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Homework 20

  • Page ID
    28877
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    1.73,2.55


    Q 1.73

    Question 1:
    The atomic mass for Helium is 4.003 amu and the mass spectrum for Helium shows a spike at this mass. Meanwhile, the atomic mass for Bromine is 79.904 amu; however, the mass spectrum for Bromine does not show a peak at that mass. What causes the difference?

    Solution:

    1. The best way to start this problem is to recall what the atomic mass means in the context of the problem. This is the standard atomic mass, which is the weighted average of all the isotopes. "Weighted" in this situation means that the abundance of each isotope factors in to how it factors in to the overall the standard atomic mass for that particular element.
    2. Next step is to look at Helium. Helium has two isotopes and is stable at either; however, in atmospheric values, Helium is found most abundantly as 4He (99.999863%) and much less commonly found as 3He (0.000137%). Consequently, 3He does not have a strong influence on the standard atomic mass of Helium or much expression on the mass spectrum which measures relative abundance.
    3. Examining Bromine reveals two major isotopes for that element as well. Both 79Br as well as 81Br are rather prevalent (79Br with 50.69% and 81Br with 49.31%). This means that their relative abundance is high enough that they would make a significant impact on both the mass spectrum as well as the standard atomic weight of Bromine.
    4. To connect the dots, the dichotomy between the two elements comes from the abundance of their isotopes. Bromine's two isotopes almost equally comprise the atomic mass so the atomic mass they produce does not match up with an actual relatively abundant isotope of Br so therefore the line for that particular atomic mass is absent on the mass spectrum for Br. Helium, meanwhile, is primarily influenced by the 3He because of its incredibly high abundance and therefore the mass spectrum and the standard atomic mass are very similar.

    Q 2.55

    Question 2:

    Find the mass (in grams) of each of the following:

    a. 2.1 x 1022 Fe molecules b. 2.9 x 1023 Al molecules
    c. 3.9 x 1022 Ca molecules d. 4.3 x 1022 Ga molecules

    Solution:

    1. Firstly, because the question requires the same steps for each part, the best way to begin would be to break down the steps necessary into a basic equation:

    \[\frac{number\; of\; molecules}{1}|\frac{1\; mol}{6.022\cdot 10^{23}molecules}|\frac{g}{1\; mol}\]

    This equation is formed by the fact there are 6.022•1023 molecules per mole. Next, the g/mol is the atomic weight of the respective elements in order to finalize the conversion from molecules to grams.

    1. Next is to fill in the numbers for each part and each element.

    a.

    \[\frac{2.1\cdot 10^{22}Fe\; molecules}{1}|\frac{1\; mol}{6.022\cdot 10^{23}molecules}|\frac{55.845\; g}{1\; mol}\]

    =1.9 g

    b.

    \[\frac{2.9\cdot 10^{23}Al\; molecules}{1}|\frac{1\; mol}{6.022\cdot 10^{23}molecules}|\frac{26.982\; g}{1\; mol}\]

    =13.0 g ( from 12.99 g)]

    c.


    \[\frac{3.9\cdot 10^{22}Ca\; molecules}{1}|\frac{1\; mol}{6.022\cdot 10^{23}molecules}|\frac{40.078\; g}{1\; mol}\]

    =2.6 g

    d.


    \[\frac{4.3\cdot 10^{22}Ga\; molecules}{1}|\frac{1\; mol}{6.022\cdot 10^{23}molecules}|\frac{69.723\; g}{1\; mol}\]

    =5.0 g ( from 4.98 g)


    Homework 20 is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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