Mole Concept | Calculation of Moles
Calculation of Number of Moles of atoms
Atomic mass:
Atomic mass is defined as mass of one atom. Its unit is a.m.u or gram/mole. For example, if atomic mass of an atom is 'x' a.m.u or gram/mole, then it means that the mass of one mole of this atom is 'x' grams.
Example
Atomic mass of Ca = 40 a.m.u. It means mass of one mole Ca or 6.022 x 1023 Ca atoms is 40 grams.
Note
One mole is numerically equal to Avogadro constant or 6.022 x 1023.
- We can either use unitary method to calculate the number of moles of atoms or we can use the below formula:
\[Number\ of\ moles\ of\ atoms = {Mass \over Atomic\ mass}\]
Calculation of Number of Moles of molecules
Molecular mass:
Molecular mass is defined as mass of one molecule. Its unit is a.m.u or gram/mole. For example, if molecular mass of a molecule is 'x' a.m.u or gram/mole, then it means that the mass of one mole of this molecule is 'x' grams.
Example
Molecular mass of CaCO3 = 100 a.m.u. It means mass of one mole CaCO3 or 6.022 x 1023 CaCO3 molecules is 100 grams.
- We can either use unitary method to calculate the number of moles of molecules or we can use the below formula:
\[Number\ of\ moles\ of\ molecules = {Mass \over Molecular\ mass}\]
Questions
Calculate the number of moles in 10 grams of Ca. Atomic mass of Ca = 40 u.
Method 1 (Unitary Method)
\[1\ mole\ of\ Ca\ has\ mass = 40\ grams.\]
\[Number\ of\ moles\ in\ 40\ grams\ Ca = 1\]
\[Number\ of\ moles\ in\ 1\ gram\ Ca = {1 \over 40}\]
\[Number\ of\ moles\ in\ 10\ grams\ Ca = {1 \times 10 \over 40} = {0.25\ moles}\]
\[Number\ of\ moles\ of\ Ca\ atoms = {Mass \over Atomic\ mass}\]
\[Number\ of\ moles\ of\ Ca\ atoms = {10\ grams \over 40\ grams/mole}\]
\[Number\ of\ moles\ of\ Ca\ atoms = 0.25\ moles\]
For 5 gram of U238, find: (a) number of moles of U (b) Number of atoms of U.
(a) Number of moles of U can be calculated as:
\[238\ gram\ of\ U\ makes\ 1\ mole.\]
\[1\ gram\ of\ U\ makes\ {1 \over 238} mole.\]
\[5\ gram\ of\ U\ makes\ {1 \times 5 \over 238} moles = {5 \over 238} moles.\]
(b) Number of atoms of U can be calculated as:
\[238\ grams\ of\ U\ has\ 6.022 \times 10^{23} atoms.\]
\[1\ gram\ of\ U\ has\ {6.022 \times 10^{23} \over 238} atoms.\]
\[5\ gram\ of\ U\ has\ {6.022 \times 10^{23} \times 5 \over 238} = {1.27 \times 10^{22}} atoms.\]
For 10 grams of CaCO3, calculate:
(a) Number of moles of CaCO3.
(b) Number of molecules of CaCO3.
(c) Number of moles of Ca.
(d) Number of moles of C.
(e) Number of moles of O.
(f) Mass of Ca.
(g) Mass of C.
(h) Mass of O.
(a) Number of moles of CaCO3 is calculated as:
\[Number\ of\ moles\ of\ CaCO_3 = {Mass \over Molecular\ mass} = {10 \over 100} = 0.1\]
(b) Number of molecules of CaCO3 is calculated as:
\[Number\ of\ CaCO_3\ molecules = 0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22}\]
(c) Number of moles of Ca is calculated as:
\[ 1\ mole\ of\ CaCO_3\ contains\ 1\ mole\ Ca. \]
\[∴\ 0.1\ mole\ of\ CaCO_3\ contains\ 0.1\ mole\ Ca.\]
(d) Number of moles of C is calculated as:
\[ 1\ mole\ of\ CaCO_3\ contains\ 1\ mole\ C. \]
\[∴\ 0.1\ mole\ of\ CaCO_3\ contains\ 0.1\ mole\ C.\]
(e) Number of moles of O is calculated as:
\[ 1\ mole\ of\ CaCO_3\ contains\ 3\ moles\ O. \]
\[∴\ 0.1\ mole\ of\ CaCO_3\ contains\ 0.3\ mole\ O.\]
(f) Mass of Ca is calculated as:
\[Mass\ of\ Ca = moles \times Atomic\ mass = 0.1 \times 40 = 4\ grams\]
(g) Mass of C is calculated as:
\[Mass\ of\ C = moles \times Atomic\ mass = 0.1 \times 12 = 1.2\ grams\]
(h) Mass of O is calculated as:
\[Mass\ of\ O = moles \times Atomic\ mass = 0.3 \times 16 = 4.8\ grams\]
Calculate the mass of 1000 molecules of CaCO3.
\[Mass\ of\ 1\ CaCO_3\ molecule = 100\ u\]
\[Mass\ of\ 1000\ CaCO_3\ molecules = 1000 \times 100\ u\]
\[Mass\ of\ 1000\ CaCO_3\ molecules = 1000 \times 100 \times {1.67 \times 10^{-24}}\]
\[Mass\ of\ 1000\ CaCO_3\ molecules = 1.67 \times 10^{-19} grams\]
Calculate the mass of 1020 molecules of CaCO3.
\[Mass\ of\ 1\ CaCO_3\ molecule = 100\ u\]
\[Mass\ of\ 10^{20}\ CaCO_3\ molecules = 10^{20} \times 100\ u\]
\[Mass\ of\ 10^{20}\ CaCO_3\ molecules = 10^{20} \times 100 \times {1.67 \times 10^{-24}}\]
\[Mass\ of\ 10^{20}\ CaCO_3\ molecules = 1.67 \times 10^{-2} grams\]
Calculate the mass 5.5 moles of CaCO3.
\[Mass\ of\ 5.5\ moles\ of\ CaCO_3 = Moles \times Molecular\ mass\]
\[Mass\ of\ 5.5\ moles\ of\ CaCO_3 = 5.5 \times 100 = 550\ grams\]
Calculate the number of molecules in 5 grams of CaCO3.
\[Number\ of\ moles\ of\ CaCO_3 = Moles \times Molecular\ mass\]
\[Number\ of\ moles\ of\ CaCO_3 = {5 \over 100} = 0.05 \]
\[Number\ of\ molecules\ of\ CaCO_3 = 0.05 \times 6.022 \times 10^{23}\]
\[Number\ of\ molecules\ of\ CaCO_3 = 3.011 \times 10^{22}\]
Calculate the number of moles of O atoms in 10 grams of O2.
\[Number\ of\ moles\ of\ O_2 = Moles \times Molecular\ mass\]
\[Number\ of\ moles\ of\ O_2 = {10 \over 32} \]
\[1\ mole\ of\ O_2\ contains\ 2\ moles\ of\ O\ atoms.\]
\[ {10 \over 32}\ mole\ of\ O_2\ contains\ {2 \times 10 \over 32} = {5 \over 8}\ moles\ of\ O\ atoms.\]
Calculate the number of moles of 10 mL H2O. Density of H2O is 1 g/mL.
Density of H2O is 1 g/mL.
It means 1 mL of H2O has mass = 1 gram.
∴ 10 mL of of H2O has mass = 10 gram.
\[Moles\ of\ H_2O = {Mass \over Molecular\ mass}\]
\[Moles\ of\ H_2O = {10 \over 18} = {5 \over 9} \]
Calculate the number of moles of 100 mL ethanol (CH3CH2OH), whose density is 0.9 g/mL. Molecular mass of ethanol = 46 u.
Density of ethanol is 0.9 g/mL.
It means 1 mL of ethanol has mass = 0.9 gram.
∴ 100 mL of of ethanol has mass = 100 x 0.9 = 90 gram.
\[Moles\ of\ ethanol = {Mass \over Molecular\ mass}\]
\[Moles\ of\ ethanol = {90 \over 46} = {45 \over 23} \]
Hence, the number of moles of 100 mL ethanol (CH3CH2OH), whose density is 0.9 g/mL is \({45 \over 23}\)