Question

What is the number of atoms of each element in 6.3 g of nitric acid \[(HN{O_3})\] ? (Atomic mass of
  \[H = 1u\] , Atomic number of \[N = 14u\] , Atomic mass of \[O = 16\]

Answer 1

Hint: To solve this question, follow these steps: first calculate the number of moles of \[HN{O_3}\] present. Then multiply the number of moles with Avogadro’s number to find the number of molecules. Now multiply this value to atomicity of individual constituent elements.

Formula used:
 Number of moles of \[HN{O_3} = \dfrac{{Weight\,\,of\,\,the\,\,given\,\,sample}}{{molecular\,\,weight\,\,of\,\,HN{O_3}}}\]
Number of molecules = (Avogadro’s number) (number of moles of \[HN{O_3}\]

Complete Step-by-Step Answer:
Before we move forward with the solution of this question, let us discuss some important concepts.
1)Mole: a mole is a physical quantity which represents the amount of mass of the substance required to have a collective of \[6.022 \times {10^{23}}\] atoms of the given substance. Mole is a widely used unit for calculating the amount of matter of a substance. One mole of any substance weighs about the same as the molecular mass of that substance.
Now moving back to the question, we have been given a sample of 6.3 grams of \[HN{O_3}\] . We know that the atomic weight of hydrogen is 1 amu, atomic weight of nitrogen is 14amu and the atomic weight of oxygen is 16amu. Hence the molecular weight of \[HN{O_3}\] can be calculated as:
Molecular weight of \[HN{O_3}\] = (no. of atoms of hydrogen) (atomic weight of hydrogen) +
                      (no. of atoms of nitrogen) (atomic weight of nitrogen) +
                      (no. of atoms of oxygen) (atomic weight of oxygen)
                     \[ = \left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( {14} \right) + \left( 3 \right)\left( {16} \right)\]
                    = 63 \[g/mol\]
Now, to calculate the number of moles of \[HN{O_3}\] present in the given sample, we use the formula:
Number of moles of \[HN{O_3} = \dfrac{{Weight\,\,of\,\,the\,\,given\,\,sample}}{{molecular\,\,weight\,\,of\,\,HN{O_3}}}\] \[ = \dfrac{{Weight\,\,of\,\,the\,\,given\,\,sample}}{{molecular\,\,weight\,\,of\,\,HN{O_3}}}\] \[ = \dfrac{{6.3}}{{63}}\] = 0.1 moles
Now, the number of atoms/molecules present in 1 mole of a substance is represented by a quantity known as Avogadro’s Number. This quantity is equivalent to \[6.02 \times {10^{23}}\] atoms per mole of a substance. Since, we have 0.1 moles of \[HN{O_3}\] , the number of molecules of \[HN{O_3}\] present are:
Number of molecules = (Avogadro’s number) (number of moles of \[HN{O_3}\]
                      \[ = (6.02 \times {10^{23}})(0.1)\])
                      \[ = 0.602 \times {10^{23}}\] molecules
Hence, there are \[0.602 \times {10^{23}}\] molecules of \[HN{O_3}\] present in a 6.3 gram sample of \[HN{O_3}\] .
  \[HN{O_3}\] has 1 hydrogen, 1 nitrogen and 3 oxygen atoms. Hence the atomicity of these elements in the given compound are 1, 1, 3 respectively.
Hence, the total number of atoms of each element are:
No of atoms of \[H = 0.1 \times 1 \times 6.02 \times {10^{23}} = {6.0210^{22}}\]
No of atoms of \[N = 0.1 \times 1 \times 6.02 \times {10^{23}} = 6.02 \times {10^{22}}\]
No of atoms of \[O = 0.1 \times 1 \times 6.02 \times {10^{23}} = 18.069 \times {10^{22}}\]

Note: Mole concept simplifies the mass relation among reactants and products such that we can base our calculation on the coefficients (numbers of molecules involved in the reaction). At the same time, mass or the quantity of substance is on lab scale in grams.