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CHEM 1A: Challenge Problem Set 2

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CHEM 1A: Challenge Problem Set 2 CHEM 1A: Challenge Problem Set 2 1. Shown below, the mass composition data of carbon dioxide is compared to another oxide of carbon. Using the Law of Multiple Proportions, what is a possible formula for the second substance? Compound Mass of Carbon (g) Mass of Oxygen (g) Substance 1: Carbon Dioxide (CO2) 1.000 2.667 Substance 2: UNKNOWN 1.000 0.888 The Law of Multiple Proportions is used to compare the masses of one element (in this case oxygen) as it varies in two different compounds, relative to a fixed amount of another element (in this case carbon). To do this, start by evaluating the proportion below: 푀푎푠푠 푂 푖푛 퐶푂2 2.667 푔 푂 푀푎푠푠 퐶 푖푛 퐶푂 1.000 푔 퐶 2 = ≈ 3.00 푀푎푠푠 푂 푖푛 푈푛푘푛표푤푛 0.888 푔 푂 푀푎푠푠 퐶 푖푛 푈푛푘푛표푤푛 1.000 푔 퐶 Now reflect on what this ratio tells you. In this case it indicates: if the two substances had the SAME AMOUNT OF CARBON, the carbon dioxide would have 3 times as much oxygen. In other words, for every carbon in the unknown, there must be 3 times less oxygen as there is in CO2. This implies a formula of CO2/3. Since formulas must express the smallest whole number ratio of atoms, this ratio of atoms can be multiplied by 3 to scale the formula to integer numbers: (CO2/3)x3 = C3O2 2. Rubidium has two naturally occurring isotopes, Rubidium-85 (relative mass 84.9118 amu) and Rubidium-87 (relative mass 86.9092 amu). If rubidium has an average atomic mass of 85.47 amu, what is the abundance of each isotope (in percent, %)? Be sure to report the correct number of significant figures. 3. Radon gas, a radioactive Noble gas, is typically the single largest contributor to an individual's background radiation dose, and is the most variable from location to location. Despite its short lifetime, some radon gas from natural sources can accumulate to far higher than normal concentrations in buildings, especially in low areas such as basements and crawl spaces due to its high density. Radon-222 undergoes the following series of radioactive decays: What is the atomic number and mass number for the element produced at the end of this decay chain? Three α decays: each results in the release of a helium nucleus, and a smaller atom. Atomic number is reduced by 2 & mass number is reduced by 4 for each α decay: 222 4 218 86푅푛 → 2퐻푒 + 84푃표 218 4 214 84푃표 → 2퐻푒 + 82푃푏 214 4 210 82푃푏 → 2퐻푒 + 80퐻푔 Two β- decays: 1 neutron is converted into 1 high energy electron and 1 proton. By converting one nucleon (a neutron) into another nucleon (a proton), the mass number stays constant but the atomic number is increased by 1: 210 0 210 80퐻푔 → −1푒 + 81푇푙 210 0 210 81푇푙 → −1푒 + 82푃푏 Then finish the decay chain to arrive at the final product with a mass number = 202 & an atomic number = 80. 210 4 206 82푃푏 → 2퐻푒 + 80퐻푔 206 0 206 80퐻푔 → −1푒 + 81푇푙 206 0 206 81푇푙 → −1푒 + 82푃푏 206 4 ퟐퟎퟐ 82푃푏 → 2퐻푒 + ퟖퟎ푯품 ALTERNATE SOLUTION: There’s always more than one way to solve a problem. In this case, you could consider the result of all decays collectively to find the final product: 푪풉풂풏품풆 풊풏 푨풕풐풎풊풄 푵풖풎풃풆풓 푪풉풂풏품풆 풊풏 푴풂풔풔 푵풖풎풃풆풓 ퟓ 휶 풅풆풄풂풚풔: ퟓ ∙ −2 = −ퟏퟎ ퟓ ∙ −4 = −ퟐퟎ ퟒ 휷 풅풆풄풂풚풔: ퟒ ∙ +1 = +ퟒ ퟒ ∙ 0 = ퟎ -------------------------------------------------------------------------------------------------------------- 푭풊풏풂풍 푷풓풐풅풖풄풕: ퟖퟔ − ퟏퟎ + ퟒ = ퟖퟎ ퟐퟐퟐ − ퟐퟎ + ퟎ = ퟐퟎퟐ .
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