH3 Step 2 Step 3 Calculator: 5.4 ÷ 2 X 2 ÷ 3 X 17 = 31 g 1 mol N2 3 mol H2 2 mol NH3 1 mol N2 3 mol H2 2 mol NH3 Mole-Mole Calculations Practice Problems • How many moles of ammonia are produced when 3 moles of nitrogen reacts with hydrogen? N2(g) + 3H2(g) 2NH3(g) • How many grams of ammonia are produced when 84 grams of nitrogen reacts with hydrogen? Other Stoichiometric Calculations • In a typical stoichiometric problem, the given quantity is first converted to moles. Then the mole ratio from the balanced equation is used to calculate the number of moles of the wanted substance. Finally, the moles are converted to any other unit of measurement related to the unit mole, as the problem requires. – How many molecules of NH3 are produced when 5.4 grams of hydrogen react with excess nitrogen? N2(g) + 3H2(g) 2NH3(g) 1. Convert 5.40 grams H2 into moles 2. Change the moles of hydrogen to moles of ammonia 3. Change the moles of ammonia to molecules of ammonia 5.40 g H2 X 1 mol H2 2.0 g H2 Step 1 X 2 mol NH3 X 3 mol H2 Step 2 6.02x1023 molecules NH3 1 mol NH3 = 1.08x1024 NH3molecules Step 3 Calculator: 5.4 ÷ 2 X 2 ÷ 3 X 6.02x1023 = 1.08x1024 Limiting and Excess Reagents: • In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms. • An example of excess and limiting reagents is making muffins. You have more than enough flour, butter, water, and eggs. However, you only have one cup of sugar. The number of muffins you can make will be limited by how much sugar you have on hand. Therefore, the sugar is the limiting ingredient in this recipe and the other ingredients are the excess ingredients. A chemist often has the same situation, with balanced • Abut chemist’s recipe ischemical a balanced equations. equation such as: N2 (g) + 3H2 (g) 2NH3 (g) Limiting Reagent: • To show how to find the limiting and excess reagent in an actual equation, we will use the formula for ammonia: N2 (g) + 3H2 (g) 2NH3 (g) • Before this reaction takes place, nitrogen and hydrogen are in a 2:3 ratio. One molecule (mole) of N2 reacts with three molecules (moles) of H2 to make two molecules of NH3. When this happens, all of the H has been used up, and the reaction stops. One molecule of leftover nitrogen is • Therefore, in this reaction, only the hydrogen molecule is left. used up, and consequently, it is the limiting reagent, or the reagent that determines the amount of product that can be made in the reaction. More on next page Limiting and Excess Reagents • In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product formed. – Limiting reagent is the reactant that determines the amount of product that can be formed by a reaction – Excess reagent is the reactant that is not completely used up in a chemical reaction 2Na + Cl2 2NaCl Cl Cl Cl Cl Na Cl Cl Na Cl Cl Na Cl Cl Na Cl Cl Na Na Cl Cl Na Cl Cl Na Na Na Identify the limiting reagent and the excess reagent from the above reaction about to occur Before reaction Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl + 2Na + Cl2 2NaCl Na Na Na Na Na Na Na Na Na Na After reaction In this reaction, sodium (Na) is the limiting reagent because it is entirely used up in the reaction. Chlorine Gas (Cl2) is the excess reagent because some chlorine gas is left over after the reaction Limiting Reagent Problem • Copper reacts with sulfur to form copper(I) sulfide according to the following balanced equation. 2Cu(s) + S(s) Cu2S(s) • What is the limiting reagent when 80.0 grams of copper react with 25.0 grams of sulfur? 1. Convert mass to moles: 1 mole Cu = 1.26 mole Cu 63.5g Cu 1 mole S = 0.779 mole S 25.0g S X 32.1g S 80.0g Cu X 2. Determine how many moles of sulfur you need to react with 1.26 moles of Cu 1.26 mole Cu X 1 mole S = 0.630 mole S 2 mole Cu 3. Because 0.630 moles of S is less than the 0.779 moles of sulfur present in the reaction, sulfur is in excess and Cu is limiting Limiting Reagent Problem • Copper reacts with sulfur to form copper(I) sulfide according to the following balanced equation. 2Cu(s) + S(s) Cu2S(s) • What is the maximum number of grams of Cu2S that can be formed when 80.0 grams of Cu react with 25.0 grams of S? 1. 2. Convert mass to moles: 1 mole Cu 80.0g Cu X 63.5g Cu 1 mole S 25.0g S X 32.1g S = 0.779 mole S X 1 mole S 2 mole Cu = 0.630 mole S Determine the limiting and excess reagents • 4. = 1.26 mole Cu Determine how many moles of sulfur you need to react with 1.26 moles of Cu 1.26 mole Cu 3. Italicized numbers are molar masses Because 0.630 moles of S is less than the 0.779 moles of sulfur present in the reaction, sulfur is in excess and Cu is limiting Calculate the maximum amount of product formed using the limiting reagent Cu 1.26 mol Cu X 1 mole Cu2S 2 mole Cu X 159.1 g Cu2S 1 mole Cu2S = 100 grams Cu2S Percent Yield: • In your classroom, not everyone gets 100% on everything. The grade you receive on a quiz or test is a ratio of the number of questions answered correctly and the number of total questions on the exam. Chemists use similar calculations when the product from a chemical reaction is •less Thethan theoretical expected yield based is one on tool the equation. used in finding the percent yield of the equation. The theoretical yield is the maximum amount of product that could be formed from given amounts of reactants. •Another tool is the amount of product that actually forms in your lab when the reaction is carried out that is called the actual yield. • The percent yield is the ratio of the actual yield to the theoretical yield shown as a percent. Continued on next page Percent Yield Continued: • The formula to find percent yield is: Percent yield= actual yield theoretical yield 100 • The actual yield of the chemical reaction is mostly less than the yield and therefore the yield is less than 100% for the most part. • The percent yield is a measure of the efficiency of a reaction carried out in the laboratory. (basically the measure of how messy the experiment was in the lab) • In order to perform this procedure, remember that you MUST have both the theoretical and actual yields. Percent Yield Example: • What is the percent yield if 13.1 g CaO is actually produced when 24.8g CaCO3 is heated?: CaCO3 (S) CaCO (S) + CO2 (g) 1. State your knowns: actual yield= 13.1 g CaO theoretical yield= 13.9 g CaO 2. Calculate: percent yield =13.1 g CaO 13.9 g CaO = 3. Check your answer 94.2% 100 Theoretical yield example: • Calcium carbonate, which is found in seashells, is decomposed by heating. The balanced equation for this reactionCaCO is: 3 (S) CaO (S) + CO2 (g) What is the theoretical yield? 1. State your knowns: Mass of CaCO3= 24.8 g 2. Calculate: 1 mol CaCO3 1 mol CaO 56.1 g CaO 24.8 g CaCO3100.1 g CaCO31 mol CaCO3 1 mol CaO = 3. Check your answer 13.9g CaO Percent Yield • No process is one hundred percent efficient • Because no process is one hundred percent efficient, calculated yields from a chemical reaction are always greater than the actual yield • The actual yield of a chemical reaction divided by the calculated yield (theoretical yield) multiplied by 100% is the percent yield of a chemical reaction % Yield = Actual Yield ÷ Theoretical Yield X 100% • The percent yield is a measure of the efficiency of a reaction carried out in the laboratory. Percent Yield Problem • Copper reacts with sulfur to form copper(I) sulfide according to the following balanced equation. 2Cu(s) + S(s) Cu2S(s) • If 95 grams of Cu2S forms from the reaction of 80.0 grams of Cu with 25.0 grams of S, what was the percent yield of the reaction? 1. Convert mass to moles: 2. Determine how many moles of sulfur you need to react with 1.26 moles of Cu 3. Determine the limiting and excess reagents 4. Calculate the maximum amount of product formed (Theoretical yield) using the limiting reagent (Cu) 1.26 mol Cu X 5. 1 mole Cu2S 2 mole Cu Italicized 159.1 g Cu2S numbers are X = 100 grams Cu2S molar masses 1 mole Cu2S Divide the actual yield by the theoretical yield and multiply by 100% 95 grams Cu2S 100 grams Cu2S X 100% = 95%