Chemistry · Physics

Atomic Structure and Mass

308 Questions

Atomic structure and mass focus on the composition of atoms, including protons, neutrons, electrons, and isotopes. This topic is essential for the chemistry sections in engineering and civil services competitive exams. Review these questions to practice calculating atomic mass, neutron count, and isotopic distributions.

Atomic mass unit calculationsNeutron and proton countsIsotopic distribution averagesLaw of triadsRest energy of atoms

Atomic Structure and Mass Questions

Multiple choice chemistry the language of chemistry atomic weight and molecular weight masses of atoms and molecules chemical analysis and formulae

Among the following quantities:
(i) mass number
(ii) average mass of a Carbon atom in amu
(iii) the charge of nucleus in amu and
(iv) mass of a Carbon -12 atom in grams

  1. only (i) is whole number

  2. only (i) and (iii) are whole number

  3. only (i) and (iv) are whole numbers

  4. only (iii) and (iv) are whole numbers
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

(1) The mass number is always the whole number because the mass number is the total number of neutrons and protons in the nucleus of the atom and an atom never contains fractional or an integer number of protons or neutrons.

(2) Average mass of carbon atom in $amu$ is $12.011u$. It is in decimal and can be written as fractional form.
(3) Charge on the nucleus in $amu$:- Charge on the nucleus is equal to the number of protons and the number of protons are always whole.
(4) Mass of carbon atom= $\cfrac {Mass\quad of\quad carbon}{Avogadro\quad number}$
$=\cfrac {12g}{6.022\times 10^{23}}$
$=1.99\times 10^{-23}$ grams
It is not a whole number.

Multiple choice chemistry the language of chemistry atomic weight and molecular weight masses of atoms and molecules chemical analysis and formulae

Which of the following is/are correct?

  1. One atomic mass unit is a mass unit equal to exactly one -twelfth (1/12th) the mass of one atom of carbon-12.

  2. One atomic mass unit is a mass unit equal to exactly one - sixteenth (1/16th) the mass of one atom of oxygen-16.

  3. The relative atomic mass of the atom of an element is defined as the average mass of the atom, as compared to 1/12h the mass of one carbon-12 atom.

  4. None of the above.
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

The following are correct. One atomic mass unit is a mass unit equal to exactly one -twelfth (1/12th) the mass of one atom of carbon-12.
One atomic mass unit is a mass unit equal to exactly one - sixteenth (1/16th) the mass of one atom of oxygen-16.
The relative atomic mass of the atom of an element is defined as the average mass of the atom, as compared to 1/12h the mass of one carbon-12 atom.
For example, the mass of one carbon-12 atom is 12 amu. The mass of one magnesium-24 atom is 24 amu. The mass of one calcium-40 atom is 140 amu.

Multiple choice chemistry the language of chemistry atomic weight and molecular weight masses of atoms and molecules chemical analysis and formulae

Oxygen occurs in nature as a mixture of isotopes $^16O$, $^17O$ and $^18O$ having atomic masses of 15.995 u, 16.999 u and 17.999 u and relative abundance of 99.763%, 0.037%, and 0.200% respectively. What is the average atomic mass of oxygen?

  1. 15.999 u

  2. 16.999 u

  3. 17.999 u

  4. 18.999 u
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
 Average atomic mass of an element existing in different isotopes is given by:
$M _{ avg }=\dfrac { \sum _{ i=1 }^{ n }{ { M } _{ i }{ A } _{ i } }  }{ \sum _{ i=1 }^{ n }{ A _{ i } }  } $
where $M _i=$atomic mass of an isotope with relative abundance of $A _i$
Given:$M _1=15.995 u,A _1=99.763,M _2=16.999 u, A _2=0.037,M _3= 17.999 u, A _3=0.200$
on subtitutiing we get:
${ M } _{ avg }=\dfrac { 15.995\times 99.763+16.999\times 0.037+17.999\times 0.200 }{ 99.763+0.037+0.200 } $
${ M } _{ avg }=15.999\ u$
option A is correct
Multiple choice chemistry the language of chemistry atomic weight and molecular weight masses of atoms and molecules chemical analysis and formulae

For every ,one $^{37}Cl$ isotope there are three $^{35}Cl$ isotopes, in a sample of chlorine. What will be the average atomic mass of chlorine?

  1. 35

  2. 37

  3. 35.5

  4. 35.6
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Average atomic mass of an element existing in different isotopes is given by:
$M _{ avg }=\dfrac { \sum _{ i=1 }^{ n }{ { M } _{ i }{ A } _{ i } }  }{ \sum _{ i=1 }^{ n }{ A _{ i } }  } $
where $M _i=$atomic mass of an isotope with relative abundance of $A _i$
Given:$M _1=37 u,A _1=1,M _2=35 u, A _2=3$
on subtitutiing we get:
${ M } _{ avg }=\dfrac { 37\times 1+35\times 3 }{ 1+3 } $
${ M } _{ avg }=35.5\ u$
option C is correct
Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

There are two isotopes of an element with atomic mass $z$. Heavier one has atomic mass $z+2$ and lighter one has $z-1$, then an abundance of lighter one is:

  1. $66.6\%$
  2. $96.7\%$
  3. $6.67\%$
  4. $33.3\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Abundance of an isotope can be calculated as :


Let abundance of isotope with atomic mass $Z+2$ is x
Then abundance of isotope $Z-1$ is $1-x$

$(Z+2)x+(Z-1)(1-x)\quad =Z$

On solving this, we get
$x=\dfrac { 1 }{ 3 } $
Therefore,% abundance of isotope with atomic no.$Z+2$ is =$=\dfrac { 1 }{ 3 } \times 100=33.33$%
Therefore,% abundance isotope with atomic no. $Z-1$ is =$100-33.33=67.66%$%
Therefore isotope with atomic $Z-1$ is higher in % abundance

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

$ _{ 17 }^{ 35 }{ Cl }$ and $ _{ 17 }^{ 37 }{ Cl }$ are two isotopes of chlorine. If average atomic mass is $35.5$ then ratio of these two isotopes is:

  1. $35 : 37$
  2. $1 : 3$
  3. $3 : 1$
  4. $2 : 1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solution:- (C) $3 : 1$

Let $x$ and $y$ be the fraction of ${ _{17}^{35}{Cl}}$ and ${ _{17}^{37}{Cl}}$ in ${ _{17}^{35.5}{Cl}}$.
Therefore,
Average at. mass $= \cfrac{x \times 35 + y \times 37}{x + y}$
$35.5 = \cfrac{35x + 37y}{x+y}$
$\Rightarrow 35.5 x + 35.5 y = 35x + 37y$
$\Rightarrow 35.5 x - 35x = 37 y - 35.5 y$
$\Rightarrow 0.5 x = 1.5 y$
$\Rightarrow \cfrac{x}{y} = \cfrac{1.5}{0.5} = \cfrac{3}{1}$
Hence the ratio of ${ _{17}^{35}{Cl}}$ and ${ _{17}^{37}{Cl}}$ in ${ _{17}^{35.5}{Cl}}$ is $3 : 1$.

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

Chlorine has two naturally occurring isotopes, $^{35}Cl$ and $^{37}Cl$. If the atomic mass of Cl is 35.5 the ratio of natural abundance of $^{35}Cl$ and $^{37}Cl$ is closest to :

  1. $ 3 : 5 $
  2. $3 : 1$
  3. $2 : 5 $
  4. $4 : 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let x be the fraction of Cl-35. Then 35x + 37(1-x) = 35.5. 35x + 37 - 37x = 35.5. -2x = -1.5. x = 0.75. The ratio of Cl-35 to Cl-37 is 0.75 / 0.25 = 3:1.

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

If an element $z$ having atomic weight $x$ exists in two isotopes of mass number $(x-1)$ and $(x+2)$ then, the percentage abundance of heavier isotopes is:    

  1. $25\%$
  2. $66.6\%$
  3. $33.3\%$
  4. $20\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the percentage of the heavier isotope (mass x + 2) be p, and the lighter isotope (mass x - 1) be 100 - p. The average atomic weight is given by ((x - 1)(100 - p) + (x + 2)p) / 100 = x. Solving this equation: 100x - 100 + (x - 1)(-p) + xp + 2p = 100x, which simplifies to -100 + p + 3p = 0, giving 3p = 100, so p = 33.3 percent.

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

The average atomic mass of copper is $63.546$ amu .Natural copper consists of two iostopes: $^{63} Cu$ and $^{65} Cu$.Their natural abundances are $69.09\%$ and $30.91\%$ respectively. If the mass of $^{63} Cu$ isotope is $62.9298$ amu ,What is the mass of $^{65} Cu$ isotope?

  1. 64.9000

  2. 65.1233

  3. 64.9233

  4. 65.1933
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Average mass = (mass1 * abundance1) + (mass2 * abundance2). 63.546 = (62.9298 * 0.6909) + (mass2 * 0.3091). 63.546 = 43.4779 + 0.3091 * mass2. 20.0681 = 0.3091 * mass2. mass2 = 64.9243.

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

 Rhenium (Re) consists of $37.1$% $185$ Re and $62.9$% $187$ Re. Calculate the relative atomic mass?

  1. $185.6$
  2. $185.9$
  3. $186.3$
  4. $186.1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Relative atomic mass is calculated as (abundance1 * mass1 + abundance2 * mass2) / 100. For Rhenium: (37.1 * 185 + 62.9 * 187) / 100 = (6863.5 + 11762.3) / 100 = 18625.8 / 100 = 186.258, which rounds to 186.3.

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

The relative atomic mass of an atom is:

  1. measured in atomic mass units (u)

  2. based on the mass of 1 atom of carbon-12

  3. different for different isotopes of an element

  4. all of the above are true
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Relative atomic mass is the mass of an atom measured relative to 1/12th the mass of 1 atom of C-12 isotope which is also known as atomic mass unit or amu(u).