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Name Lab Section _Date The Mole Concept and Atomic Weights The purpose of this activity is to better understand the concepts of relative atomic mass, counting by weighing and the mole. Percent composition and average atomic mass are included. Part I. Relative Atomic Masses and the Mole - Early Method When John Dalton proposed his atomic theory, he stated that the atoms of each element had a characteristic mass. He carriedoutexperimentstodetermine the relative atomic massofeachelement. To dothis, he had to establishastandard because a single atom was too small to weigh. The standardhe chose was that the mass of hydrogen would be setequal to 1.000. Inasimple experiment, Dalton would measure the grams of anelement such as sulfurthatreacted with 1.00 gramofhydrogen. Forsulfur, the reactingmasswas found to be 32.0grams, and so 32.Owas the relative mass of sulfur with respect to the standard hydrogen. (Note: The current standard for atomic mass is the most abundant isotope of carbon, C-12, with an assigned mass of exactly 12.000 amu.) The following activity will demonstrate how the relative mass method works. 1. Weighfive of the red beads to three decimal places. Be sure totare out the mass of the plastic cup. Record the mass in the table below. 2. Weigh five colorless beads. Record mass. Weigh five green beads. Record mass. 3. Set the 'Relative Mass' of the red beads equal to 1.000. Calculate the relative mass of the colorless and green beads by dividing their mass by the mass of the red beads. 4. Now, weigh 1.000 gram of red beads. Try to get the mass as close as possible to 1.000 gram. Count the number of red beads in this mass and record it. Be as precise as possible 5. For the colorless beads, weigh out a mass in grams equal to its relative mass. Count the number of colorless beads in this mass. Repeat this for the green beads. Bead size Mass (g) Relative Mass Number of beads in gram- relative mass bike Small.(red) 1.469 1.000 Yellow Medium (colorless) 4 beads 3 " 3.973 5. 89 10.89 , Large (green) 7.459 Page 1 of

Answer 1

2. Panty. Further calculations Calculate the number of fe alone in the measurement uncertainty concentrimity 1 0.001g imol Fe 55.85 g 6.02x10 2 atoms 55.859 26:02K.10 23 alams 0.001 6:0281029 aroms x 0.001g 55.85 _ 1.08 x 10 19 atoms fe 3. 1 mol of U-238 has mass of 238.05 g 6.02X1023 atoms of U-238 : 238.059 35,000 atoms of U-238 : 238.05 g - X 35,000 abang 6-028 X10 23 alomas - = 1994.01.34020 g - 1384.0I XIO Pant. I _. Compounds and mdar masses on formula weights __ | The smallest panhole of a covalent compound is called a molecule - The smallest particle of an conic compound is called a ion. 2. a t I Calculate the molar marses. ₂ Hu lmol. N . 14:09lmol lmolux La /mol molar mans of N₂ H - 2 mol 14. 14 28 +4 - 330 g/mol + 4.1 b. Ca(OH)2 4 mol Ca = 40.08 g/mol I mal o = 16.00 g/mol molt = 'g/mol

i molar mass of Ca(OH)2 = 1x 40.08 + 2x16 + Qal = 40.08 + 32 + 2 --- 9408 g/mol c. Ca(NO3)2 I mol ca = 40.08 g/mol I mol N :1 gmol Imol o = 16 g/mol :. molon mass of Ca(NO3)2 = 1X40008 + 2X 14 +6X16 240.08 + 28 t 96 = 164608 g/mol 3. Molan maus ef_Ny Hy: 32 g/mol mol of Ng Hy = 6.02 x 10 23 molecules. 6 32 g of N₂ My = 6:02 x 10 23 molecules 40:09 of No Hy : 602X1023 x40 328 = 2.52 X 102 molecules i mol Ne Hy : 32 g/mol 6.02 x 10 23 molecules 105 X.c020 mdeeules 329 32 x 15 xlo 20 6.02X1023 = 7.97 X 10 - 3 a mol of Ca(OH)2 = 74-08 g/mol in lose mol of Facebo 74608 x16 - (012g 1 mol of Calon) 5 miles of ca.com, a atoms of H 2x1.5 avons of H - 3. atoms of H

Pant vì :...cuemical ferraules and parent composibion by mas y of element in compound -(no of atoms of element in formula) batomie mans) Josumula mars Abomiomass of cam 40.08 4- Abensic mass of N 14:09 Atomic mass of o = 16.0 g Fonomiela mass of.Ca(NO3)2 2 IX 40.08 + 2x14:0. + 6 X 16 2 164.08 % of ca 1840.08 164.08 24-43 % 4-N 2XL 164.08 - 17.06 % - Tokal --- ---- _68162 96 164.08 7646 59.51% ce + 2 0 24:43_4_+ 17.06% + 58.51%. 100 . As shoxon above, te of Nitreges in Ca(NO3)2 = 17.06% 1 pound of Ca(NO3)2 453.59 g · Grams of nitrogen contained in pound of Ca(NO3)2 2 17.06% of 453.59 g : 17.06 x 453.59 - 77.389 element (no. of males of abon O % composihon m olan mass of element molar mass of compound