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Chapter 6 Chemical Composition Problem: You work at a jelly bean store. Someone comes in and asks for 1000 jelly beans. What would you do so you don’t have to take the time to count exactly 1000 jelly beans? If the average mass of a jelly bean is 5grams, how would you fulfill the customer’s order? In this problem I told you the average mass of a jelly bean. If you had to figure out the average mass of a jelly bean, how would you do it? If you weigh 10 jelly beans and find the following results, what would the average jellybean weigh? Problem 2: A new customer comes in and says that she needs jellybeans for one of her kids and mints for the other kid. She doesn’t care how much candy she gets, as long as she has equal numbers of jellybeans and mints. You happen to know that every jelly bean weighs 5 grams and every mint weighs 15grams. You put a scoop of jellybeans on the scale and it weighs 500grams. How many grams of mints should you get? There are several strategies. -use ratios -use equivalence statements I. Atomic Masses: Counting by Weighing Atoms are really small. The mass of a carbon atom is 1.99 x 10-23 grams. It doesn’t make sense to weigh atoms in grams because the number is so small. Scientist came up with a new unit of mass for measuring the mass of atoms. I. Atomic Masses: Counting by Weighing A. Atomic Mass Unit (amu): A unit of mass that is used to measure the mass of atoms. 1. 1 amu = 1.66 x 10-24 grams B. Average Atomic Mass: Tells the average mass of an atom of an element. (it’s like knowing the average mass of a jelly bean.) 1. To find atomic mass, look on the periodic table. a. 2. Example: The average atomic mass of carbon = 12.01amu Average atomic mass can be used as an equivalence statement. (1 carbon atom = 12.01amu) a) b) What is the mass in amu’s of 198atoms of carbon.? How many carbon atoms are in 2.50 x 1019 amu’s of carbon.? c) d) e) f) g) What is the mass of 75 atoms of aluminum? How many nitrogen atoms are in 1387amu’s of nitrogen? What is the mass of 1 molecule of H2O? How many molecules of water are in 630.70amu’s of water. What is the mass in amu’s of 45molecules of CH4 Back to Jellybeans and Mints -Remember that a jellybean in 5grams and a mint is 15grams. 1. For each of the following, fill in the blank so that the number of jellybeans and mints will be the same. a. 15 5g Jellybeans and _____g mints. b. 15 tons of mints. 5 tons of jellybeans and _____ c. 15 tons of mints. 5lbs of jellybeans and _____ It doesn’t matter what unit you are using, the mass of mints will always be 3 times larger than the mass of jellybeans. (or a ratio of 5 to 15) 105 35g of jellybeans and _______g of mints 35g x 3 = 105 or 15 35g jellybeans x 5 = 105g mints 30 tons of mints and ______tons of jellybeans. 10 30tons ÷ 3 = 10tons or 30 tons mints x 5 15 = 10 tons jellybeans Lets apply this to atoms 1. For each of the following, fill in the blank so that the number of aluminum atoms and copper atoms will be the same. a. 26.98amu’s Al and _____amu’s Cu. 63.55 b. 26.98g Al and 63.55 _____g Cu. 2. How many atoms were we referring to in letter a? 1 of each 3. How many atoms were we referring to in letter b? ??????????? We know that when we look at the periodic table 12.01g Carbon 16.00g Oxygen 63.55g Copper 14.01g Nitrogen 26.98g Aluminum They all have the same number of atoms. Atoms are too small to count. Atoms are also to small for us to weigh in small quantities. It would take 500 million trillion carbon atoms to show up on our scales. So 500 million trillion is a small quantity of carbon. Chemists have come up with a way to weigh large quantities of atoms. C. The MOLE!!!!!!!!: it is defined as the number of atoms in 12.01g of carbon. -It could also be defines as the number of atoms in 14.01g of nitrogen -It could also be defines as the number of atoms in 16.00g of Oxygen -It could also be defines as the number of atoms in 35.45g of chlorine. 63.55 -It could also be defines as the number of atoms in _________g of copper 26.98 It could also be defines as the number of atoms in _______g of aluminum C. The MOLE!!!!!!!!: it is defined as the number of atoms in 12.01g of carbon. 1. Avagadro’s Number: The actual number of objects in a mole of something. a) Avagadro’s Number = 6.022 x 1023 b) 1 mole = 6.022 x 1023 (A very important equivalence) This number is so large that A mole of seconds is 4 million times as long as the earth has existed (according to science). A mole of marbles would cover the earth 50 miles deep. That’s how large a mole is. However, a mole of carbon atoms fits in a beaker. That’s how small atoms are. 2. Think of a mole as being like a dozen, but a whole lot larger. • • 1 dozen = 12 of something 1 mole = 6.022x 1023 of something 3. Determining numbers of atoms (use avagadro’s number) a. 5.78moles of carbon = _______atoms? b. 5.78moles of oxygen = _______atoms? c. 5.78 moles of marbles = _______marbles? d. 9.38 x 1023 copper atoms = _________moles? e. 4.98 x 1024 molecules of water = _________ moles? D. Molar Mass(M.M.): The mass of one mole of a substance. 1. The same # as the average atomic mass a. b. Example: 1 mole C = 12.01g Example: 1 mole H2O = 18.02g c. What is the molar mass of NH3 ? d. What is the mass of one mole of CH4 ? e. What is the mass of 6.022 x 1023 atoms of chlorine? 2. Molar Mass as an equivalence statement. M.M. = 1mole = 6.022 x 1023 units The M.M. of carbon is 12.01g/mol. So 12.01gC = 1mole And 12.01g C = 6.022 x 1023 atoms 12.01gC = 1mole And 12.01g C = 6.022 x 1023 atoms How many atoms are in 35.6g of carbon? 35.6g C x 6.02210^23 12.01 = 2.14 x 1025 atoms C How much does 5.98moles of carbon weigh? 12.01 5.98 moles C x 1 = 71.82g C How many moles of water are in 27.6g of H2O? 1 2 27.6g H2O x 18.02 2 = 1.53mol H2O II. % Composition A. Mass Percent: Tells the percentage of each element in a compound. (based on mass) Mass % = 1 1 ℎ x 100% What are the mass percent's of each element in ethanol (C2H6O)? 24.02 Mass 2 mol C = 12.01g x 2 = 24.02g % C= 46.08 x 100% =52.1% Mass 6 mole H = 1.01g x 6 = 6.06g % H= 46.08 x 100% =13.2% Find the mass % of each element in water. 6.06 16.00 Mass 1 mole O = 16.00 x 1 = 16.00g % O= x 100% = 34.7% 46.08 Mass 1 mole Compound = 46.08g III. Formulas of Compounds A. Molecular Formula: The actual ratio of atoms in a compound. 1. Example: C6H12O6 shows a ratio of 6 carbon to 12 hydrogen to 6 oxygen B. Empirical Formula: The smallest whole number ratio of atoms in a compound. 1. Examples: Molecular Formula C6H12O6 C3H9 C2H5O4 Empirical Formula 2. Calculating an Empirical Formula a) Steps 1. 2. 3. 4. Determine how many grams of each element are present. Calculate how many moles of each element are present. Calculate the ratio of the moles. (Divide by the smallest) Write the formula that shows the ration from step 3. Example: What is the empirical formula of a compound that contains 74.8% C and 25.2% H