How To Calculate Mass In Chemical Reactions


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1 We have used the mole concept to calculate mass relationships in chemical formulas Molar mass of ethanol (C 2 H 5 OH)? Molar mass = 2 x x x = g/mol Mass percentage of carbon in ethanol? % C = 2 x x 100 % = % We can also use the mole concept to calculate mass relationships in chemical reactions is the study of mass relationships It requires a balanced equation The coefficients in a balanced chemical equation represents how many moles of one reactant are needed to react with other reactants. It also shows how many moles of product will be formed. Chemical equation relates moles of reactants to moles of products The equation DOES NOT directly relate the masses of reactants and products Consider the reaction 3 H 2 + N 2 2 NH 3 Balanced? Coefficients in a balanced equation number of moles 3 H 2 + N 2 2 NH 3 The coefficients in a balanced chemical equation can be used to relate the number of moles of each substance involved in a reaction. Molar ratios 3 molecules 1 molecules 2 molecules 300 molecules 100 molecules 200 molecules 3(6.02x10 23 ) molecules 6.02x10 23 molecules 2(6.02x10 23 ) molecules 3 moles 1 mole 2 moles mol reactant mol reactant mol product mol reactant mol product mol product 1
2 For the reaction: N H 2 2 NH 3 If you have 1.0 mole of H 2, how many moles of NH 3 can you produce? 3 different s can be written (why 3?) N H 2 2 NH 3 Note: Make sure your equation is balanced! 1 mol N 2 1 mol N 2 3 mol H 2 3 mol H 2 2 mol NH 3 2 mole NH 3 MOLAR RATIOS = CONVERSION FACTORS Given: 1.0 mol H 2 Find: mol NH 3 Conversion factor: # mol NH 3 = 1.0 mol H 2 x 2 mol NH 3 3 mole H 2 If you have 1.0 mole of H 2, how many moles of N 2 will be required to completely react all of the H 2? N H 2 2 NH 3 = 0.67 mol NH 3 Given: 1 mol H 2 Find: mol N 2 Conversion factor: 2
3 # moles N 2 = 1.0 mol H 2 x 1 mol N 2 3 mol H 2 How many moles of N 2 are needed to produce 0.50 moles of NH 3? N H 2 2 NH 3 = 0.33 mol N 2 Given: 0.5 mol NH3 Find: mol N 2 Conversion factor: From a balanced chemical equation we get the number of moles of reactants and products BUT # mol N 2 = 0.50 mol NH 3 x 1 mol N 2 2 mol NH 3 = 0.25 mol N 2 We don t measure out moles in the lab! Chemists use a balance to measure the mass of a substance used or produced in a reaction. How can you determine the mass of reactants or products? 3
4 Use the molar mass to convert from moles to grams The number of grams of a substance per mole Mass (g) Compound A X Mass (g) Compound B Molar mass Molar mass Moles Compound A Molar ratio Moles Compound B For the following reaction: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) How many grams of NH 3 would form if 2.11 moles of N 2 reacted with excess H 2? Given: 2.11 mol N 2 Find: g NH 3 Conversion factors:, molar mass g NH 3 = 2.11 mol N 2 x 2 mol NH 3 x g NH 3 1 mol N 2 1 mol NH 3 = 71.9 g NH 3 How many grams of N 2 are required to react completely with 9.47 grams of H 2? N 2 (g) + 3 H 2 (g) 2 NH 3 (g) CH O 2 CO H 2 O Answer the following questions: How many moles of O 2 are required to react with 1.72 moles of CH 4? Given: 9.47g H 2 Find: g N 2 Conversion factors: molar mass, How many grams of H 2 O will form when 1.09 moles of CH 4 react with excess O 2? g N 2 = 9.47 g H 2 x 1 mol H 2 x 1 mol N 2 x 28.01g N g H 2 3 mol H 2 1 mol N 2 How many grams of O 2 must react with excess CH 4 to produce 8.42 grams of CO 2? = 43.9 g N 2 4
5 Metallic iron reacts with oxygen to form iron(iii) oxide Balanced eqn: 4 Fe + 3 O 2 2 Fe 2 O 3 Calculate the grams of iron needed to produce 5.00 g of product. Given: 5.0 g Fe 2 O 3 Find: g Fe Conversion factors: molar masses Strategy: 4 Fe + 3 O 2 2 Fe 2 O 3 molar mass grams Fe 2 O 3 moles Fe 2 O 3 grams Fe moles Fe molar mass g Fe = 5.0 g Fe 2 O 3 x 1 mol Fe 2 O 3 x = 3.5 g Fe g Fe 2 O 3 x 4 mol Fe x g Fe 2 mol Fe 2 O 3 1 mol Fe Molar Mass of Fe 2 O 3 = 2 (55.85 g/mole) + 3 (16.0 g/mole) = g Fe 2 O 3 /mole 5
6 The steel industry relies on the reaction between iron(iii) oxide and carbon to produce iron and CO 2. 2Fe 2 O 3 3C 4Fe + 3CO 2 What mass of iron can be obtained from 454 g of iron(iii) oxide? Ethane (C 2 H 6 ) burns in oxygen to form CO 2 and water 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O What mass of ethane is required to produce 100. g of water? What mass of carbon is required to react with 454 g of iron(iii) oxide? What mass of CO 2 is formed along with the 100. g of water? Making Turkey Sandwiches Suppose you were going to make turkey sandwiches: slices bread + 1 slice cheese + 1 slice turkey 1 sandwich Making Turkey Sandwiches 2 slices bread + 1 slice cheese + 1 slice turkey 1 sandwich You have 8 slices of bread and 20 slices of turkey and 20 slices of cheese. How many sandwiches you can make using the above recipe? We can only make 4 sandwiches because we don t have enough bread! Bread = limiting reagent or limiting reactant You have 4 slices of cheese and 20 slices of turkey and 20 slices of bread. How many sandwiches you can make using the above recipe? Limiting reactant? 6
7 Similar situations occur in chemical reactions when one of the reactants is used up before the others. No further reaction can occur The excess reactant(s) are leftovers. 2 H 2 (g) + O 2 (g) 2 H 2 O (l) If we react 10 moles of H 2 with 7 moles of O 2, not all of the O 2 will react because we will run out of H 2 first! For 10 moles of H 2 we need only 5 moles of O 2! Limiting reactant: the reactant that is completely consumed in a reaction determines or limits the amount of product formed. To determine which reactant is the limiting reagent: Compare the number of moles of each reactant needed with the number of moles of each reactant available OR Calculate the number of grams of product that each reactant could form Reactant that forms the least amount of product will be the limiting reagent. 2 H 2 (g) + O 2 (g) 2 H 2 O (l) 7
8 If 10.0 grams of H 2 are mixed with 75.0 grams of O 2, which reactant is the limiting reagent? 2 H 2 (g) + O 2 (g) 2 H 2 O (l) Method 1 Step 1: Convert mass to moles Moles H 2 = 10.0 g H 2 x 1 mole = 4.95 mol H 2 available 2.02 g Moles O 2 = 75.0 g O 2 x 1 mole = 2.34 mol O 2 available 32.0 g Step 2 Step 3 Pick H 2 and find the moles O 2 needed to react with all of the H 2 moles O 2 needed = 4.95 mol H 2 x 1 mol O 2 2 mol H 2 = 2.48 moles O 2 Compare the # moles O 2 needed to # moles O 2 available. O 2 needed = 2.48 moles O 2 O 2 available = 2.34 moles O 2 Less O 2 is available than we need to react with all of the H 2. O 2 will run out first. Therefore, O 2 is the limiting reagent. 8
9 Method 2 Step 1: Calculate the amount of product each reactant could form. g H 2 O from H 2 = 10.0 g H 2 x 1 mol H 2 x 2 mol H 2 O 2.02 g H 2 2 mol H 2 x 18.0 g H 2 O = 89.1 g H 2 O 1 mol H 2 O g H 2 O from O 2 = 75.0 g O 2 x 1 mol O 2 x 2 mol H 2 O 32.0 g O 2 1 mol O 2 x 18.0 g H 2 O = 84.4 g H 2 O 1 mol H 2 O Limiting reagent is the one that produces the least product. O 2 is the 10.0 g H 2 could produce 89.1 g H 2 O limiting 75.0 g O 2 could produce 84.4 g H 2 O reagent The Limiting Reactant The Limiting Reactant Suppose we have a mixture of 20.0 g of CH 4 and grams of O 2. Which is the limiting reactant? CH O 2 CO H 2 O CH O 2 CO H 2 O Now suppose that the 20.0 grams of CH 4 reacts with the grams of O 2. What mass of the unused reactant is left over when all the CH 4 has been consumed? 9
10 The Limiting Reactant Reaction Yield H 2 + C CH 4 Suppose that 20.0 g of H 2 reacts with 40.0 g of carbon Which reactant is limiting? What mass of CH 4 could form when combining these reactants? How much of the unused reactant is left over? We have been calculating how much product would form for various chemical reactions The amount we calculate is usually higher than what we would obtain in lab This is not a violation of the law of conservation of matter Why are the two values different? Reaction Yield Reaction Yield Often, the actual yield is less than the theoretical yield: reactants may not react completely i.e. the reaction does not go to completion byproducts may form unwanted side reactions (competing reactions) difficulty isolating and purifying the desired product The mass of product obtained in the reaction is the actual yield The maximum amount possible, calculated from the limiting reagent principle, is the theoretical yield The percentage yield is defined as: Actual yield Theoretical yield x 100 = percentage yield 10
11 Reaction Yield Reaction Yield A chemist isolates g of product from a reaction that has a calculated theoretical yield of g. What is the percentage yield? Lime (CaO) is produced by heating CaCO 3 according to the following equation: CaCO 3 (s) CaO(s) + CO 2 (g) Suppose we heat 510 grams of CaCO 3 and isolate 235 grams of CaO. What is the percentage yield? 11
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