Question

calculate the atomic mass of each element described below. then use the periodic table to identify each element.
32.

isotope	mass (amu)	percent abundance
⁶⁵x	64.928	30.83

33.

isotope	mass (amu)	percent abundance
³⁷x	36.966	24.23

Explanation:

Step1: Recall atomic - mass formula

The atomic mass ($A$) of an element with two isotopes is given by $A = m_1\times p_1 + m_2\times p_2$, where $m_1$ and $m_2$ are the masses of the isotopes and $p_1$ and $p_2$ are their percent - abundances (expressed as decimals).

Step2: Calculate atomic mass for problem 32

For the first set of isotopes in problem 32:
$p_1=0.6917$, $m_1 = 62.930$ amu, $p_2 = 0.3083$, $m_2=64.928$ amu.
$A_1=62.930\times0.6917+64.928\times0.3083$
$A_1 = 62.930\times0.6917=43.527681$
$A_1 = 64.928\times0.3083 = 19.9172024$
$A_1=43.527681 + 19.9172024=63.4448834\approx63.44$ amu. Looking at the periodic table, the element with an atomic mass close to 63.44 amu is Copper (Cu).

Step3: Calculate atomic mass for problem 33

For the second set of isotopes in problem 33:
$p_1 = 0.7577$, $m_1=34.969$ amu, $p_2=0.2423$, $m_2 = 36.966$ amu.
$A_2=34.969\times0.7577+36.966\times0.2423$
$A_2=34.969\times0.7577 = 26.4960113$
$A_2=36.966\times0.2423=8.9578618$
$A_2=26.4960113+8.9578618 = 35.4538731\approx35.45$ amu. Looking at the periodic table, the element with an atomic mass close to 35.45 amu is Chlorine (Cl).

Answer:

For problem 32: Atomic mass is approximately 63.44 amu, and the element is Copper (Cu).
For problem 33: Atomic mass is approximately 35.45 amu, and the element is Chlorine (Cl).