Question

1. calculate the atomic mass of the unknown element. then identify the element.

isotope mass (amu) percent abundance operations element:
¹⁸⁴x 184.953 37.40%
¹⁸⁶x 186.956 62.60%

1. calculate the atomic mass of the unknown element. then identify the element.

isotope mass (amu) percent abundance operations element:
¹¹²x 112.904 4.30%
¹¹⁴x 114.904 95.70%
show all calculation setups. adjust for significant figures and give units.

• protium is the most prevalent hydrogen isotope, with an abundance of 99.98%. it consists of one proton and one electron.
• deuterium is a hydrogen isotope consisting of one proton, one neutron and one electron.
• tritium is a hydrogen isotope consisting of one proton, two neutrons and one electron.

1. the term \average atomic mass\ is a average and so is calculated differently from a

ormal\ average.

1. the element copper has naturally - occurred isotopes with mass numbers of 63 and 65.

the relative abundances and atomic masses are 69.2% for a mass of 62.93 amu and 30.8% for a mass of 64.93 amu.

Explanation:

Step1: Recall atomic mass formula

The average atomic mass ($A_{avg}$) of an element with isotopes is calculated using the formula $A_{avg}=\sum_{i = 1}^{n}(m_i\times p_i)$, where $m_i$ is the mass of the $i$-th isotope and $p_i$ is the percent - abundance of the $i$-th isotope (expressed as a decimal).

Step2: Calculate atomic mass for problem 6

For the first isotope of problem 6 with mass $m_1 = 184.953$ amu and percent - abundance $p_1=0.3740$, and the second isotope with mass $m_2 = 186.956$ amu and percent - abundance $p_2 = 0.6260$.
$A_{avg}=(184.953\times0.3740)+(186.956\times0.6260)$
$A_{avg}=184.953\times0.3740 = 69.172422$
$A_{avg}=186.956\times0.6260=117.034456$
$A_{avg}=69.172422 + 117.034456=186.206878\approx186.21$ amu. Looking up this atomic mass in the periodic table, the element is Rhenium (Re).

Step3: Calculate atomic mass for problem 7

For the first isotope of problem 7 with mass $m_1 = 112.904$ amu and percent - abundance $p_1 = 0.0430$, and the second isotope with mass $m_2=114.904$ amu and percent - abundance $p_2 = 0.9570$.
$A_{avg}=(112.904\times0.0430)+(114.904\times0.9570)$
$A_{avg}=112.904\times0.0430 = 4.854872$
$A_{avg}=114.904\times0.9570 = 110.953128$
$A_{avg}=4.854872+110.953128 = 115.808\approx115.81$ amu. Looking up this atomic mass in the periodic table, the element is Indium (In).

Answer:

Problem 6: Atomic mass is approximately 186.21 amu, element is Rhenium (Re).
Problem 7: Atomic mass is approximately 115.81 amu, element is Indium (In).