Question

part ii - isotopes and average atomic mass (unit 2)
elements occur as a mixture of isotopes, meaning they are made up of atoms with the same atomic number but different atomic masses. isotopes of an element differ in the number of neutrons. the average atomic mass is the weighted - average of all the isotopes of an element.
example: a sample of cesium is 75% ^133cs, 20% ^132cs and 5% ^134cs. what is its average atomic mass?
answer: 0.75×133 = 99.75
0.20×132 = 26.4
0.05×134 = 6.7
add: 132.85 amu = average atomic mass
(remember: to change a percent to a decimal, divide by 100, or move the decimal point 2 places to the left.)
directions: determine the average atomic mass of the following mixtures of isotopes. use 4 significant figures.
show all work!

1. what is the atomic mass of hafnium if, out of every 100 atoms, 5 have a mass of 176, 19 have a mass of 177, 27 have a mass of 178, 14 have a mass of 179, and 35 have a mass of 180.0?
2. calculate the average atomic mass of magnesium using the following data for three magnesium isotopes.

isotope mass (u) relative abundance
mg - 24 23.985 0.7870
mg - 25 24.986 0.1013
mg - 26 25.983 0.1117

1. calculate the average atomic mass of lithium, which occurs as two isotopes that have the following atomic masses and abundances in nature: 6.017 u, 7.30% and 7.018 u, 92.70%.
2. uranium is used in nuclear reactors and is a rare element on earth. uranium has three common isotopes. if the abundance of 234u is 0.01%, the abundance of 235u is 0.71%, and the abundance of 238u is 99.28%, what is the average atomic mass of uranium?
3. titanium has five common isotopes: ^46ti(8.0%), ^47ti(7.8%), ^48ti(73.4%), ^49ti(5.5%), ^50ti(5.3%). what is the average atomic mass of titanium?

Explanation:

Step1: Recall the formula for average atomic mass

The formula for average atomic mass $A_{avg}=\sum_{i} (m_i\times p_i)$, where $m_i$ is the mass of the $i$-th isotope and $p_i$ is the relative abundance of the $i$-th isotope.

Step2: Solve for the atomic mass of hafnium

The relative abundances of the isotopes are:

• For mass $m_1 = 176$: $p_1=\frac{5}{100}=0.05$
• For mass $m_2 = 177$: $p_2=\frac{19}{100}=0.19$
• For mass $m_3 = 178$: $p_3=\frac{27}{100}=0.27$
• For mass $m_4 = 179$: $p_4=\frac{14}{100}=0.14$
• For mass $m_5 = 180.0$: $p_5=\frac{35}{100}=0.35$

$A_{avg}=(176\times0.05)+(177\times0.19)+(178\times0.27)+(179\times0.14)+(180.0\times0.35)$
$=8.8 + 33.63+48.06 + 25.06+63$
$=178.55\approx178.6$ u

Step3: Solve for the average atomic mass of magnesium

$A_{avg}=(23.985\times0.7870)+(24.986\times0.1013)+(25.983\times0.1117)$
$=18.876295+2.5310818+2.9023011$
$=24.3096779\approx24.31$ u

Step4: Solve for the average atomic mass of lithium

$p_1 = 0.0730$, $m_1=6.017$ u, $p_2 = 0.9270$, $m_2 = 7.018$ u
$A_{avg}=(6.017\times0.0730)+(7.018\times0.9270)$
$=0.439241+6.495686$
$=6.934927\approx6.935$ u

Step5: Solve for the average atomic mass of uranium

$p_1 = 0.0001$, $m_1 = 234$ u, $p_2=0.0071$, $m_2 = 235$ u, $p_3 = 0.9928$, $m_3=238$ u
$A_{avg}=(234\times0.0001)+(235\times0.0071)+(238\times0.9928)$
$=0.0234+1.6685+236.2864$
$=237.9783\approx238.0$ u

Step6: Solve for the average atomic mass of titanium

Assume the masses of the isotopes are approximately equal to their mass - numbers (since not given otherwise).
$p_1 = 0.080$, $p_2 = 0.078$, $p_3 = 0.734$, $p_4 = 0.055$, $p_5 = 0.053$
$A_{avg}=(46\times0.080)+(47\times0.078)+(48\times0.734)+(49\times0.055)+(50\times0.053)$
$=3.68+3.666+35.232+2.695+2.65$
$=47.923\approx47.92$ u

Answer:

1. $178.6$ u
2. $24.31$ u
3. $6.935$ u
4. $238.0$ u
5. $47.92$ u