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Problem : Find the number of moles in 500 grams of pure Calcium carbonate?
Solution : The numbers of moles is given by :
$$n=\frac{g}{{M}_{o}}=\frac{500}{100}=5$$
Problem : Which represents greatest mass among 100 gm of calcium, 3 g-atoms of calcium , 1 mole of calcium oxide, ${10}^{25}$ molecules of oxygen?
Solution : We convert each of the given quantity in grams for comparison :
$$100\phantom{\rule{1em}{0ex}}\text{gm of calcium}=100\phantom{\rule{1em}{0ex}}\text{gm}$$
$$3g-\text{atoms of calcium}=3X{A}_{Ca}=3X40=120\phantom{\rule{1em}{0ex}}\text{gm}$$
$$1\text{mole of calcium oxide}={M}_{CaO}\phantom{\rule{1em}{0ex}}\text{gm}=40+16=56\phantom{\rule{1em}{0ex}}\text{gm}$$
$${10}^{25}\phantom{\rule{1em}{0ex}}\text{molecules of oxygen}=\frac{{10}^{25}}{6.022X{10}^{23}}\phantom{\rule{1em}{0ex}}\text{moles}=0.166X{10}^{2}\phantom{\rule{1em}{0ex}}\text{moles}$$
$$\Rightarrow {10}^{25}\phantom{\rule{1em}{0ex}}\text{molecules of oxygen}=16.6\phantom{\rule{1em}{0ex}}\text{moles}=16.6X{M}_{{O}_{2}}\text{gm}=16.6X32=531.2\phantom{\rule{1em}{0ex}}\text{gm}\phantom{\rule{1em}{0ex}}$$
Clearly, ${10}^{25}$ molecules of oxygen represents greatest mass.
The number of moles in a given mass of a molecule is given by :
$$\text{moles(n)}=\frac{g}{{M}_{o}}$$
Since each mole has ${N}_{o}$ entities, the total numbers of entities in “n” moles is :
$$\text{Total number of molecules}\left(N\right)=n{N}_{0}=\frac{g}{{M}_{o}}X{N}_{0}$$
If we are interested to count atoms in a sample of an element, then we use “gram atom” as :
$$\text{Total number of atoms}\left(N\right)={n}_{a}{N}_{0}=\frac{g}{A}X{N}_{0}$$
Problem : If ${10}^{22}$ water molecules are removed from 100 mg of water, then find the remaining mass of water.
Solution : Here we need to find the mass of ${10}^{22}$ water molecules. We know that numbers of molecules is related to mass in gram as :
$$N=n{N}_{0}=\frac{g}{{M}_{o}}X{N}_{0}$$
$$\Rightarrow g=\frac{N{M}_{o}}{{N}_{0}}=\frac{{10}^{22}{X}_{18}}{6.022X{10}^{23}}=2.989X0.1=0.2989\phantom{\rule{1em}{0ex}}\text{gm}=2.989\phantom{\rule{1em}{0ex}}\text{mg}$$
$$\Rightarrow \text{remaining mass of water}=100-2.989=97.011\phantom{\rule{1em}{0ex}}\text{mg}$$
Chemical reactions involve dissociation and association at atomic/molecular/ionic level. The measurements of constituent entities of a reaction can be carried out using conventional mass measurements in grams or kilograms. What is the need to carry out analysis in terms of moles?
Let us have a look at the chemical reaction :
$$2{H}_{2}+{O}_{2}\to 2{H}_{2}O$$
Let us further consider that there are 2 gms of hydrogen involved in the reaction. Can we interpret the given balanced equation to predict the mass of oxygen involved in the reaction? Clearly, it would not be possible unless we know the molecular mass. In terms of molecular mass, we see that 4 a.m.u of hydrogen reacts with 32 a.m.u of oxygen. This mass correlation, however, is slightly incomprehensible to our measurement sense as a.m.u is very minute mass almost incomprehensibly too small for our perceptible laboratory measurements in grams.
We can though extend the molecular mass proportion to grams. We note that the mass ratio of two elements in the reaction is 4:32 i.e. 1:8. Hence, 2 grams of hydrogen reacts with 16 grams of oxygen. This completes the picture. But, slight discomfort is that the correspondence of mass is not directly related with the numbers of molecules (coefficients of balanced chemical equation) involved in the reaction.
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. Looking at the equation, we can say that 2 molecules of hydrogen react with one molecule of oxygen. Extending this understanding of reaction to the moles of the substance, we come to the conclusion that “2 moles of hydrogen reacts with 1 mole of oxygen”. Here, we have simply scaled the molecular relation ${N}_{o}$ times.
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