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Unit 2 Moles, Chemical Reactions, and Stoichiometry The Mole • A mole (mol) is the amount of a substance that contains as many particles as there are atoms in exactly 12 g of carbon-12. • It is a counting unit, similar to a dozen. In a dozen, there are 12 things. In a mole, there are 6.02 x 1023 things. Visual Concept Molar Mass • Molar Mass is the mass of one mole of a pure substance. • Molar mass units are g/mol. • The molar mass of an element is the same number as its atomic mass, only the units are different. • Examples: Molar mass of H = 1.0 g/mol Molar mass of O = 16.0 g/mol Formula Mass • The formula mass of any compound is the sum of the masses of all the atoms in its formula. Example: Formula mass of water, H2O: H2 = 1.0 amu x 2 = 2.0 amu. O= + 16.0 amu. 18.0 amu • A compound’s molar mass is numerically equal to its formula mass. Only the units are different. (Ex: Molar mass of H2O = 18.0 g.) Molar Masses Sample Problem Determine the molar mass of each of the following compounds: Solution: Al2 = 27.0 x 2 = 54.0 g a. Al2S3 S3 = 32.1 x 3 = + 96.3 g 150.3 g = b. Ba(OH)2 Ba O2 = 16.0 x 2 = H2 = 1.0 x 2 = 137.3 g 32.0 g + 2.0 g 171.3 g Percent Composition • The percentage by mass of each element in a compound is known as the percent composition of the compound. % of element = mass of element in compound x 100 molar mass of compound Percent Composition Sample Problem Find the percentage composition of copper(I) sulfide, Cu2S. Cu2 = 63.5 x 2 = 127.0 g Solution: S = + 32.1 g 1. Find the molar mass of Cu2S: 159.1 g 2. Find the percentage by mass of each element: 127.0 g % Cu = x 100 = 79.80% Cu 159.1 g 32.1 g %S= x 100 = 20.2% S 159.1 g Molar Mass as a Conversion Factor • The molar mass of a compound can be used as a conversion factor to convert between moles and grams for a given substance. Example: • What is the mass of 2.5 moles of H2O? molar mass of H2O = 18.0 g/mol given conversion factor 2.5 mol H2O x 18.0 g H2O = 45 g H2O 1 mol H2O Molar Mass as a Conversion Factor Sample Problem Calculate the moles in 1170 g of copper (II) nitrate. Solution: 2+ 1. Determine the correct formula: Cu (NO3) 2 Cu(NO3)2 = 63.5 g 2. Calculate the molar mass: Cu N2 = 14.0 x 2 = 28.0 g 3. Convert from g to mol: O6 = 16.0 x 6 = + 96.0 g Given 1170 g Cu(NO3)2 x Conversion factor 187.5 g 1 mol Cu(NO3)2 = 6.24 mol Cu(NO3)2 187.5 g Cu(NO3)2 Avogadro’s Number • Avogadro’s number: 6.022 1415 × 1023— the number of particles in exactly one mole of a pure substance. • Named for nineteenth-century Italian scientist Amedeo Avogadro. Visual Concept Conversions with Avogadro’s Number • Avogadro’s number can be used as a conversion factor between number of particles (atoms or molecules) and moles. Example: •How many moles of silver, Ag, are in 3.01 1023 atoms of silver? Given 3.01 x 1023 atoms Ag x Conversion factor 1 mol Ag = 0.500 mol Ag 6.02 x 1023 atoms Ag Combining Conversion Factors Sample Problem What is the mass in grams of 1.20 108 atoms of copper, Cu? nd 2 Solution: st Given 1.20 x 108 atoms Cu x 1 Conversion factor Conversion factor 1 mol Cu x 63.5 g Cu 6.02 x 1023atoms Cu 1 mol Cu = 1.27 x 10-14 g Cu Standard Molar Volume • Standard Temperature and Pressure (STP) is 0oC and 1 atm. • The Standard Molar Volume of a gas is the volume occupied by one mole of a gas at STP. It has been found to be 22.4 L. Molar Volume Conversion Factor • Standard Molar Volume can be used as a conversion factor to convert from the number of moles of a gas at STP to volume (L), or vice versa. Molar Volume Conversion Sample Problem a. What quantity of gas, in moles, is contained in 5.00 L at STP? 5.00 L x 1 mol 22.4 L = 0.223 mol b. What volume does 0.768 moles of a gas occupy at STP? 0.768 mol x 22.4 L 1 mol = 17.2 L The Mole Map • You can now convert between number of particles, mass (g), and volume (L) by going through moles. Chemical Reactions • A chemical reaction is the process by which one or more substances are changed into one or more different substances. • Signs that a chemical reaction is taking place: 1. Release of energy as heat and/or light. 2. Production of a gas. 3. Formation of a precipitate - a solid that separates from a liquid solution. 4. Color change. The Law of Conservation of Mass • In any chemical reaction, the original substances are called reactants and the resulting substances are called products. • According to the law of conservation of mass, the total mass of reactants must equal the total mass of products for any given chemical reaction. Chemical Equations • A chemical equation represents a chemical reaction using symbols and formulas. Example: 2H2O(l) 2H2(g) + O2(g) Reactant Products Diatomic Molecules • Oxygen gas (O2) is an example of an element that normally exists as a diatomic molecule. You need to memorize all seven: Balancing Equations • The final step in writing correct chemical equations is to make sure the law of conservation of mass is satisfied. – The numbers and types of atoms on both sides of the equation must be the same – this is called balancing an equation. – Equations are balanced by inserting coefficients - whole numbers that appear in front of formulas in a chemical equation. Balancing Equations Sample Problem A Balance the following equation: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) Solution: • Start with the easiest element…carbon. – Carbon is already balanced. • Next count the hydrogen atoms. – Two more hydrogen atoms are needed on the right. • Finally, count oxygen atoms. – There are 4 oxygens on the right side of the equation, but only two on the left. – Add a coefficient 2 in front of the O2 on the left. Balancing Equations Sample Problem B Balance the following equation: Al4C3(s) + 12 H2O(l) 3 CH4(g) + 4 Al(OH)3(s) Solution: • Let’s start with aluminum. – Add a coefficient 4 to Al(OH)3 on the right. • Next count the carbon atoms. – Add a coefficient 3 to CH4 on the right. • Balance the oxygen atoms. – Add a 12 to the H2O on the left. • Lastly, count the hydrogen atoms. – Hydrogen is already balanced. Writing Chemical Equations Sample Problem Write an equation for the reaction that occurs when solid copper metal reacts with aqueous silver nitrate to produce solid silver metal and aqueous copper(II) nitrate. Solution: • First, use correct formulas and symbols to write a chemical equation. • Then, balance your equation. Cu(s) + 2 AgNO3(aq) 2 Ag(s) + Cu(NO3)2(aq) Types of Chemical Reactions • There are 5 basic types of chemical reactions: 1. 2. 3. 4. 5. Synthesis Decomposition Single-Displacement Double-Displacement Combustion Synthesis Reactions • In a synthesis reaction (also called a composition reaction) two or more substances combine to form a new compound. • This type of reaction is represented by the following general equation: A+X AX Decomposition Reactions • In a decomposition reaction, a single compound breaks apart to form 2 or more simpler substances. • Decomposition is the opposite of synthesis. • This type of reaction is represented by the following general equation: AX A+X Single-Displacement Reactions • In a single-displacement reaction (also called single-replacement) one element replaces a similar element in a compound. • They often take place in aqueous solution. • This type of reaction is represented by the following general equation: A + BX AX + B Double-Displacement Reactions • In double-displacement reactions, the ions of 2 compounds exchange places in an aqueous solution to form 2 new compounds. • One of the compounds formed is usually either a precipitate, a gas, or water. • Represented by the following general equation: AX + BY AY + BX Combustion Reactions • In a combustion reaction, a hydrocarbon fuel combines with oxygen, releasing a large amount of energy in the form of light and heat. • Products of combustion reactions are always carbon dioxide and water vapor. • Example: Combustion of propane C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) Types of Reactions Sample Problem Classify each of the following reactions as a synthesis, decomposition, single-displacement, double-displacement, or combustion reaction. a. N2(g) + 3H2(g) → 2NH3(g) synthesis b. 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g) single-displacement c. 2NaNO3(s) → 2NaNO2(s) + O2(g) decomposition d. 2C6H14(l) + 19O2(g) → 12CO2(g) + 14H2O(l) combustion What is Stoichiometry? • Stoichiometry is the branch of chemistry that deals with the mass relationships of elements in a chemical reaction. • Stoichiometry calculations always start with a balanced chemical equation. Mole Ratio • A mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction Example: 2Al2O3(l) → 4Al(s) + 3O2(g) Mole Ratios: 2 mol Al2O3 4 mol Al 2 mol Al2O3 3 mol O2 4 mol Al 3 mol O2 Mole to Mole Conversions • Using the mole ratio, you can convert from moles of one substance to moles of any other substance in a chemical reaction. Example: • For the reaction N2 + 3H2 → 2NH3, how many moles of H2 are required to produce 12 moles of NH3? This is what we’re given: Use mole ratio as a conversion factor 12 mol NH3 x 3 mol H2 2 mol NH3 = 18 mol H2 Mole to Mass Conversions • To convert from moles of one substance to grams of another, you need 2 conversion factors: 1. mole ratio. 2. molar mass of the unknown. • To set up your conversion factor, always put the units you have on the bottom and the units you need on the top. Mole to Mass Conversions (continued) Example: Given the equation: 2Mg(s) + O2(g)→ 2MgO(s), Calculate the mass in grams of magnesium oxide which is produced from 2.00 mol of magnesium. Given 2.00 mol Mg x 1st C.F. Mole Ratio 2nd C.F. Molar Mass of the Unknown 2 mol MgO x 40.3 g MgO = 80.6 g MgO 2 mol Mg 1 mol MgO Mass to Mole Conversions • To convert from grams of one substance to moles of another, you need 2 conversion factors: 1. molar mass of the given. 2. mole ratio. • Mass to mole conversion factors are the inverse of mole to mass conversion factors. Mass to Mole Conversions (continued) Example: Given the equation: 2HgO(s) → 2Hg(s) + O2(g), How many moles of HgO are needed to produce 125g of O2? Given 125 g O2 1st C.F. Molar Mass of the Given x 1 mol O2 32.0 g O2 2nd C.F. Mole Ratio x 2 mol HgO 1 mol O2 = 7.81 mol HgO Mass to Mass Conversions • To convert from grams of one substance to grams of another, you need 3 conversion factors: 1. molar mass of the given. 2. mole ratio. 3. molar mass of the unknown. Mass to Mass Conversions (continued) Example: Given the equation: 2HgO(s) → 2Hg(s) + O2(g), How many grams of HgO are needed to produce 45.0 g of O2? 1st Given C.F. Molar Mass of the Given 1 mol O2 x 45.0 g O2 32.0 g O2 3rd C.F. Molar Mass of 2nd C.F. Mole Ratio the Unknown x 2 mol HgO x 216.6 g HgO = 609 g 1 mol O2 1 mol HgO HgO Volume Ratios • You can use the volume ratios as conversion factors just like you would use mole ratios. 2CO(g) + O2(g) → 2 molecules 1 molecule 2 mole 1 mole 2 volumes 1 volume 2CO2(g) 2 molecules 2 mol 2 volumes • Example: What volume of O2 is needed to react completely with 0.626 L of CO to form CO2? 0.626 L CO x 1 L O2 2 L CO = 0.313 L O2 Gas Stoichiometry Sample Problem Assume that 5.61 L H2 at STP reacts with excess CuO according to the following equation: CuO(s) + H2(g) → Cu(s) + H2O(g) a. How many moles of H2 react? 5.61 L H2 x 1 mol H2 = 0.250 mol H2 22.4 L H2 b. How many grams of Cu are produced? 5.61 L H2 x 1 mol H2 x 1 mol Cu x 63.5 g Cu = 15.9 g 22.4 L H2 1 mol H2 1 mol Cu Cu Limiting Reactants • When combining 2 or more different things to make a product, you have to stop when one of the things is used up. • For example, no matter how many tires there are, if there are only 8 car bodies, then only 8 cars can be made. Limiting Reactants (continued) • The limiting reactant is the reactant that limits the amount of product formed. • The excess reactant is the substance that is not used up completely. • Once the limiting reactant is used up, a chemical reaction will stop. Limiting Reactants (continued) Example: Silicon dioxide reacts with hydrogen fluoride according to the following equation: The limiting SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l) reactant If 6.0 mol HF is added to 4.5 mol SiO2, which makes the is the limiting reactant? Mole ratio C.F. least product. Set up 2 6.0 mol HF x equations, one for each given: 4.5 mol SiO2 x 1 mol SiF4 = 1.5 mol SiF44 4 mol HF Limiting Reactant is HF 1 mol SiF4 = 4.5 mol SiF4 1 mol SiO2 Percentage Yield • The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant (found with stoichiometry). • The actual yield of a product is the measured amount of that product obtained from a reaction (given in problem). • The percentage yield is the ratio of the actual yield to the theoretical yield, multiplied by 100. Percentage Yield Sample Problem A Given the following equation: C6H6 (l) + Cl2(g) → C6H5Cl(l) + HCl(g) When 36.8 g C6H6 react with excess Cl2, the actual yield of C6H5Cl is 38.8 g. What is the % yield of C6H5Cl? Set up a mass to mass conversion This is the Theoretical Yield 36.8 g C6H6 x 1 mol C6H6 x 1 mol C6H5Cl x 112.5 g C6H5Cl = 53.1 g 78.0 g C6H6 1 mol C6H6 1 mol C6H5Cl C6H5Cl Actual x 100 38.8 g C6H5Cl x 100 Percent = 73.1% Now use = = Yield Theoretical 53.1 g C6H5Cl the formula Percentage Yield Sample Problem B According to the following reaction : CO(g) + 2H2(g) → CH3OH(l) If the typical yield is 80.0%, what mass of CH3OH should be expected if 75.0 g of CO reacts with excess H2 gas? Set up a mass to mass conversion This is the Theoretical Yield 75.0 g CO x 1 mol CO x 1 mol CH3OH x 32.0 g CH3OH = 85.7 g 28.0 g CO 1 mol CO 1 mol CH3OH CH3OH Multiply theoretical yield by percentage to get actual yield 85.7 g CH3OH x 80.0% = 68.6 g CH3OH