Question

1. lithium has only two stable isotopes. use the sim to determine the following: a. atomic mass of lithium - 6 = _ amu b. atomic mass of lithium - 7 = _ amu c. average atomic mass of a sample containing three lithium - 6 atoms and two lithium - 7 atoms. ___ amu d. is the average atomic mass you just determined closer to the mass of lithium - 6 or lithium - 7? explain 5. describe a method to calculate the average atomic mass of the sample in the previous question using only the atomic masses of lithium - 6 and lithium - 7 without using the simulation. 6. test your method by creating a few sample mixtures of isotopes with the sim and see if your method correctly predicts the average atomic mass of that sample from only the atomic masses of the isotopes and the quantity of each isotope. use the table below to track your progress. table with columns: element, atomic mass and quantity of each isotope, average atomic mass of sample (calculate yourself), average atomic mass of sample (from simulation) model 3: nature’s mix of isotopes 1. using the sim, examine “nature’s mix of isotopes” for several different elements. if you assumed 100 total atoms in a sample, how could you relate the % values shown in the sim into a number you could use for your calculation of average atomic mass?

Explanation:

Step1: Recall atomic mass values

The atomic mass of lithium - 6 is 6.0151 amu and lithium - 7 is 7.0160 amu.

Step2: Calculate average atomic mass for the sample

The formula for average atomic mass of a sample with $n_1$ atoms of mass $m_1$ and $n_2$ atoms of mass $m_2$ is $\text{Average atomic mass}=\frac{n_1m_1 + n_2m_2}{n_1 + n_2}$. Here, $n_1 = 3$ (lithium - 6 atoms), $m_1=6.0151$ amu, $n_2 = 2$ (lithium - 7 atoms), $m_2 = 7.0160$ amu. So, $\text{Average atomic mass}=\frac{3\times6.0151+2\times7.0160}{3 + 2}=\frac{18.0453+14.032}{5}=\frac{32.0773}{5}=6.41546$ amu.

Step3: Analyze which mass it is closer to

The difference between the average atomic mass (6.41546 amu) and the mass of lithium - 6 (6.0151 amu) is $|6.41546 - 6.0151|=0.40036$ amu. The difference between the average atomic mass and the mass of lithium - 7 (7.0160 amu) is $|6.41546 - 7.0160| = 0.60054$ amu. Since $0.40036<0.60054$, the average atomic mass is closer to the mass of lithium - 6.

Step4: Describe the general method

The general method to calculate the average atomic mass of a mixture of isotopes is to use the formula $\text{Average atomic mass}=\sum_{i = 1}^{n}x_im_i$, where $x_i$ is the relative abundance (fraction) of the $i$-th isotope and $m_i$ is the atomic mass of the $i$-th isotope. In the case of discrete numbers of atoms like in part c, it's $\text{Average atomic mass}=\frac{\sum_{i = 1}^{n}n_im_i}{\sum_{i = 1}^{n}n_i}$, where $n_i$ is the number of atoms of the $i$-th isotope and $m_i$ is the atomic mass of the $i$-th isotope.

Answer:

a. 6.0151 amu
b. 7.0160 amu
c. 6.41546 amu
d. It is closer to the mass of lithium - 6 because the difference between the average atomic mass (6.41546 amu) and the mass of lithium - 6 (6.0151 amu) is 0.40036 amu, while the difference between the average atomic mass and the mass of lithium - 7 (7.0160 amu) is 0.60054 amu, and 0.40036 < 0.60054.

1. The general method to calculate the average atomic mass of a mixture of isotopes is $\text{Average atomic mass}=\sum_{i = 1}^{n}x_im_i$, where $x_i$ is the relative abundance (fraction) of the $i$-th isotope and $m_i$ is the atomic mass of the $i$-th isotope. For a sample with discrete numbers of atoms of each isotope, $\text{Average atomic mass}=\frac{\sum_{i = 1}^{n}n_im_i}{\sum_{i = 1}^{n}n_i}$, where $n_i$ is the number of atoms of the $i$-th isotope and $m_i$ is the atomic mass of the $i$-th isotope.