6.02 x x baseballs would cover the Earth to a height of several hundred miles.

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1 A stack of paper with 6 x sheets would be so tall that it would reach from here to the sun-not just once but more than a million times! 6 x baseballs would cover the Earth to a height of several hundred miles. If you could travel at the speed of light, it would take you more than 100 billion years to travel 6 x miles. For 6 x drops of water to pass over the Niagara Falls, more than 100,000 years would be required x x seconds is a length of time about 4 million times the age of the earth. A computer that can make 200 million counts per second would need almost 100 million years to count up to 6 x If 6 x dollars were divided among the six billion people on Earth, we would have enough money that each of us could spend a million dollars every minute, night and day, for as long as we lived and still have more than half left. If you spread 6 x tiny grains of sand over the entire state of California, the sand pile would be as high as a ten-story building. Hill, John W., and Doris Kolb. Chemistry for Changing Times. 9th ed. New Jersey: Prentice-Hall,

2 How was Avogadro's number determined? February 16, 2004 Chemist George M. Bodner of Purdue University explains. Contrary to the beliefs of generations of chemistry students, Avogadro s number--the number of particles in a unit known as a mole--was not discovered by Amadeo Avogadro ( ). Avogadro was a lawyer who became interested in mathematics and physics, and in 1820 he became the first professor of physics in Italy. Avogadro is most famous for his hypothesis that equal volumes of different gases at the same temperature and pressure contain the same number of particles. The first person to estimate the actual number of particles in a given amount of a substance was Josef Loschmidt, an Austrian high school teacher who later became a professor at the University of Vienna. In 1865 Loschmidt used kinetic molecular theory to estimate the number of particles in one cubic centimeter of gas at standard conditions. This quantity is now known as the Loschmidt constant, and the accepted value of this constant is x m -3. The term "Avogadro s number" was first used by French physicist Jean Baptiste Perrin. In 1909 Perrin reported an estimate of Avogadro s number based on his work on Brownian motion--the random movement of microscopic particles suspended in a liquid or gas. In the years since then, a variety of techniques have been used to estimate the magnitude of this fundamental constant. Accurate determinations of Avogadro s number require the measurement of a single quantity on both the atomic and macroscopic scales using the same unit of measurement. This became possible for the first time when American physicist Robert Millikan measured the charge on an electron. The charge on a mole of electrons had been known for some time and is the constant called the Faraday. The best estimate of the value of a Faraday, according to the National Institute of Standards and Technology (NIST), is 96, coulombs per mole of electrons. The best estimate of the charge on an electron based on modern experiments is x coulombs per electron. If you divide the charge on a mole of electrons by the charge on a single electron you obtain a value of Avogadro s number of x particles per mole. Another approach to determining Avogadro s number starts with careful measurements of the density of an ultrapure sample of a material on the macroscopic scale. The density of this material on the atomic scale is then measured by using x-ray diffraction techniques to determine the number of atoms per unit cell in the crystal and the distance between the equivalent points that define the unit cell (see Physical Review Letters, 1974, 33, 464). 2

3 MadSci Network: Chemistry Re: How was the Avagadro number determined to be 6 x 10 23? Date: Wed Apr 19 20:17: Posted By: Dan Patel, Undergraduate, Chemistry Major/Math Minor, University of Houston Area of science: Chemistry ID: Ch Message: Avogadro's number is officially defined as the number of particles in 12.0 grams of Carbon-12 (carbon that has 6 protons and 6 neutrons in its nucleus). The accepted value today is approximately *10^23, and with the equipment available to a modern chemist there are several ways to arrive at this number. A chemist could use a mass spectrometer, a precise instrument which can determine an element or compound's mass. We won't go into details, but such a device is sensitive to the number of atoms present, and it could be used for "atom counting." Of course, a chemist wouldn't count out an entire mole of atoms, but could count a small amount and then using proportions could find out how many atoms are in a mole. Also, one could use the theory behind Brownian Motion to calculate Avogadro's number. Albert Einstein did this successfully in Brownian Motion occurs when small solvent molecules collide with larger solute molecules. (You can observe this by placing a drop of milk under a microscope). Finally, one could determine Avogadro's number using x-rays. By examining how x-rays bounce off of a crystal, a chemist can tell how atoms in the crystal are spaced. By using the density of the crystal and the fact that density = mass/volume, one can calculate Avogadro's number. The actual formula used is the following: Na = (nm)/(pv) In this case Na = Avogadro's number, n is the number of atoms in a crystal cell, M is the molar mass of the element, p is the element's density, and V is the volume of a cell. A cell is the smallest repeating structure of a crystal. Also, you can check the following site for more information and additional methods for calculating Avagadro's number: Any college level text book will also contain information on at least one method of calculating Avogadro's number. 3

4 The Mole Concept Avogadro's Number No. of particles 6.02 x Gram-FormulaWeight g of NaCl g of CaCO g of PbSO 4 Gram-Molecular Weight 2.00 g of H g of O g of Cl 2 Gram-Atomic weight g of C atoms g of Fe atoms g of S atoms 4

5 6 C 12 no. of p+ no. of p + + no. of n o 6 p+ = (6)( x 10-24) = x grams 6 e- = (6)( x 10-28) = x grams 6 no = (6)( x 10-24) = x grams x grams %p+= x grams/ x grams x 100= 49.95% %e-= x grams/ x grams x 100=.027% %no= x grams/ x grams x 100= 50.02% % of the atom s mass or weight comes from the protons and neutrons. Thus, 2.00 x grams x 1/12 = 1.66 x grams = 1 µ (atomic mass unit) 2.00x10 23 g =1.66x10 24 g x = NA (Avogadro s Number) Then, 1/NA = 1/6.02 x = 1.66 x grams = 1 µ (atomic mass unit) 5

6 1 6.02x10 23 =1.66x10 24 g 12 g of C-12/12 µ = 1 mol. = NA = 6.02 x C 12 µ 12g 12µ =1mol.= N A = 6.02x10 23 atoms Therefore, 1 mol. = 6.02 x atoms of C-12 = 12 grams of C- 12 6

7 Name Date Chemistry Chapter 11 MOLES Fill in the chart below. Substance Atomic Weight or Formula Weight (µ) Mass of 1 mole of substance Number of Atoms or Formula Units in Substance 1. Au 2. B 3. Zn 4. Fe 5. He 6. Cd 7. Br 8. K 9. Al 10. S 11. O H 2 O 7

8 Substance Atomic Weight or Formula Weight (µ) Mass of 1 mole of substance Number of Atoms or Formula Units in Substance 13. NH CH FeO 16. H 2 O CO NaCl 19. HCl 20. B 2 S 3 8

9 Answers 1-20 Substance Atomic Weight or Formula Weight (µ) Mass of 1 mole of substance Number of Atoms or Formula Units in Substance 1. Au u g 6.02 x B u g 6.02 x Zn u g 6.02 x Fe u g 6.02 x He 4.00 u 4.00 g 6.02 x Cd u g 6.02 x Br u g 6.02 x K u g 6.02 x Al u g 6.02 x S u g 6.02 x O H 2 O 13. NH 3 2 x 15.99= u (2x1) u 14 + (3x1)= u g 6.02 x g 6.02 x g 6.02 x

10 14. CH FeO 16. H 2 O CO NaCl 19. HCl 20. B 2 S (4x1)= u = u (2x1) + (2x15.99)= u (2x15.99)= u = u = u (2x10.82) + (3x32.06)= u g 6.02 x g 6.02 x g 6.02 x g 6.02 x g 6.02 x g 6.02 x g 6.02 x

11 Name Date Chemistry Chapter 11 MOLES AND AVAGADRO'S NUMBER Calculate the mass in grams for each of the following. Moles of Substance Mass of Substance moles of Na moles of Ca moles of Mg moles of Al moles of H moles of B moles of Zn moles of O 2 Calculate the number of moles for each of the following. Mass of Substance Moles of Substance grams of Li grams of Ne grams of Ca 11

12 grams of Zn grams of Br grams of Fe grams of Cl grams of NaCl Calculate the number of atoms or ions for each of the following. Moles of Substance Number of Atoms or ions of substance moles of Na atoms mole of N atoms moles of K atoms grams of Na grams of S grams of Ca moles of Mg +2 ions moles of Cl -1 ions 12

13 Answers Moles of Substance Mass of Substance moles of Na 68.9 g moles of Ca 100 g moles of Mg 122 g moles of Al 108 g moles of H 3.03 g moles of B 45.8 g moles of Zn g moles of O g Mass of Substance Moles of Substance grams of Li 3.6 mol grams of Ne 2.97 mol grams of Ca mol grams of Zn mol. 13

14 grams of Br mol grams of Fe mol grams of Cl mol grams of NaCl 1.71 mol. Moles of Substance Number of Atoms or ions of substance moles of Na atoms 1.2 x atoms mole of N atoms 6.02 x atoms moles of K atoms 1.84 x atoms grams of Na 1.2 x atoms grams of S 4.69 x atoms grams of Ca 1.5 x atoms moles of Mg +2 ions 3.01 x ions moles of Cl -1 ions x ions 14

15 Name Date Chemistry Chapter 11 Fill in the chart. Substance Atomic Wgt. Formula Wgt. Molecular Wgt. (µ) Mass No. of particles (atoms, ions, molecules) Moles 45. Ba 4.22 x Zn H H 2 SO C 6 H 12 O x O NaCl CO CaO 6.02 x NH 4 OH C 8 H x AgNO

16 Answers Substance Atomic Wgt. Formula Wgt. Molecular Wgt. (µ) Mass No. of particles (atoms, ions, molecules) Moles 45. Ba x Zn x H x H 2 SO x C 6 H 12 O x O x NaCl x CO x CaO x NH 4 OH x C 8 H x AgNO x

17 Name Date Chemistry Chapter 11 Moles, AMU S, Avagadro s Number When using the atomic weights on the periodic table, use the weight to the nearest.01 place. Fill in the chart. 1 mol. = 6.02 x µ = 1.66 x g Substance AMU S of the substance (µ) Mass of 1 atom or molecule Mass of the substance No. of particles (atoms, ions, molecules) Moles (g/µ) 57. KF x NaCl X H 2 SO C 6 H 12 O /2 O x

18 Answers Substance AMU S of the substance (µ) Mass of 1 atom or molecule Mass of the substance No. of particles (atoms, ions, molecules) Moles (g/µ) 57. KF x x NaCl x x H 2 SO x x C 6 H 12 O x x /2 O x x

19 Name Date Chemistry Chapter 11 Moles, AMU S, Avagadro s Number Fill in the chart. 1 mol. = 6.02 x µ = 1.66 x g Substance AMU S of the substance (µ) Mass of 1 atom or molecule Mass of the substance No. of particles (atoms, ions, molecules) Moles (g/µ) 62. C x NaCl H 2 O x CO g 66. Bi x KF O x

20 Answers Substance AMU S of the substance (µ) Mass of 1 atom or molecule Mass of the substance No. of particles (atoms, ions, molecules) Moles (g/µ) 62. C x x NaCl x x H 2 O x x CO x g 6.02 x Bi x x KF x x O x x

21 Name Date Chemistry Chapter 11 Dilution Problems C 1 V 1 = C 2 V 2 V 2 V 1 = water added Answer the following questions. Show work! 69. How many milliliters of an 18M H 2 SO 4 solution is needed to prepare 500 ml of a 5M solution? ans: ml 70. How many milliliters of water must be added to 620 ml of a M solution of HNO 3 to obtain a 9780 ml of a 1M solution? ans: 9160 ml 71. If 250 ml of 2M solution of HCl is diluted with water to a.5m solution, what is the new volume of the solution? ans: 1000 ml 72. What is the molarity of a solution which has a volume of 1500 ml if it was obtained by diluting 250 ml of a 6M solution of H 2 SO 4? ans: 1M 21

22 Name Date Chemistry Chapter 11 Molarity Quiz Molarity = moles of solute L of solution moles = mass awt or mwt Show all work. 73. Calculate the molarity of a 1 L solution that contains 100 grams of NaCl. Answer: 1.71M 74. If 50 grams of NaOH are used to prepare a 3M NaOH solution, what volume of solution can be made? Answer:.416 L 75. Calculate the mass of solute required to make 2.50 L of a 2M KCl solution. Answer: g 22

23 Name Date Chemistry Chapter 11 Molarity Molarity = moles of solute L of solution moles = mass awt or mwt Show all work. 76. Calculate the molarity of a 1.5L solution that contains 200 grams of MgCl 2. Answer: 1.4M 77. Calculate the molarity of a.5l solution that contains 10 grams of NaOH. Answer:.5M 78. If 160 grams of NaOH are used to prepare a 2M NaOH solution, what volume of solution can be made? Answer: 2L 79. Calculate the mass of solute required to make.750l of a 2.5M NaCl solution. Answer: 110 g 23

24 The Mole: The Chemistry Counting Unit Reactions: None Procedure: 1. Calibrate the triple beam balance. 2. Obtain a beaker of beans. Record the beaker number in Table Pour the contents of beaker on to the pan of the triple beam balance. Mass all the beans and record mass in grams in Table Pour all the beans back into the beaker. 5. Remove 50 beans from the beaker and place on the triple beam balance pan. Record the number of beans removed from the beaker in Table Mass the sample of 50 beans removed from the beaker and record mass in grams in Table Put the sample of beans that were removed from the beaker back into the beaker. 8. Record the name of your counting unit in Table 1. You may be creative here!! B# 9. Calculate the number of beans in the beaker by: xc = F. Record the number in Table 1. " D$ 10. Return the beaker of beans. Results: Table 1 A. Beaker number B. Mass of the all the beans in the beaker C. Number of beans removed from the beaker D. Mass of the beans removed from beaker E. Name of the counting unit F. Calculation of grains or beans in the container by (B/D) x C = F Calculations: 24