The Mole Chapters 10 & 11

What is a mole?

What is a mole?

moleA unit of measurement for molecules # of particles of any chemical substance 1 mole = 6.02x10 23 particles This is known as Avagodro’s # 1mole of Carbon = 6.02x10 23 atoms of Carbon 1mole of water = 6.02x10 23 molecules of H 2 O 1mole of glucose = 6.02x10 23 molecules of C 6 H 12 O 6 1mol of popcorn kernels = 6.02x10 23 kernels (enough to bury the entire US under 9 feet of popcorn kernels!) What is a mole in CHEMISTRY?

Practical Application of the Mole = mass of 1 mole of a substance in grams = average atomic mass given in periodic table SubstanceMolar Mass Carbon12.0 g Magnesium24.3 g Copper63.5 g Lead207.2 g

= sum of the atomic masses of each element represented in formula Practical Application of the Mole CompoundFormulaFormula Mass WaterH2OH2O2(1) + 1(16) = 18g Calcium HydroxideCa(OH) 2 1(40) + 2(16) + 2(1) = 74g Iron (III) ChlorideFeCl 3 1(55.8) + 3(35.4) = 162.0g Potassium NitrateKNO 3 1(39) + 1(14) + 3(16) = 101g

Learning Check Prozac, C 17 H 18 F 3 NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. Find its molar mass. 17 x 12 =204g 18 x 1 =18g 3 x 19 =57g 1 x 14 =14g 1 x 16 =16g 309g

Calculations with Molar Mass molar mass Grams Moles

Calculations with Molar Mass Practice Example #1 Aluminum is often used for the structure of light- weight bicycle frames. How many grams of Al are in 3.00 moles of Al? GivenSolving For 3.00 moles of Al_____ g Al 3.00 moles of Al27 g of Al 1 mole of Al = 81g of Al

Learning Check The artificial sweetener aspartame (Nutra- Sweet) formula C 14 H 18 N 2 O 5 is used to sweeten diet foods, coffee and soft drinks. How many moles of aspartame are present in 225 g of aspartame? 225g aspartame1 mole aspartame 294g aspartame = 0.76 mol aspartame

molar mass Avogadro’s # Grams Moles Particles Calculations with Molar Mass

Multistep Practice Problem #1 How many atoms of Cu are present in 35.4 g of Cu? = 3.4 X atoms Cu 35.4 g Cu1 mol Cu6.02 X atoms Cu 63.5 g Cu1 mol Cu

How many atoms of K are present in 78.4 g of K? Multistep Practice Problem #2 = 1.21 x atoms K 78.4 g K1 mol K6.02 X atoms K 39.0 g K1 mol K

What is the mass (in grams) of 1.20 X molecules of glucose (C 6 H 12 O 6 )? Multistep Practice Problem #3 = 358 g glucose 1.20x10 24 molecules of C 6 H 12 O 6 1 mol glucose180 g glucose 6.02 X molecules1 mole glucose

How many atoms of O are present in 78.1 g of oxygen gas? Multistep Practice Problem # g O 2 1 mol O X molecules O 2 2 atoms O 32.0 g O 2 1 mol O 2 1 molecule O 2 = 2.94 x atoms of O

Gases at STP (Standard Temperature & Pressure) with the same number of particles will occupy the same volume Calculations with Molar Volume 22.4L One mole of any gas at STP = 22.4L

Calculations with Molar Volume Molar Volume Practice Problem A student fills a 1.0L flask with carbon dioxide (CO 2 ) at standard temperature and pressure. How many molecules of gas are in the flask? 1.0 L CO 2 1 mol CO x molecules of CO L CO 2 1 mol CO 2 = 2.69 x molecules CO 2

Empirical & Molecular Formulas The mass of each element in a compound compared to the total mass Percent Hydrogen of 1 mole of H 2 O % H = 2.0 g H = 11% 18 g H 2 O % Composition Using Experimental Data KnownsUnknowns 8.2 g = 60.3% Mg Mass Mg = 8.2 g % Mg = 13.6 g Mass O = 5.4 g % O = 5.4 g = 39.7% O Total Mass = 13.6 g 13.6 g

What is the empirical formula of a compound containing 25.9% N and 74.1% O? Step 1: Assume 100 g of the compound (only necessary if given percents to start) 25.9% N = 25.9 g N 74.1% O = 74.1 g O Step 2: Convert grams to moles25.9 g N1 mol N = 1.85 mol N 14.0 g N 74.1 g O1 mol O = 4.63 mol O 16.0 g O Step 3: Divide both mole amounts by the smaller of the 2 numbers 1.85 mol N = 1 mol N mol O= 2.50 mol O 1.85 Step 4: Multiply both to obtain whole numbers 1 mol N x 2 = 2 mol N 2.5 mol O x 2=5 mol O N2O5N2O5 Calculating Empirical Formulas

Calculating Molecular Formulas Step 1: Step 1: Calculate the empirical formula N 2 O 5 (answer from previous problem) Step 2: Step 2: Based on the empirical formula, calculate the formula mass 2x14 = 28 5x16 = g Step 3: Step 3: Divide the given molar mass by the calculated formula mass If the molar mass of the compound is 324g, what is the molecular formula? 324g/108g = 3 Step 4: Multiply the subscripts in the empirical formula by the answer found in #3 N 6 O 15

Stoichiometry: Mole-Mole A balanced formula indicates the number of moles of each compound. Ex: 2HCl + Zn  ZnCl 2 + H 2 2 moles HCl+1 mole Zinc yield 1 mole Zinc Chloride +1 mole hydrogen gas Based on this formula, how many moles of HCl are needed to react with 2.3 moles of Zn? 2.3 mol Zn2 mol HCl = 4.6 mol HCl 1 mol Zn

Stoichiometry: Mass-Mass In a thermite reaction, powdered aluminum reacts with iron (III) oxide to produce aluminum oxide and molten iron. What mass of aluminum oxide is produced when 2.3g of aluminum reacts with iron (III) oxide? Balanced Equation Stoichiometry Al + Fe 2 O 3  Al 2 O 3 + Fe g Al1 mol Al 1 mol Al 2 O 3 102g Al 2 O 3 27g Al 2 mol Al 1 mol Al 2 O 3 = 4.3g Al 2 O 3

Stoichiometry: Mass-Volume Sodium Bicarbonate (NaHCO 3 ) can be used to extinguish a fire. When heated, it decomposes into sodium carbonate, carbon dioxide and water. If a sample contains 4.0g NaHCO 3, what volume of CO 2 gas is produced? Balanced Equation Stoichiometry NaHCO 3  Na 2 CO 3 + H 2 O + CO g NaHCO 3 1 mol NaHCO 3 1 mol CO L CO 2 84g NaHCO 3 2 mol NaHCO 3 1 mol CO 2 =.53L CO 2

Limiting Reactants limiting reactant The limiting reactant in any equation would produce a smaller amount of product is completely used up first determines the amount of product made Which reactant in an equation is the limiting reactant? Identify the limiting reactant when 1.7 g of sodium reacts with 2.6L of chlorine gas at STP to produce sodium chloride. Step 1Write the balanced equation Step 2Determine the mass of product that would be produced from each given reactant quantity. 2 Na + Cl 2  2 NaCl 1.7g Na1mol Na2 mol NaCl58g NaCl = 4.3g NaCl 23g Na2 mol Na1 mol NaCl 2.6L Cl 2 1 mol Cl22 mol NaCl58g NaCl = 13g NaCl 22.4L Cl 2 1 mol Cl 2 1 mol NaCl

Percent Yield Percent Yield = Actual yield X 100 Expected yield Expected yield is determined from a stoichiometry problem. Actual yield is the amount of product measured or recovered. A piece of copper with a mass of 5.00g is placed in a solution of silver (I) nitrate containing excess AgNO 3. The silver metal produced has a measured mass of 15.2g. What is the percent yield for this reaction? Step 1Write balanced equation Step 2Find expected mass of product Cu + 2 AgNO3  2 Ag + Cu(NO3)2 5.00g Cu1mol Cu2 mol Ag107.9g Ag = 17.0g Ag 63.5g Cu1 mol Cu1 mol Ag Percent Yield=Actual yield =15.2g Ag X 100= 89.4% Expected yield17.0g Ag