Question

name____ period__ date____ directions: solve each problem. circle your final answer or write your answer on the line provided. 1. what are the atomic number, mass number, and symbol for the element that has 45 protons and 58 neutrons? write the shorthand symbol for each neutral atom based on the characteristics listed. 2. 19 electrons and 20 neutrons 3. bromine atom with a mass number of 80 4. 20 protons and 22 neutrons 5. 146 neutrons with a mass number of 238 6. magnesium occurs in nature in three isotopic forms: ^24mg: 23.985 amu (78.70% abundance) ^25mg: 25.983 amu (11.17% abundance) ^26mg: 24.986 amu (10.13% abundance) calculate the atomic mass of mg from this data. follow the directions given for each section (7 & 8) 7. use complete sentences to answer each question. 1. boron has two isotopes, boron - 10 and boron - 11. which is more abundant given that the atomic mass of boron is 10.81 amu? justify your answer. 8. solve each problem. show all work in order to receive credit. circle your final answer. 3. element x has two naturally occurring isotopes. the isotope with a mass of 16.012 amu (^16x) has a relative abundance of 21.91 percent. the isotope with a mass of 18.009 amu (^18x) has a relative abundance of 78.09 percent. calculate the atomic mass of element x.

Explanation:

Step1: Recall atomic number and mass number concepts

Atomic number ($Z$) is equal to number of protons. Mass number ($A$) is sum of protons and neutrons.

Step2: For element with 45 protons and 58 neutrons

Atomic number $Z = 45$. Mass number $A=45 + 58=103$. The element with $Z = 45$ is rhodium, symbol $Rh$.

Step3: For atom with 19 electrons and 20 neutrons

In a neutral atom, number of electrons = number of protons. So $Z = 19$, $A=19 + 20 = 39$. The element is potassium, symbol $K$, shorthand symbol $^{39}_{19}K$.

Step4: For bromine atom with mass number 80

Bromine has atomic number $Z = 35$, shorthand symbol $^{80}_{35}Br$.

Step5: For atom with 20 protons and 22 neutrons

$Z = 20$, $A=20+22 = 42$. The element is calcium, symbol $Ca$, shorthand symbol $^{42}_{20}Ca$.

Step6: For atom with 146 neutrons and mass number 238

$Z=A -$ number of neutrons. So $Z = 238-146 = 92$. The element is uranium, symbol $U$, shorthand symbol $^{238}_{92}U$.

Step7: Calculate atomic mass of Mg

Atomic mass ($M$) of an element is the weighted - average of the masses of its isotopes.
$M=(23.985\times0.7870)+(25.983\times0.1170)+(24.986\times0.1013)$
$M = 23.985\times0.7870=18.876295$
$M_2=25.983\times0.1170 = 3.039011$
$M_3=24.986\times0.1013=2.531082$
$M=18.876295 + 3.039011+2.531082=24.446388\approx24.45$ amu

Step8: Determine more abundant boron isotope

Let the abundance of boron - 10 be $x$, then the abundance of boron - 11 is $1 - x$.
Atomic mass of boron $M = 10x+11(1 - x)=10.81$
$10x+11 - 11x=10.81$
$-x=10.81 - 11=-0.19$
$x = 0.19$ and $1 - x=0.81$. So boron - 11 is more abundant.

Step9: Calculate atomic mass of element X

$M=(16.012\times0.2191)+(18.009\times0.7809)$
$M_1=16.012\times0.2191 = 3.508239$
$M_2=18.009\times0.7809=14.062228$
$M=3.508239+14.062228 = 17.570467\approx17.57$ amu

Answer:

1. Atomic number: 45, Mass number: 103, Symbol: $Rh$
2. $^{39}_{19}K$
3. $^{80}_{35}Br$
4. $^{42}_{20}Ca$
5. $^{238}_{92}U$
6. 24.45 amu
7. Boron - 11 is more abundant. Justification: Let the abundance of boron - 10 be $x$, solving the equation $10x + 11(1 - x)=10.81$ gives $x = 0.19$ (abundance of boron - 10) and $1 - x = 0.81$ (abundance of boron - 11).
8. 17.57 amu