Question

atomic mass (route)

1. example: the natural distribution of the isotope (percent abundance) of lutetium (lu) is 97.41% ^175lu at a mass of 174.9408 amu and the rest is ^176lu at a mass of 175.9428 amu. calculate the atomic mass of lutetium.

0.9741
100% - 97.41% = 2.59%
0.9741×170.94 = 166.49?
0.0259×174.91 = 4.535 171.02 amu

1. example: chlorine has two natural isotopes 75.78% is chlorine - 35 at 34.96885272 u and 24.22% is chlorine - 37 at 36.96590262 u. what is the atomic mass of chlorine?

(1) 24.22% by 100% = 0.2422×36.96590262 u=
(2) 75.78% by 100% = 0.7578×34.96885272 u=
(1) 8.9531414 → 8.953
(2) 26.499396 → 26.50
8.953 + 26.50 = 35.45 amu

1. example 3: the percent abundances of isotopes of potassium in nature are as follows:

(a) 93.2581% of k - 39 weighing 38.9637074 amu
(b) 0.0117% of k - 40 weighing 39.9639992 amu
(c) 6.7302% of k - 41 weighing 40.9618254 amu
(1) 93.2581% by 100 = 0.932581×38.9637074 = 36.3393
(2) 0.0117% by 100 = 0.000117×39.9639992 = 0.00457
(3) 6.7302% by 100 = 0.067302×40.9618254 = 2.7568
36.3393+0.00457 + 2.7568 = 39.10067 amu

Explanation:

Step1: Recall atomic - mass formula

The atomic mass of an element is calculated as the sum of the products of the relative abundance of each isotope and its mass. The formula is $M=\sum_{i = 1}^{n}x_im_i$, where $x_i$ is the relative abundance of the $i$ - th isotope and $m_i$ is the mass of the $i$ - th isotope.

Step2: Solve for lutetium

For lutetium:
The relative abundance of $^{175}Lu$ is $x_1 = 0.9741$ and its mass $m_1=174.9408\ amu$. The relative abundance of $^{176}Lu$ is $x_2=1 - 0.9741=0.0259$ and its mass $m_2 = 175.9428\ amu$.
The atomic mass $M_1=x_1m_1 + x_2m_2=0.9741\times174.9408+0.0259\times175.9428$
$=170.797+4.557 = 175.904\ amu$

Step3: Solve for chlorine

For chlorine:
The relative abundance of $^{35}Cl$ is $x_1 = 0.7578$ and its mass $m_1 = 34.96885272\ amu$. The relative abundance of $^{37}Cl$ is $x_2=0.2422$ and its mass $m_2 = 36.96590262\ amu$.
The atomic mass $M_2=x_1m_1+x_2m_2=0.7578\times34.96885272+0.2422\times36.96590262$
$=26.499 + 8.953=35.452\ amu$

Step4: Solve for potassium

For potassium:
The relative abundance of $^{39}K$ is $x_1 = 0.932581$, mass $m_1 = 38.9637074\ amu$; relative abundance of $^{40}K$ is $x_2=0.000117$, mass $m_2 = 39.9639992\ amu$; relative abundance of $^{41}K$ is $x_3=0.067302$, mass $m_3 = 40.9618254\ amu$.
The atomic mass $M_3=x_1m_1+x_2m_2+x_3m_3$
$=0.932581\times38.9637074+0.000117\times39.9639992+0.067302\times40.9618254$
$=36.3393+0.0046+2.7568 = 39.099\ amu$

Answer:

For lutetium, the atomic mass is approximately $175.90\ amu$. For chlorine, the atomic mass is approximately $35.45\ amu$. For potassium, the atomic mass is approximately $39.10\ amu$.