Question

The molecular mass of \[{H_2}S{O_4}\] is \[98amu\]. Calculate the number of moles of each element in \[294g\] of \[{H_2}S{O_4}\].

Answer 1

Hint: We must know that one gram is equal to \[6.022{\text{ }}X{\text{ }}{10^{23}}amu\]. Also one mole of molecules is equal to \[6.022{\text{ }}X{\text{ }}{10^{23}}\] molecules.




Complete step by step solution:
We are provided with the molecular mass of \[{H_2}S{O_4}\] which is \[98amu\]. The unit \[amu\] stands for atomic mass unit, it is defined as 1/12th the mass of the carbon-12 atom and is used to describe the mass of atomic and subatomic particles.
We know that one mole of a substance is equal to \[6.022{\text{ }}X{\text{ }}{10^{23}}\] molecules. Similarly one gram is equal to\[6.022{\text{ }}X{\text{ }}{10^{23}}amu\]. Hence,
1 mole of \[{H_2}S{O_4}\] has a mass of 98 grams. Also, one mole of \[{H_2}S{O_4}\] contains 2 mole of hydrogen each weighing 1g each, one mole of Sulphur weighing 32g and four moles of Oxygen each weighing 16g. When we total them we get $2 \times 1{\text{g + 32g + 4}} \times {\text{16g = 98g}}$which is equal to one mole of \[{H_2}S{O_4}\].
 Now, we are having \[294g{\text{ }} of {\text{ }}{H_2}S{O_4}\],
\[294g{\text{ }}of{\text{ }}{H_2}S{O_4}\]is having $\dfrac{{294}}{{98}} = 3$ moles of \[{H_2}S{O_4}\].
From above we know that 1 mole of \[{H_2}S{O_4}\]contains two mole of Hydrogen, one mole of sulphur, four mole of oxygen.
So, in case of 3 moles of \[{H_2}S{O_4}\], 6 mole of Hydrogen, 3 moles of Sulphur, 12 moles of oxygen will be present.



Note: We must know that both molar mass and molecular mass are different things, molar mass the mass of one mole of the substance and is denoted using g/mol whereas molecular mass is the mass of one molecule of the substance denoted by Dalton or \[amu\].