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:
2.667 𝑔 𝑂
𝑀𝑎𝑠𝑠 𝑂 𝑖𝑛 𝐶𝑂2
1.000 𝑔 𝐶
𝑀𝑎𝑠𝑠 𝐶 𝑖𝑛 𝐶𝑂2
=
≈ 3.00
0.888 𝑔 𝑂
𝑀𝑎𝑠𝑠 𝑂 𝑖𝑛 𝑈𝑛𝑘𝑛𝑜𝑤𝑛
1.000 𝑔 𝐶
𝑀𝑎𝑠𝑠 𝐶 𝑖𝑛 𝑈𝑛𝑘𝑛𝑜𝑤𝑛
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: (CO 2/3)x3 =
C 3O 2
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
86𝑅𝑛
→ 42𝐻𝑒 + 218
84𝑃𝑜
218
84𝑃𝑜
→ 42𝐻𝑒 + 214
82𝑃𝑏
214
82𝑃𝑏
→ 42𝐻𝑒 + 210
80𝐻𝑔
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
80𝐻𝑔
210
81𝑇𝑙
→
→
0
−1𝑒
0
−1𝑒
+ 210
81𝑇𝑙
+ 210
82𝑃𝑏
Then finish the decay chain to arrive at the final product with a mass number = 202 & an atomic number =
80.
210
82𝑃𝑏
→ 42𝐻𝑒 + 206
80𝐻𝑔
206
80𝐻𝑔
→
206
81𝑇𝑙
206
82𝑃𝑏
→
0
−1𝑒
0
−1𝑒
+ 206
81𝑇𝑙
+ 206
82𝑃𝑏
→ 42𝐻𝑒 + 𝟐𝟎𝟐
𝟖𝟎𝑯𝒈
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:
𝟓 𝜶 𝒅𝒆𝒄𝒂𝒚𝒔:
𝟒 𝜷 𝒅𝒆𝒄𝒂𝒚𝒔:
𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝑨𝒕𝒐𝒎𝒊𝒄 𝑵𝒖𝒎𝒃𝒆𝒓 𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝑴𝒂𝒔𝒔 𝑵𝒖𝒎𝒃𝒆𝒓
𝟓 ∙ −2 = −𝟏𝟎
𝟓 ∙ −4 = −𝟐𝟎
𝟒 ∙ +1 = +𝟒
𝟒∙0=𝟎
-------------------------------------------------------------------------------------------------------------𝑭𝒊𝒏𝒂𝒍 𝑷𝒓𝒐𝒅𝒖𝒄𝒕:
𝟖𝟔 − 𝟏𝟎 + 𝟒 = 𝟖𝟎
𝟐𝟐𝟐 − 𝟐𝟎 + 𝟎 = 𝟐𝟎𝟐