VI. Organic Compounds = carbon containing compounds cinnamaldehyde

VI. Organic Compounds = carbon containing compounds cinnamaldehyde A. Hydrocarbons = only carbon and hydrogen B. Functionalized Hydrocarbons = contain other elements + C + H I. Chemical Stoichiometry [Chapter 4.1-4.3] A. Balanced equations provide the relationship between the relative numbers of reacting molecules and product molecules 2 CO + O2 2 CO2 1. 2 CO molecules react with 1 O2 molecules to produce 2 CO2 molecules 2. 2 moles CO molecules react with 1 mole O2 molecules to produce 2 moles CO2 molecules 3. Since moles can be converted to masses, we can determine the mass ratio of the reactants and products as well a. 2 moles CO = 1mole O2 = 2 moles CO2 b. 1 mole CO = 28.01 g, 1 mole O2 = 32.00 g, and 1 mole CO2 = 44.01 g c. 2(28.01) g CO = 1(32.00) g O2 = 2(44.01) g CO2 B. Procedure for calculating Masses of Reactants and Products in Reactions 1. Write a balanced equation for the reaction 2. Convert the known mass of a given reactant or product to moles

3. Use the balanced equation to find the moles of the unknown compound 4. Convert the moles of unknown to grams of unknown C. Example: Determine the number of grams of Carbon Monoxide required to react with 48.0 g Oxygen, and determine the mass of Carbon Dioxide produced 1. Write the balanced equation 2 CO + O2 2 CO2 2. Use the coefficients to find the mole relationship 2 moles CO = 1 mol O2 = 2 moles CO2 3. Determine the Molar Mass of each 1 mol CO = 28.01 g 1 mol O2 = 32.00 g 1 mol CO2 = 44.01 g 4. Use the molar mass of the given quantity to convert it to moles 1 mol O2 48.0 g O 2 x 1.5 mol O 2 32.00 g 5. Use the mole relationship to convert the moles of the given quantity to the moles of the desired quantity 1.5 mol O2 x 2 mol CO

3 mol CO 1 mol O 2 1.5 mol O2 2 mol CO2 x 3 mol CO 2 1 mol O 2 6. Use the molar mass of the desired quantity to convert the moles to mass 44.01 g 28.01 g 3 mol CO 2 x 132 g CO2 3 mol CO x 84.0 g CO 1 mol CO2 1 mol CO D. Example: What mass of O2 will react with 96.1g of propane (C3H8)? C3H8 + 5O2 -------> 3CO2 + 4H2O 1 mol C 3 H 8

2.18 mol C 3 H 8 96.1 g C 3 H 8 44 . 1 g C H 3 8 5 mol O 2 10.9 mol O 2 2.18 mol C 3 H 8 1 mol C H 3 8 32.00 g O 2 349 g O 2

10.9 mol O 2 1 mol O 2 E. Example: Lithium hydroxide absorbs carbon dioxide to form lithium carbonate and water. How much CO2 can be absorbed by 1.00 kg LiOH? II. Limiting and Excess Reactants A. Definitions 1. A reactant which is completely consumed is called a limiting reactant 2. A reactant not completely consumed in a reaction is called an excess reactant 3. The maximum amount of a product that can be made when the limiting reactant is completely consumed is called the theoretical yield B. Example: CH4 + H2O -------> 3H2 + CO 1.

The stoichiometry is supposed to be 1:1 for methane and water 2. Water is the limiting reactant 3. Methane is the excess reactant 4. The theoretical yield is 9H2 5. The theoretical yield is 3CO C. Determine the Mass of Carbon Dioxide produced when 48.0 g of Oxygen reacts with 56.0 g of Carbon Monoxide 1. Write the balanced equation 2 CO + O2 2 CO2 2. Use the coefficients to find the mole relationship 2 moles CO = 1 mol O2 = 2 moles CO2 3. Determine the Molar Mass of each 1 mol CO = 28.01 g 1 mol O2 = 32.00 g

1 mol CO2 = 44.01 g 4. Determine the moles of each reactant 1 mol O2 48.0 g O2 x 1.50 moles O2 32.00 g 1 mol CO 56.0 g CO x 2.00 moles CO 28.01 g 5. Determine the number of moles of reactant A needed to react with reactant B 1 moles O2 2.00 moles CO x 1.00 moles O2 2 mole CO 6. Compare the calculated number of moles of reactant A to the number of moles given of reactant A a. If the calculated moles is greater, then A is the Limiting Reactant; if the calculated moles is less, then A is the Excess Reactant b. The calculated moles of O2 (1.00 moles) is less than the given 1.50 moles, therefore O2 is the excess reactant

7. Use the limiting reactant to determine the moles of product, then the mass of product 2 moles CO2 44.01 g CO2 2.00 moles CO x x 88.0 g CO2 2 mole CO 1 mol CO2 D. 25.0 kg N2 and 5.00 kg H2 are reacted to make how much NH 3? 1. N2(g) + 3H2(g) -------> 2NH3(g) 25,000 g N2 x 1 mol N2 893 moles N2 28.0 g 5,000 g H2 x 1 mol H2 2,480 moles H2 2.016 g 1 moles N2 2,480 moles H 2 x 827 moles needed of N2

3 mole H 2 2. 827 mol N2 are needed and we are given 893 mol N2 a. H2 is the limiting reagent b. N2 is the excess reagent 3. Calculate the amount of product formed from the limiting reagent 2,480 moles H 2 x 2 moles NH 3 3 mole H 2 x 17.0 g NH 3 1 mol NH 3 28,000 g NH 3 28.0 kg NH 3 III. Percent Yield A. Most reactions do not go to completion 1. The amount of product made in an experiment is called the actual yield 2. The percentage of the theoretical yield that is actually made is called the percent yield Actual Yield

Percent Yield = x 100% Theoretical Yield B. Example: CO and H2 combine to form CH3OH. What is the theoretical yield of CH3OH if 68.5 kg CO and 8.60 kg H2 were reacted? What is the percent yield if 35.7 kg was the actual yield of CH3OH? 1. 68,500 g CO x CO + 2H2 -------> CH3OH 1 mol CO 2,440 moles CO 28.02 g 8,600 g H2 x 1 mol H2 4,270 moles H2 2.016 g 1 moles CO 4,270 moles H 2 x 2,135 moles needed of CO

2 mole H 2 2. 2,135 mol CO are needed and we are given 2,440 mol CO a. H2 is the limiting reagent b. CO is the excess reagent 3. Calculate the amount of product formed from the limiting reagent 4,270 moles H 2 x 1 moles CH 3 OH 2 mole H 2 x 32.04 g CH 3 OH 1 mol CH 3 OH 68,600 g CH 3 OH 68.6 kg CH 3 OH 4. The Theoretical Yield of CH3OH is 68.6 kg 5. The Percent Yield of CH3OH = ? 35.7 kg Percent Yield x 100% 52.0% 68.6 kg