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Chapter 6: Moles, Molar Mass, Percent Composition and Formulas From moles to mass and to the moon! AMU (Atomic Mass Units) The mass of Carbon-12 is 12 AMU. But wait, when I look on the periodic table, the atomic mass is listed as 12.01078 AMU??? WHY? Why, cruel world? WwWWwwhhHHHhhyyYYYYyy??? 6.1 Atoms and Moles Amedeo Avogadro Avogadro’s Number 6.022 x 1023 Avogadro discovered that there are 6.022 x 1023 atoms in 1 gram of hydrogen. Count Lorenzo Romano Amedeo Carlo Avogadro di Quaregna e Cerreto Be able to explain and use the concept of the “mole” This number is called a “mole.” The word “mole” is just like the word “dozen”. Dozen means “12”. You can have a dozen of anything. You can also have a mole of anything. Hmmm… I shall call 6.022 x 1023... a “mole”. Yes…that has a nice ring to it. So How Big is a “MOLE” Ummm… NO! Don’tout be Here it is written Aaaiiie cruel now… e 602,200,000,000,000,000,000,000 That’s 602 billion groups of a trillion! Let’s just do an example with paper clips. If you have a mole of paper clips and made them into a chain, how many times could you go to the moon and back with your chain? Assume a paper clip ( still folded) is about 3 cm long. To find the total distance of the paper clips we use the following equation: 3cm 6.02 x10 clips 24 cm 1.806 x 10 x clip 1 mole mole 23 Notice the unit “clips” cancels!!! Isn’t that Great… Anyone… Anyone see the greatness??? Man I love Conversions! The moon is 382,171 km from Earth, so to the moon and back would be 764,342 km. So we need to convert our cm into km… … oh how fun… this is a metric conversion This of course is a “2-step conversion” because both units have a prefix 1m 1 km 19 1.806 x10 cm 1 . 806 x 10 km 100 cm 1000m 24 I love conversions! We’re almost done!!! 1 trip and back 13 1.806 x10 km ( ) 2.36 x10 trips 764,342km 19 That’s 23 trillion trips!! Mole-tastic! Marshmallow example: A bed of marshmallows covering the U.S. would be 776 miles deep Convert moles to # of atoms How many atoms are in 3.2 mol potassium (K)? Remember: 1 mol = 6.02 x 1023 atoms This can be written as a conversion factor: 6.02 10 23 atoms 1 mol 6.02 10 23 atoms 19.264 x 10 23 atoms of K 3.2 mol K 1 mol 1.9 x10 24 atoms of K How do we use the “Mole” in chemistry? The atomic mass of an element is the grams of 1 mole of that atom Why do chemists use moles? It’s fun. It’s impossible to count atoms with your hands. You can easily measure the mass (in grams) of a chemical. Atomic mass = grams of 1 mole of this element, Cobalt Convert moles of an atom to grams I need 2.0 moles of copper (Cu) for an experiment. How many grams is that? Atomic mass of Cu = 63.55 g/mol (round to 2 decimals) “mol” is the abbreviation of “Mole”… I know it’s only one letter different… chemists!!! 63.55 g 2.0 mol Cu 127.01 1 mol 130g Cu (2 sig figs! ) Converting grams to moles I have 302 grams of silver (Ag). How many moles of silver do I have? Step 1: Atomic mass of Ag = 107.87 g/mol Step 2: Calculate 1 mol 2.79966 mol 302g Ag 107.87 g 2.80 mol (3 sig figs) 6.2 Molar Mass and Percent Composition Atomic Mass = mass of one mole of an atom Molar Mass= mass of one mole of a substance Calculate Molecular Weights Example: Calculate the Molecular Weight (MW) of RbI2 Step 1: Assume you have 1 mole of this molecule and determine how much each element weighs from the periodic table. Step 2: Determine how many of each element you have Step 3: Add all the masses together Step 1: Find how much each element weighs from the periodic table Rb is atomic # 37. How much does each mole of Rb weigh? 85.47 grams/mol Rb I is atomic # 53. How much does it weigh? 126.90 g/mol I Step 2: Determine how many of each element you have Look at the formula: We have 1 “Rb” atom and 2 “I” atoms RbI2 Step 3: Add all the masses together You will need to show this work: (85.47 g/mol Rb) (1 Rb) 85.47 g/mol plus (126.90 g/mol) (2 I) 253.80 g/mol Because the units are the same we can add these two numbers together, so… 253.80 g/mol + 85.47 g/mol = 339.27 g/mol 339.27 g/mol is the “molar mass” Converting from moles of a compound to grams Example: I need 3.00 mol NaCl for an experiment. How many grams is that? Step 1: Find the molar mass Molar mass = 22.09g/mol + 35.45g/mol = 57.54 g/mol Step 2: Use the molar mass like a conversion factor. 57.54 g 3.00 mol 172.62 g 1 mol 173g NaCl Converting from grams of a compound to moles Example: How many moles are in 10.0 g of Na2SO4? Step 1: Find the molar mass. Molar mass = 142.1 g/mol Step 2: Use the molar mass like a conversion factor. You need “grams” on the bottom of the fraction. 1 mol 2 moles of Na 2SO4 = 10.0 g 7.04 10 mol 142.1 g 6.3 Formulas of Compounds Calculate “percent composition” Just like any other % stuff x 100 Percent Compositio n total stuff Stuff = grams of elements Calculate “percent composition” Ex: calculate % of Cu and S in Cu2S stuff x 100 Percent Compositio n total stuff Stuff = grams Cu (63.55 g/mol Cu)(2 mol Cu) = 127.1g Cu Total stuff = grams Cu + grams S = 127.1 g + 32.07 g = 159.17 g = 159.2 g 127.1 g x 100 79.84% 159.2 g You should be able to… Identify an “empirical formula” and a “molecular formula” Empirical formula – simplest ratio of atoms of each element in a compound (whole #’s only) Molecular formula – actual # of atoms of each element in a compound Molecular H2O2 C3H6 N2O3 C2H6 C3H9 Empirical HO CH2 N2O3 ? ? Using % composition to determine a formula Law of Definite Composition – Any amount of a pure compound will always have the same ratio of masses for the elements that make up that compound Ex: H2O is always 88.9% O and 11.1% H by mass Only the simplest formula (ratio) can be found… in other words, you can only find empirical formulas Using % composition to calculate the formula Process is as follows: 1.Calculate % by mass of each element 2.Determine mass of each element Easy if you use 100 g of the chemical 3.Use mass to find the # of moles of each element 4.Find the smallest ratio of the atoms ÷ the number of moles of each element by the element with the smallest # of moles Round to the nearest whole # Example A molecule is 75% C & 25% H. Calculate the empirical formula. Using 100g total = 75g C and 25 g H Calculate moles of each = H 1 mol 1 mol 6.2 mol 25mol 75g C 25 g H 12.01g 1.01g Ratio = 6.2C : 25H, simplify by ÷ each by 6.2. Whole number only!! C 6.2 C 6.2 1 25 H 6.2 Final ratio ≈ 1C : 4H so CH4 4 C H Percent 75% 25% 100 g total Moles 75g 25g Ratio 75g C 1 mol 12.01g 1.01g 6.2 mol 25 mol 6.2 6.2 25 6.2 = 1 Formula 25 g H 1 mol = 4 CH4 Find the molecular formula Ex: C3H6O2 is an empirical formula for a chemical. The molar mass of the compound is 148 g/mol. What is the molecular formula of the compound?? Point: The ratio of C:H:O will always be what the empirical formula shows Steps 1. Calculate the empirical formula mass 2. Calculate molar mass/empirical formula mass 3. Multiply your subscripts by that #. Steps 1. Calculate the empirical formula mass: C3H6O2 mass = (3)(12.01) + (6)(1.01) + (2)(16.00) = 148.09 g/mol 2. Calculate (molar mass)/(empirical formula mass) Round to a WHOLE number. 148 2 74 3. Multiply the subscripts of the empirical formula by that number. C3x2H6x2O2x2 = C6H12O4