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# notes_Chapter_3 - Chapter 3 Chapter Stoichiometry Chapter 3...

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Chapter 3 Chapter 3 Stoichiometry Stoichiometry

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Chapter 3 - Stoichiometry 3.1 Atomic Masses 3.2 The Mole 3.3 Molar Mass 3.4 Percent Composition of Compounds 3.5 Determining the Formula of a Compound 3.6 Chemical Equations 3.7 Balancing Chemical equations 3.8 Stoichiometric Calculations: Amounts of Reactants and Products 3.9 Calculations Involving a Limiting Reactant
Reaction of zinc and sulfur.

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Figure 3.1: Mass spectrometer
Chemists using a mass spectrometer to analyze for copper in blood plasma. Source: USDA Agricultural Research Service

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A herd of savanna-dwelling elephants Source: Corbis
Figure 3.2: Relative intensities of the signals recorded when natural neon is injected into a mass spectrometer.

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Figure 3.3: Mass spectrum of natural copper
Atomic Definitions II: AMU, Dalton, 12C Std. Atomic mass Unit (AMU) = 1/12 the mass of a carbon - 12 atom on this scale Hydrogen has a mass of 1.008 AMU. Dalton (D) = The new name for the Atomic Mass Unit, one dalton = one Atomic Mass Unit on this scale, 12C has a mass of 12.00 daltons. Isotopic Mass = The relative mass of an Isotope relative to the Isotope 12C the chosen standard. Atomic Mass = “Atomic Weight” of an element is the average of the masses of its naturally occurring isotopes weighted according to their abundances.

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Isotopes of Hydrogen 1 1 H 1 Proton 0 Neutrons 99.985 % 1.00782503 amu 2 1 H (D) 1 Proton 1 Neutron 0.015 % 2.01410178 amu 3 1 H (T) 1 Proton 2 Neutrons -------- ---------- The average mass of Hydrogen is 1.008 amu 3H is Radioactive with a half life of 12 years. H 2 O Normal water “light water “ mass = 18.0 g/mole , BP = 100.000000C D 2 O Heavy water mass = 20.0 g/mole , BP = 101.42 0C
Element #8 : Oxygen, Isotopes 16 8 O 8 Protons 8 Neutrons 99.759% 15.99491462 amu 17 8 O 8 Protons 9 Neutrons 0.037% 16.9997341 amu 18 8 O 8 Protons 10 Neutrons 0.204 % 17.999160 amu

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Calculating the “Average” Atomic Mass of an Element 24Mg (78.7%) 23.98504 amu x 0.787 = 18.876226 amu 25Mg (10.2%) 24.98584 amu x 0.102 = 2.548556 amu 26Mg (11.1%) 25.98636 amu x 0.111 = 2.884486 amu ___________ amu With Significant Digits = __________ amu Problem: Calculate the average atomic mass of Magnesium! Magnesium Has three stable isotopes, 24Mg ( 78.7%); 25Mg (10.2%); 26Mg (11.1%).
Calculate the Average Atomic Mass of Zirconium, Element #40 Zirconium has five stable isotopes: 90Zr, 91Zr, 92Zr, 94Zr, 96Zr. Isotope (% abd.) Mass (amu) (%) Fractional Mass 90Zr (51.45%) 89.904703 amu X 0.5145 = 46.2560 amu 91Zr (11.27%) 90.905642 amu X 0.1127 = 10.2451 amu 92Zr (17.17%) 91.905037 amu X 0.1717 = 15.7801 amu 94Zr (17.33%) 93.906314 amu X 0.1733 = 16.2740 amu 96Zr (2.78%) 95.908274 amu X 0.0278 = 2.6663 amu ____________ amu With Significant Digits = ___________ amu

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Problem: Calculate the abundance of the two Bromine isotopes: 79Br = 78.918336 g/mol and 81Br = 80.91629 g/mol , given that the average mass of Bromine is 79.904 g/mol. Plan: Let the abundance of 79Br = X and of 81Br = Y and X + Y = 1.0 Solution: X(78.918336) + Y(80.91629) = 79.904 X + Y = 1.00 therefore X = 1.00 - Y (1.00 - Y)(78.918336) + Y(80.91629) = 79.904 78.918336 - 78.918336 Y + 80.91629 Y = 79.904 1.997954 Y = 0.985664 or Y = 0.4933 X = 1.00 - Y = 1.00 - 0.4933 = 0.5067 %X = % 79Br = 0.5067 x 100% = 50.67% = 79Br %Y = % 81Br = 0.4933 x 100% = 49.33% = 81Br
LIKE SAMPLE PROBLEM 3.2 During a perplexing dream one evening you come across 200 atoms of Einsteinium. What would be the total mass of this substance in grams?

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