Question

Calculate no of atoms of each type in 1.8g glucose [Give ans in terms of NA]

Answer 1

Hint: Molecular mass of glucose is $180g/mol$. One mole of atoms/molecules will contain $6.022 \times {10^{23}}$ atoms/molecules respectively. This constant number is known as the Avogadro number (${N_A}$)

Formulae used:
$n = \dfrac{w}{M}$
Where $n$ is the number of moles, $w$ is the given mass and $M$ is the molecular mass.

Complete step by step answer:
Firstly, we need to find the number of moles of glucose given. As we know, the number of moles is the ratio between given mass and molecular mass. Therefore,
$n = \dfrac{w}{M}$
Where $n$ is the number of moles, $w$ is the given mass and $M$ is the molecular mass. Substituting our values, we get:
$n = \dfrac{{1.8}}{{180}} = 0.01moles$
As we know, the molecular formula of glucose is ${C_6}{H_{12}}{O_6}$. Therefore, in every molecule of glucose, there are 6 atoms of carbon, 12 atoms of hydrogen and 6 atoms of oxygen. Thus, in every mole of glucose, the number of atoms of each type will be the product of Avogadro number and the number of atoms of that type in one molecule of glucose. For example,
Number of atoms of carbon in one mole of glucose is $6 \times {N_A}$
$6 \times 6.022 \times {10^{23}} = 36.132 \times {10^{23}}$ atoms
Similarly, we can find the number of atoms of hydrogen per mole of glucose as:
$12 \times 6.022 \times {10^{23}} = 72.264 \times {10^{23}}$ atoms
And lastly, number of atoms of oxygen in one mole of glucose:
$6 \times 6.022 \times {10^{23}} = 36.132 \times {10^{23}}$ atoms
These are the values for one mole of glucose, i.e., 180g of glucose. We are required to find the number of atoms in 1.8g (0.01 moles, as we found earlier). Therefore, we multiply the values obtained above with 0.01 to get the number of atoms in 1.8g. And to obtain the values in terms of ${N_A}$, we divide it with the Avogadro constant.
Number of carbon atoms in 1.8g glucose is
$\dfrac{{36.132 \times {{10}^{23}} \times n}}{{{N_A}}} = \dfrac{{36.132 \times {{10}^{23}} \times 0.01}}{{6.022 \times {{10}^{23}}}}$
On simplifying, we get number of carbon atoms = $0.06{N_A}$
Similarly, number of hydrogen atoms in 1.8g glucose is
$\dfrac{{72.264 \times {{10}^{23}} \times n}}{{{N_A}}} = \dfrac{{72.264 \times {{10}^{23}} \times 0.01}}{{6.022 \times {{10}^{23}}}}$
Number of hydrogen atoms = $0.12{N_A}$
Number of oxygen atoms in 1.8g glucose is
$\dfrac{{36.132 \times {{10}^{23}} \times n}}{{{N_A}}} = \dfrac{{36.132 \times {{10}^{23}} \times 0.01}}{{6.022 \times {{10}^{23}}}}$

On simplifying, we get number of oxygen atoms = $0.06{N_A}$

Additional Information: You may wonder as to why we multiplied with the Avogadro number first and then divided with it to get the final answer. This was done just so that the process of how we came to the conclusion became clear. Indeed, if you are asked to give the answer in terms of ${N_A}$ we can directly write the result by multiplying the number of moles with the number of atoms of the particular type in the molecule.

Note: Note that one mole of any quantity, regardless of its nature, will always contain the same amount of entities, which is equal to the avogadro number. It is just like saying that a dozen bananas or a dozen books means twelve entities of the respective object. Thus, in this question, we can also get the answer by directly multiplying the number of moles of glucose with the number of atoms of each type, as that would give us the answer in terms of atoms per mole, which is the same as number of atoms expressed in terms of the avogadro number.