ࡱ > bjbj .( d d j j : : : : : N N N 8 , $ N / $ , ' h : n : : n n n : : : n n n n 0T 4 : n 0 / n A n A n n : n $ n / A j : Section 6.1: Team Learning Worksheet Name ____________ 1. Why cant we just count atoms directly instead of counting them by weighing them? Why must we count atoms at all? That is, why is it important to know how many atoms we have in a sample? 2. The average mass of a carbon atom is 12.011. Assuming you could pick up one carbon atom, what is the likelihood that you would randomly get one with a mass of 12.011? Explain. 3. Chlorine exists mainly as two isotopes, 37Cl and 35Cl. Which is more abundant? How do you know? Section 6.2: Team Learning Worksheet 1. Explain the difference between the terms atomic mass and molar mass. 2. A 0.821-mol sample of a substance composed of diatomic molecules (of the same atom) has a mass of 131.3 g. Identify this molecule. 3. How many molecules of water are there in a 5.00-g sample of water? How many hydrogen atoms are there in this sample? 4. Consider separate samples of carbon dioxide and ammonia, each with the same mass. Which sample contains the greater number of molecules? How many times greater? Answers 6.1 1. Atoms are too small to count directly, and even if we could see them, there are too many in a small sample to count. In order to know the formula for a compound, the relative numbers of each type of atom in the compound must be known. 2. There is a 0% chance of finding a carbon atom with a mass of 12.011. The number 12.011 is an average mass of carbon isotopes, and no carbon atom has that mass. 3. The 35Cl isotope is more abundant. The average atomic mass for chlorine given on the periodic table is 35.45. The value 35.45 is closer to 35 than 37. 6.2 1. The term atomic mass refers to the mass of one atom, while the term molar mass refers to the mass of 1 mol of a substance (either an element or a compound). The atomic masses listed on the periodic table are actually average atomic masses. The units for atomic masses are amus, and the units for molar masses are grams (thus, when we say that the molar mass of water, for example, is 18.016 g, we are saying that 1 mol of water has a mass of 18.016 g). 2. The molecule is Br2. Since mass and mol values are given, the molar mass of the substance could be calculated as follows: EMBED Equation.3 = 159.9 g/mol Because the molecule is diatomic (with the same atoms), we know that the formula of the molecule must be X2. Thus the mass of each atom is (159.9/2) = 79.95, which is the atomic mass of bromine. 3. There are 1.67 1023 molecules of water and 3.34 1023 hydrogen atoms in 5.00 g of water. 5.00 g H2O EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 =1.67 EMBED Equation.3 1023 molecules H2O 1.67 EMBED Equation.3 1023 molecules H2O EMBED Equation.3 EMBED Equation.3 = 3.34 EMBED Equation.3 1023 hydrogen atoms 4. The sample of ammonia contains the greater number of molecules, by a factor of 2.58. A lower molar mass means that a molecule of ammonia has a lower mass than a molecule of carbon dioxide. In order to have equal masses of each, there must be a greater number of ammonia molecules than carbon dioxide molecules. To determine the factor assume any mass, determine the number of moles, and divide. For example, assume we have 100.0 g of each: 100.0 g CO2 EMBED Equation.3 EMBED Equation.3 = 2.27 mol CO2 100.0 g NH3 EMBED Equation.3 EMBED Equation.3 = 5.87 mol NH3 EMBED Equation.3 = 2.58 = EMBED Equation.3 $ % / 8 9 : > ! " B C E A l m f g * + ƽƱƥƖƄsƥƱƱƥ !j hf hf CJ EHUaJ #jm> hf hf CJ UVaJ j hf hf CJ UaJ hf hf CJ H*aJ hf hf CJ H*aJ hf hf CJ hf hf CJ aJ hf h+^ H*hl} h+^ CJ aJ hl} CJ aJ hl} hl} CJ aJ hl} h+^ - 9 : C D L^ `L h^h` gdl} C D E M Q @ A U V gdf h^h`gdf h^h`gdl} gdl} h^h` " I J U gdf h^hgdf h^h`gdf ^gdf + - . A B C E X Y Z [ \ ] p q r t ԏ}l[O hf hf CJ H*aJ !j hf hf CJ EHUaJ !j hf hf CJ EHUaJ #jN)I hf hf CJ UVaJ !jC hf hf CJ EHUaJ !j hf hf CJ EHUaJ #j>4I hf hf CJ UVaJ !j hf hf CJ EHUaJ #jE]u> hf hf CJ UVaJ j hf hf CJ UaJ hf hf CJ aJ 2 3 4 5 7 9 * + ǶǙve !j3 hf hf CJ EHUaJ !j hf hf CJ EHUaJ #j>4I hf hf CJ UVaJ !j hf hf CJ EHUaJ hf hf CJ H*aJ !jP hf hf CJ EHUaJ #jE]u> hf hf CJ UVaJ j hf hf CJ UaJ hf hf CJ aJ hf hf CJ H*aJ "+ , - . / B C D E S T a c d w x y z { | ±ԥԥԓp_ԥM #jЂu> hf hf CJ UVaJ !j hf hf CJ EHUaJ #jƎ> hf hf CJ UVaJ !j hf hf CJ EHUaJ #jE]u> hf hf CJ UVaJ hf hf CJ H*aJ !j hf hf CJ EHUaJ #j> hf hf CJ UVaJ hf hf CJ aJ j hf hf CJ UaJ !j hf hf CJ EHUaJ ±ԭ h+^ j h+^ Uhf !j hf hf CJ EHUaJ #ju> hf hf CJ UVaJ hf hf CJ aJ j hf hf CJ UaJ !j hf hf CJ EHUaJ gdf , 1h/ =!"#$% D d l J C A ? " 2 y o+k"Ph` U D `!M o+k"Ph` xMJAgF