Question

1. write the atomic symbol (symbol notation) for the two isotopes of uranium(u), whose atomic number is 92. one isotope has 142 neutrons, and the other isotope has 146 neutrons.
2. calculate the average atomic mass of the element iron(fe) using the following data: isotope % abundance iron - 54 6% iron - 56 92% iron - 57 2%
3. calculate the average atomic mass of the element nitrogen(n) using the following data: isotope % abundance nitrogen - 14 95% nitrogen - 15 3% nitrogen - 16 2%
4. calculate the average atomic mass of the element iodine(i) using the following data: isotope % abundance iodine - 127 80% iodine - 126 17% iodine - 128 3%

Explanation:

Step1: Recall atomic symbol notation

The atomic symbol notation is $_{Z}^{A}X$, where $Z$ is the atomic number, $A$ is the mass - number ($A =$ number of protons+number of neutrons), and $X$ is the chemical symbol. Uranium has an atomic number $Z = 92$.

Step2: Calculate mass - number for the first uranium isotope

For the first uranium isotope with 142 neutrons, the mass - number $A_1=92 + 142=234$. So the atomic symbol is $_{92}^{234}U$.

Step3: Calculate mass - number for the second uranium isotope

For the second uranium isotope with 146 neutrons, the mass - number $A_2=92+146 = 238$. So the atomic symbol is $_{92}^{238}U$.

Step4: Recall formula for average atomic mass

The formula for the average atomic mass $M$ of an element with isotopes $X_1,X_2,\cdots,X_n$ having masses $m_1,m_2,\cdots,m_n$ and percent abundances $p_1,p_2,\cdots,p_n$ is $M=\sum_{i = 1}^{n}m_i\times\frac{p_i}{100}$.

Step5: Calculate average atomic mass of iron

For iron:
\[

$$\begin{align*} M_{Fe}&=54\times\frac{6}{100}+56\times\frac{92}{100}+57\times\frac{2}{100}\\ &=\frac{54\times6 + 56\times92+57\times2}{100}\\ &=\frac{324+5152 + 114}{100}\\ &=\frac{5590}{100}\\ &=55.9 \end{align*}$$

\]

Step6: Calculate average atomic mass of nitrogen

For nitrogen:
\[

$$\begin{align*} M_N&=14\times\frac{95}{100}+15\times\frac{3}{100}+16\times\frac{2}{100}\\ &=\frac{14\times95+15\times3 + 16\times2}{100}\\ &=\frac{1330+45+32}{100}\\ &=\frac{1407}{100}\\ &=14.07 \end{align*}$$

\]

Step7: Calculate average atomic mass of iodine

For iodine:
\[

$$\begin{align*} M_I&=127\times\frac{80}{100}+126\times\frac{17}{100}+128\times\frac{3}{100}\\ &=\frac{127\times80+126\times17 + 128\times3}{100}\\ &=\frac{10160+2142+384}{100}\\ &=\frac{12686}{100}\\ &=126.86 \end{align*}$$

\]

Answer:

1. First uranium isotope: $_{92}^{234}U$
2. Second uranium isotope: $_{92}^{238}U$
3. Average atomic mass of iron: $55.9$
4. Average atomic mass of nitrogen: $14.07$
5. Average atomic mass of iodine: $126.86$