Mole & Molar Calculations - Lecture 3

The Mole & Molar CHEM10 Stoichi 14 ometry Lecture- 3  When you have a dozen of something, no matter what you are talking about, you have 12 of that thing. -A dozen donuts - a dozen eggs -a dozen cars, -a dozen chemistry problems are all 12 of that mat...

The Mole & Molar CHEM10 Stoichi 14 ometry Lecture- 3  When you have a dozen of something, no matter what you are talking about, you have 12 of that thing. -A dozen donuts - a dozen eggs -a dozen cars, -a dozen chemistry problems are all 12 of that material.  The mole is the exact same concept. You can have: -a mole of eggs - a mole of donuts - a mole of cars - a mole of chemistry problems. However a mole is MUCH bigger than 12...it is 6.022 x 1023. Obviously a mole of chemistry problems is way too many problems. The mole is typically used for counting or grouping very very small things, like atoms  a M O LE (mol) 1. The mole is an SI unit that allows us to count # of: - Atoms - Representati molecules - formula units ve without actually seeing them particles 2. 1 mol = 6.02 x 1023 representative particles 3. The number is called “Avogadro’s number” in honor of the Italian scientist Amadeo Avogadro 7  A DOZEN MASS of a dozen limes = MASS of a dozen cars  A MOLE CARBON COPPER MASS of a mole carbon = MASS of a mole copper 6.02 x 1023 C atoms = 6.02 x1023 Cu atoms Avogadro’s number can be used as a conversion factor: # atoms/ molecul es 6.02 # X moles 1023 10 If 1.0 mole of sodium has 6.02 X 1023 atoms, calculate the number of atoms in 2.0 moles of sodium # atoms/ molecul es 6.02 # X moles 1023 11 HOW DO YOU DETERMINE THE MASS OF A MOLE ? Average mass of an atom = atomic mass in amu Average mass of 1 mole of atoms = atomic mass expressed in grams (molar mass) Average Molar mass Mass of 1 mol  Atomic mass in grams 1 mole atoms 6.022 × 1023 atoms Atomic mass in 6.022 × 1023 = atoms grams  Molar Mass of Elements from Periodic Table Molar mass is the atomic mass expressed in grams. 1 mole of Ag 1 mole of C 1 mole of S = 107.9 g of Ag = 12.0 g of C = 32.1 g of S  Molar Mass of Elements from Periodic Table Molar mass is the atomic mass expressed in grams. Molecular Mass of Carbon = 12 amu Molar Mass of Carbon = 12 grams The molar mass of a compound can be determined by adding the molar masses of all of the atoms in its formula.  What is the molar mass of CH4? 1 C = 12 g/mole 4 H = 4 x 1g/mole = 4 g/mole molar mass of CH4 = 12+4 =16 g/mole Therefore, 1 mole of CH4 = 16 g 18  Calculations Using Molar Mass Molar mass conversion factors are fractions (ratios) written from the molar mass. relate grams and moles of an element or compound. Conversion factor: grams moles gra ms # Mola mole r s Mas Calculations involving moles To calculate the amount of moles given the mass we will use the following relationship: Amount of moles = mass (g) molecular mass (or weight) Note: mass is a property of matter, weight is the measurement of mass with a scale. We can only measure weight in chemistry, therefore we often see the notation Molecular Weight for molecular mass.  Converting Mass to Moles of Compound NaCl A box of table salt, NaCl, contains 737 g of NaCl. How many moles of NaCl are in the box? gra ms # Mola mole r s Mas  Converting Mass to Moles of Compound NaCl A box of table salt, NaCl, contains 737 g of NaCl. How many moles of NaCl are in the box? Step 1 State the given and needed quantities. Given: 737 g of NaCl Need: moles of NaCl Step 2 Write a plan to convert grams to moles. molar mass Grams of NaCl Moles of NaCl  Converting Mass to Moles of Compound NaCl A box of table salt, NaCl, contains 737 g of NaCl. How many moles of NaCl are in the box? Molar mass of NaCl = 23g + 35.5g =58.5g/mole Moles of NaCl = Given weight of NaCl = 737 g Molar mass of NaCl 58.5g/mole = 12.6 moles How many moles of iron does 25.0 g of iron represent? Atomic mass iron = 55.85 gra ms # Mola mole r s Mas s How many moles of iron does 25.0 g of iron represent? Atomic mass iron = 55.85 g/mole n= grams/molar weight n=25g/55.85g/mole gra ms n=0.448 mole # Mola mole r s Mas s  Molar Calculations: Mass to Atoms Plan: Mass in g  # moles  # atoms # atoms/ gra molecul ms es # 6.02 # Mola X mole 1023 mole r s s Mas How many iron atoms are contained in 25.0 grams of iron? Atomic mass iron = 55.85 Conversion sequence: grams Fe → atoms Fe gra ms # Mola mole r s Mas s How many iron atoms are contained in 25.0 grams of iron? Atomic mass iron = 55.85 Conversion sequence: grams Fe → atoms Fe n=0.448 mole No. of atoms = n × 6.022 × gra 1023 ms = 0.448 × 6.022 × 1023 = 2.7 × 1023 # Mola mole r s Mas s What is the mass of 3.01 x 1023 atoms of sodium (Na)? Molar mass Na = 23 g/mol # atoms/ molecul es 6.02 # X moles 1023 What is the mass of 0.365 moles of tin? Atomic mass tin = 118.7 Conversion sequence: moles Sn → grams Sn In dealing with diatomic elements (H2, O2, N2, F2, Cl2, Br2, and I2), distinguish between one mole of atoms and one mole of molecules.  Calculate the molar mass of 1 mole of H atoms. 1 H = 1(1.01 g) = 1.01 g Calculate the molar mass of 1 mole of H2 molecules. 2 H = 2(1.01 g) = 2.02 g  Molar Mass of Compounds  1 mol compound = 6.022×1023 formula units compound  Molar mass of compound = mass of 1 mol compound The molar mass of a compound is the sum of the atomic masses of each atom in the compound What is the molar mass of CO2? 1C 1(12.01g) 2O 2(16.00g) CO2 44.01g/ mol  Molar Mass of Compounds What is the molar mass of Al2(CO3)3? atomic mass Al 26.98 2Al 2(26.98g) C 12.01 3C 3(12.01g) O 16.00 9O 9(16.00g) Al2(CO3)3 233.99g/mol Proble ms  Garlic Allyl sulfide, C6H10S, is a compound that has the odor of garlic. How many moles of C6H10S are in 225 g?  Solution Step 1 State the given and needed quantities. Step 2 Write a plan to convert grams to moles. Find molar mass Convert Grams of C6H10S Moles of C6H10S How many moles of benzene, C6H6, are present in 390.0 grams of benzene? How many grams of (NH4)3PO4 are contained in 2.52 moles of (NH4)3PO4? The molar mass of (NH4)3PO4 is 149.12 g.  Calculations What is the mass of 3.61 moles of CaCl2? atomic mass a. 3.61 g Ca 40.08 b. 272 g Cl 35.45 c. 2.17 × 1024 g d. 401 g  Calculations How many moles of HCl are contained in 18.2 g HCl? a. 1.00 mol atomic mass H 1.01 b. 0.500 mol Cl 35.45 c. 0.250 mol d. 0.125 mol Percent Composition of Compounds  Percent composition of a compound is the mass percent of each element in the compound. H 2O 11.19% H by mass 88.79% O by mass If the formula of a compound is known, a two-step process is needed to calculate the percent composition. Step 1 Calculate the molar mass of the formula. Step 2 Divide the total mass of each element in the formula by the molar mass and multiply by 100. total mass of the element x 100 = percent of the element molar mass Calculate the percent composition of hydrosulfuric acid H2S. Step 1 Calculate the molar mass of H2S. 2 H = 2 x 1.01g = 1 S = 1 x 32.07 g = 32.07 g 2.02 g 34.09 g Calculate the percent composition of hydrosulfuric acid H2S. Step 2 Divide the mass of each element by the molar mass and multiply by 100.  2.02 g H  H:   (100) = 5.93%  34.09 g   32.07 g S  S:  (100)  94.07%  34.09  Percent Composition of Compounds – Known Chemical Formula 1. Determine the molar mass of Na2CO3 2(22.99 g Na) + 1(12.01 g C) + 3(16.00 g O) = 105.99 g/mol Na2CO3 2. Find ratio of the mass of each element to the mass of the compound. 2  22.99 g Na x100  43.38% Na 105.99 g Na2CO3 112.01 g C x100  11.33% C 105.99 g Na2 CO3 316.00 g O x100  45.29% O 105.99 g N a2 CO3  Percent Composition From Experimental Data Percent composition can be calculated from experimental data without knowing the composition of the compound. Step 1 Calculate the mass of the compound formed. Step 2 Divide the mass of each element by the total mass of the compound and multiply by 100. A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition. Step 1 Calculate the total mass of the compound 1.52 g N 3.47 g O 4.99 g = total mass of product A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition. Step 2 Divide the mass of each element by the total mass of the compound formed.  1.52 g N  (100) = 30.5%  4.99 g   3.47 g O  (100) = 69.5%  4.99   Calculating Empirical Formulas To calculate an empirical formula, you need to know: 1. The elements present in the compound 2.The atomic masses of each element (from the Periodic Table) 3. The ratio (by mass or %) of the combined elements Calculating Empirical Formulas Strategy to Calculate an Empirical Formula: 1. Assume a starting mass of the compound (usually 100.0 g) and express the mass of each element in grams. 2. Convert g of each element to mol using molar mass. (These numbers may or may not be whole numbers.) 3. Divide each of the mole amounts from Step 2 by the smallest mole amount. The new numbers are the subscripts in the empirical formula.  Calculating Empirical Formulas Calculate the empirical formula for a compound that contains 11.19% H and 88.79% O. Step 1 Find amounts of each element In a 100.0 g sample, there are 11.19 g H and 88.79 g O Step 2 Convert g to moles using element molar masses 1 mol 11.19 g × H = 11.10 mol H 1.008 g H H 1 mol 88.79 g O O = 5.549 mol × 16.00 g O O  Calculating Empirical Formulas Step 3 Convert to whole numbers by dividing by the smallest mole amount. 11.10 mol H = 2.000 5.549 mol O 5.549 mol O = 1.000 5.549 mol O Empirical formula is H2O  Stoichiometry Stoichiometry deals with the quantitative relationships between the reactants and products in a balanced chemical equation. 1N2(g) + 3I2(s)  2NI3(s) 1 mol N2 + 3 mol I2  2 mol NI3 Mole ratios come from the coeffi cients in the balanced equation: 3 mol I2 3 mol I2 1 mol N2 1 mol N2 are the The 3 other possibilities 2 mol NI3 of these inverse 2 molratios. NI3  Let’s think Which of these statements is not true about the reaction? 1N2(g) + 3I2(s)  2NI3(s) a. 1 mole of nitrogen is needed for every 3 moles of iodine b. 1 gram of nitrogen is needed for every 3 grams of iodine c. Both statements are true  Calculating (molar) ratios Calculate the number of moles of NI3 that can be made from 5.50 mol N2 in the reaction: 1N2(g) + 3I2(s)  2NI3(s) Plan 5.50 mol N2  mol NI3  Practice How many moles of HF will be produced by the complete reaction of 1.42 moles of H2 in the following equation? H2 + F2 2HF a. 0.710 b. 1.42 c. 2.00 d. 2.84  Stoichiometry Convert g to moles Convert moles to g Steps to calculate mass or moles from mass or moles  Mole-Mole Calculations How many moles of Al are needed to make 0.0935 mol of H2? 2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g)  Mole-Mole Calculations How many moles of HCl are needed to make 0.0935 mol of H2? 2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g) = 0.187 m  Practice! How many moles of H2 are made by the reaction of 1.5 mol HCl with excess aluminum? 2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g) a. 0.75 mol b. 3.0 mol c. 6.0 mol d. 4.5 mol  Practice! How many moles of carbon dioxide are produced when 3.00 moles of oxygen react completely in the following equation? C3H8 + 5O2 3CO2 + 4H2O a. 5.00 mol b. 3.00 mol c. 1.80 mol d. 1.50 mol 9-66  Mole-Mass Calculations What mass of H2 (2.02 g/mol) is made by the reaction of 3.0 mol HCl? 2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g)  Mole-Mass Calculations How many moles of HCl are needed to completely consume 2.00 g Al (26.98g/mol)? 2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g)  Mole-Mass Calculations What mass of Al(NO3)3 (213g/mol) is needed to react with.093 mol Na2CO3? 3Na2CO3(aq) + 2Al(NO3)3(aq)  Al2(CO3)3(s) + 6NaNO3(aq)  13 g Al(NO3 )3  Mole-Mass Calculations How many moles of Al2(CO3)3 are made by the reaction of 3.45g Na2CO3 (105.99 g/mol)? 3Na2CO3(aq) + 2Al(NO3)3(aq)  Al2(CO3)3(s) + 6NaNO3(aq)  0.0109 mol Na2 CO3  Practice! How many moles of oxygen are consumed when 38.0g of aluminum oxide (101.96 g/mol) are produced in the following equation? 4Al(s) + 3O2(g) 2Al2O3(s) a. 0.248 b. 0.559 c. 1.50 d. 3.00  Mass-Mass Calculations Now we will put it all together. What mass of Br2 (159.80 g/mol) is needed to completely consume 7.00 g Al (26.98 g/mol)? 2Al(s) + 3Br2(l)  2AlBr3(s)  Mass-Mass Calculations What mass of Fe2S3 (207.91g/mol) can be made from the reaction of 9.34 g FeCl3 (162.20 g/mol) with excess Na2S? 2FeCl3(aq) + 3Na2S(aq) Fe2S3(s) + 6NaCl(aq)  5.99 g Fe 2S3  Practice! What mass (g) of oxygen (O2 = 32 g/mol) is consumed when 54.0g of water (H2O = 18.02 g/mol) is produced in the following equation? 2H2 + O2 2H2O a. 0.167 g b. 0.667 g c. 1.50 g d. 47.9 g  Practice! What mass of H2O (18.02 g/mol) is produced when 12.0g of HCl (36.51 g/mol) react completely in the following equation? 6HCl + Fe2O3 2FeCl3 + 3H2O a. 2.97 g b. 39.4 g c. 27.4 g d. 110. g