Question

now predict how many atoms of lithium - 7 and lithium - 6 are needed to reach the average atomic mass from the periodic table (6.944).

1. for 1 atom of lithium - 6, i will need ______ atoms of lithium - 7 for an average atomic mass of 6.944

switch to the slider button instead of the bowl so that you can select more than 10 atoms to verify your prediction. set lithium - 6 to 1. then move the slider until the average is close to 6.944. record what the actual number needed.

1. for 1 atom of lithium - 6, i needed ______ atoms of lithium - 7 for an average atomic mass of 6.944.

super challenge

1. chlorine has two possible isotopes. you must use at least one atom of each isotope. what combination of the number of atoms and isotopes will come closest to the actual average atomic mass of 35.45? show your combination below. what does this mean about the percent abundance of these isotopes? (remember, the 2 percent values must add up to 100%)

cl - 35 ____ atoms ____ %
cl - 37 ____ atoms ____ %

Explanation:

Step1: Set up the average - atomic - mass formula for lithium

The formula for average atomic mass is $\text{Average Atomic Mass}=\frac{m_1n_1 + m_2n_2}{n_1 + n_2}$, where $m_1$ and $m_2$ are the masses of the isotopes and $n_1$ and $n_2$ are the number of atoms of the isotopes. For lithium, $m_1 = 6$ (mass of Lithium - 6), $m_2=7$ (mass of Lithium - 7), and the average atomic mass is 6.944. Let the number of Lithium - 6 atoms $n_1 = 1$ and the number of Lithium - 7 atoms be $x$. Then $6.944=\frac{6\times1 + 7\times x}{1 + x}$.

Step2: Cross - multiply the equation

$6.944(1 + x)=6 + 7x$. Expand the left - hand side: $6.944+6.944x=6 + 7x$.

Step3: Rearrange the equation to solve for $x$

Subtract $6.944x$ from both sides: $6.944=6 + 7x-6.944x$. Combine like terms: $6.944=6 + 0.056x$. Then subtract 6 from both sides: $0.944 = 0.056x$.

Step4: Solve for $x$

$x=\frac{0.944}{0.056}\approx16.86$. Rounding to the nearest whole number, we get $x = 17$.

For the chlorine problem:
Let the number of Cl - 35 atoms be $a$ and the number of Cl - 37 atoms be $b$. The average atomic mass formula is $35.45=\frac{35a + 37b}{a + b}$. Cross - multiply: $35.45(a + b)=35a + 37b$. Expand: $35.45a+35.45b=35a + 37b$. Rearrange: $35.45a-35a=37b - 35.45b$. $0.45a = 1.55b$. $\frac{a}{b}=\frac{1.55}{0.45}=\frac{31}{9}$.
The percent abundance of Cl - 35 is $\frac{31}{31 + 9}\times100\%=\frac{31}{40}\times100\% = 77.5\%$ and the percent abundance of Cl - 37 is $\frac{9}{31+9}\times100\%=\frac{9}{40}\times100\% = 22.5\%$. The number of Cl - 35 atoms can be 31 and the number of Cl - 37 atoms can be 9.

Answer:

1. 17
2. 17
3. Cl - 35: 31 atoms, 77.5%

Cl - 37: 9 atoms, 22.5%