The Mole Concept with Examples

Atomic Mass Unit with Examples

Since atoms are too small particles we can not measure their weights with normal methods. Thus, scientist find another way to measure mass of atoms, molecules and compounds. They approve one atom of carbon isotopes 6C12 as 12 atomic mass unit. Mass of every elements expressed in terms of atomic mass unit is called relative atomic mass. We can also calculate relative molecule mass with same method; adding individual atomic masses of elements gives us relative molecular mass. For example;

1 H atom is 1 amu (amu=atomic mass unit)

1 Ca atom is 40 amu

1 Mg atom is 24 amu

1 H2O molecule includes 2 hydrogen atoms and one oxygen atom;

(2.1)+(16)=18 amu

Example: Which one of the following molecules has greatest relative molecular mass.

I. CO

II. SO2

III. Fe2(SO4)3

IV. CaCO3

Solution:

I. One CO molecule includes one C atom and one O atom

Molecular mass of CO=(1.12)+(1.16)=28 amu

II. One mole SO2 includes one S atom and two O atoms.

Molecular mass of SO2=(1.32)+(2.16)=64 amu

III. One mole Fe2(SO4)3 includes 2 Fe atoms 3 S atoms and 12 O atoms.

Molecular mass of Fe2(SO4)3=(2.56)+(3.32)+(12.16)=400 amu

IV. One mole CaCO3 molecule includes one Ca atom, one C atom and 3 O atoms.

Molecular mass of CaCO3=(1.40)+(1.12)+(3.16)=100 amu

Thus; Fe2(SO4)3 has greater molecular mass

Including C, most of the elements have isotopes. We must consider atomic masses of all isotopes while writing it in periodic table. Example given below shows how to calculate average atomic mass of elements having isotopes.

MassX=M(X1).%X1/100+M(X2).%X1/100+…

where; MassX is the average mass of X element

M(X1) and M(X2) are masses of isotopes

%X1 and %X2 are Percentages of atomic masses of X element in nature.

Example: Relative atomic mass of one element is 44,1 amu and it has two isotopes.If one of the isotopes has atomic mass 42 amu and percentage of it is 30%, find the atomic mass of second isotope.

Solution:

If one of the isotopes has 30% of atomic mass, other isotope has 70% of atomic mass.

MassX=M(X1).%X1/100+M(X2).%X1/100+…

44,1=42.30/100+M(X2).70/100

M(X2)=45 amu

The Mole Concept and Avogadro’s Number

A concept used for measure amount of particles like atoms, molecules. Number of atoms in the 6C12 element is equal to 1 mole. Number of particles in 1 mole is called Avogadro’s number; 6,02.1023.

1 mole atom contains 6,02x1023 atoms

1 mole molecule contains 6,02x1023 molecules

1 mole ion contains 6,02x1023 ions

Mole=Number of Particles/Avogadro’s Number

Example: Which ones of the following statements are true for 2 moles CO2 compound.

I. Contains 1,204x1023 CO2 molecules

II. Contains 2 mole C atom

III. Contains 3,612x1024 atom

(Avogadro’s Number=6,02x1023)

Solution:

I. 1 mole CO2 contains 6,02x1023CO2 molecules

2 mole CO2 X CO2 molecules

__________________________________________

X=1,204x1023 CO2 molecules

I is true

II. 1 mole CO2 contains 1 mole C atom

2 mole CO2 contains Y mole C atom

_______________________________________

Y=2 mole C atoms

II is true

III. 1 mole CO2 contains 3.6,02x1023 atom

2 mole CO2contains                Z atom

______________________________________

Z=3,612x1024atoms

III is also true

Example: Find the mole of molecule including 1,204x1023 NH3.

Solution:

1 mole NH3 contains             6,02x1023 molecule

X mole NH3 contains             1,204x1023molecule

___________________________________________

x=0,2 mole NH3 molecule

We can solve this problem using formula given above;

Mole=Number of Particles/Avogadro’s Number

Mole=1,204x1023/ 6,02x1023=0,2 mole

The Mole Concept Exams and Problem Solutions