Question

complete the following table for isotopes.
symbol # protons # neutrons atomic number mass number
28 32
21 40
$_{29}^{63}cu$

1. a.) uranium has two isotopes that can be found in nature; u - 235 (235.044 amu) and u - 238 (238.051 amu). calculate the percentage of naturally - occurring uranium that is u - 235.

b.) state the number of protons, neutrons, and electrons in u - 235, and give the symbol in standard isotope notation.
c.) in order to use uranium as a fuel in a nuclear reactor or nuclear weapon, it must be enriched, which means increasing the percentage of u - 235 in the sample. in commercial nuclear reactors, uranium is enriched to 3 - 5% u - 235. if a sample of nuclear fuel is enriched to 4.0% u - 235, what is the atomic mass of the sample of enriched uranium?

1. hypothetical element, \hy\, is comprised of two isotopes. one isotope is measured and found to have a natural abundance of 25.55% and a mass of 43.0025 amu. calculate the mass of the other isotope, given that the atomic mass of hy is 44.9005 amu.
2. silicon has three stable isotopes. the major isotope is si - 28, with a mass of 27.9769 amu and percent abundance of 92.223%. the other two isotopes are si - 29 (28.9765 amu) and si - 30 (29.9738 amu). what are their percent abundances?

Explanation:

4.

Step1: First row

The atomic number is equal to the number of protons. So atomic number $Z = 28$. The mass number $A$ is the sum of protons and neutrons, $A=28 + 32=60$. The element with atomic number 28 is nickel, symbol $Ni$. So the isotope is $^{60}_{28}Ni$.

Step2: Second row

The mass number $A = 40$ and number of neutrons $n = 21$. The number of protons $p=A - n=40 - 21 = 19$. The element with atomic number 19 is potassium, symbol $K$. So the isotope is $^{40}_{19}K$.

Step3: Third row

For $^{63}_{29}Cu$, the number of protons $p = 29$, the number of neutrons $n=63 - 29 = 34$, the atomic number $Z = 29$ and the mass number $A = 63$.

5. a.

Step1: Set up equation

Use the average - atomic - mass formula based on the abundances of U - 235 and U - 238.

Step2: Expand and simplify

Expand the equation and combine like terms to solve for $x$ (the percentage of U - 235).

5. b.

Step1: Determine protons

The atomic number gives the number of protons for uranium.

Step2: Determine neutrons

Subtract the atomic number from the mass number to get the number of neutrons.

Step3: Determine electrons

For a neutral atom, the number of electrons equals the number of protons.

5. c.

Step1: Use weighted - average formula

Multiply the mass of each isotope by its percentage (in decimal form) and sum them up.

6.

Step1: Set up equation

Use the average - atomic - mass formula with the known abundance and mass of one isotope and the unknown mass of the other isotope.

Step2: Solve for $m$

Isolate the variable $m$ to find the mass of the other isotope.

7.

Step1: Set up relationships

Set up the relationship between the abundances of Si - 29 and Si - 30 and use the average - atomic - mass formula.

Step2: Substitute and solve

Substitute $y = 0.07777 - x$ into the average - atomic - mass equation and solve for $x$ (abundance of Si - 29), then find $y$ (abundance of Si - 30).

Answer:

4.

Symbol	# Protons	# Neutrons	Atomic Number	Mass Number
$^{40}_{19}K$	19	21	19	40
$^{63}_{29}Cu$	29	34	29	63

5. a.

Let the percentage of U - 235 be $x$, then the percentage of U - 238 is $(1 - x)$. The average atomic mass of uranium is approximately 238.0289 amu.
We use the formula for average atomic mass: $235.044x+238.051(1 - x)=238.0289$
$235.044x+238.051 - 238.051x=238.0289$
$- 3.007x=238.0289 - 238.051$
$-3.007x=- 0.0221$
$x=\frac{0.0221}{3.007}\approx0.74\%$

5. b.

For U - 235, the atomic number of uranium is 92. So the number of protons $p = 92$, the number of neutrons $n=235 - 92 = 143$, and for a neutral atom, the number of electrons $e = 92$. The symbol in standard isotope notation is $^{235}_{92}U$.

5. c.

If the sample is 4.0% U - 235 and 96.0% U - 238, the average atomic mass of the enriched uranium is:
$0.04\times235.044+0.96\times238.051$
$=9.40176+228.52896$
$=237.93072$ amu

6.

Let the mass of the other isotope be $m$. The natural abundance of the first isotope is $25.55\%=0.2555$, so the natural abundance of the second isotope is $1 - 0.2555 = 0.7445$.
Using the formula for average atomic mass: $43.0025\times0.2555+m\times0.7445 = 44.9005$
$10.98713875+0.7445m=44.9005$
$0.7445m=44.9005 - 10.98713875$
$0.7445m=33.91336125$
$m=\frac{33.91336125}{0.7445}\approx45.55$ amu

7.

Let the percent abundance of Si - 29 be $x$ and the percent abundance of Si - 30 be $y$.
We know that $x + y=100 - 92.223=7.777\%$ or $x + y = 0.07777$ in decimal form.
The average atomic mass of silicon is 28.0855 amu. Using the formula for average atomic mass:
$27.9769\times0.92223+28.9765x + 29.9738y=28.0855$
Substitute $y = 0.07777 - x$ into the above - equation:
$27.9769\times0.92223+28.9765x+29.9738(0.07777 - x)=28.0855$
$25.8069+28.9765x+2.3341 - 29.9738x=28.0855$
$28.141 - 0.9973x=28.0855$
$-0.9973x=28.0855 - 28.141$
$-0.9973x=-0.0555$
$x=\frac{0.0555}{0.9973}\approx0.0557 = 5.57\%$
$y=7.777 - 5.57 = 2.207\approx2.21\%$