# Mole concept  (Page 3/4)

 Page 4 / 4

For a general reaction of the type :

$xA+yB\to {A}_{x}{B}_{y}$

We say that “x moles of A reacts with y moles of B”.

This is a very powerful deduction popularly known as “mole concept”. It connects the measurement of mass neatly with the coefficients which are whole numbers. Further as moles are connected to mass in grams, measurement of quantities can be easily calculated in practical mass units like grams.

Problem : Find the mass of KCl and oxygen liberated when 24.5 grams of $KCl{O}_{3}$ is heated. Assume 100 % decomposition on heating. Given, ${M}_{K}=39$ .

Solution : On heating $KCl{O}_{3}$ decomposes as :

$2KCl{O}_{3}\to 2KCl+3{O}_{2}$

In order to apply mole concept, we need to find the moles of $KCl{O}_{3}$ in the sample.

$\text{moles of}KCl{O}_{3}=\frac{24.5}{\text{molecular wt of}KCl{O}_{3}}$

Here, molecular weight of $KCl{O}_{3}$ is :

$\text{molecular wt of}KCl{O}_{3}=39+35.5+3X16=122.5$

Putting this value in the equation of moles :

$⇒\text{moles of}KCl{O}_{3}=\frac{24.5}{\text{molecular wt of}KCl{O}_{3}}=\frac{24.5}{122.5}=0.2$

Applying mole concept to the equation, we have :

$2\phantom{\rule{1em}{0ex}}\text{mole of}KCl{O}_{3}\phantom{\rule{1em}{0ex}}\stackrel{\frown}{=}\phantom{\rule{1em}{0ex}}2\phantom{\rule{1em}{0ex}}\text{moles of}KCl\phantom{\rule{1em}{0ex}}\stackrel{\frown}{=}\phantom{\rule{1em}{0ex}}3\phantom{\rule{1em}{0ex}}\text{moles of oxygen}$

$1\phantom{\rule{1em}{0ex}}\text{mole of}KCl{O}_{3}\phantom{\rule{1em}{0ex}}\stackrel{\frown}{=}\phantom{\rule{1em}{0ex}}1\phantom{\rule{1em}{0ex}}\text{moles of}KCl\phantom{\rule{1em}{0ex}}\stackrel{\frown}{=}\phantom{\rule{1em}{0ex}}\frac{3}{2}\phantom{\rule{1em}{0ex}}\text{moles of oxygen}$

$0.2\phantom{\rule{1em}{0ex}}\text{mole of}KCl{O}_{3}\phantom{\rule{1em}{0ex}}\stackrel{\frown}{=}\phantom{\rule{1em}{0ex}}0.2\phantom{\rule{1em}{0ex}}\text{moles of}KCl\phantom{\rule{1em}{0ex}}\stackrel{\frown}{=}\phantom{\rule{1em}{0ex}}\frac{0.6}{2}\phantom{\rule{1em}{0ex}}\text{moles of oxygen}$

At this point, we need to convert the mass measurement from mole to grams,

${m}_{\text{KCl}}={n}_{\text{KCl}}X{M}_{\text{KCl}}=0.2X\left(39+35.5\right)=0.2X74.5\phantom{\rule{1em}{0ex}}\text{gm}=14.9\phantom{\rule{1em}{0ex}}\text{gm}$

${m}_{{O}_{2}}={n}_{{O}_{2}}X{M}_{{O}_{2}}=\frac{0.6}{2}X32=0.6X16\phantom{\rule{1em}{0ex}}\text{gm}=9.6\phantom{\rule{1em}{0ex}}\text{gm}$

## Ideal gas and mole concept

The concept of mole is applicable to identical entities (atoms, molecules, ions). Thus, its direct application is restricted to pure substances – irrespective of its state (solid, liquid and gas). For all practical purposes, we treat mole as an alternative expression of mass. They are connected to each other via molecular weight for a given pure substance. However, measurement of gas is generally reported in terms of volume in stoichiometric calculation. We would, therefore, need to convert the volume of the gas to mass/mole.

If the gas in question is ideal gas, then we can easily connect volume to mole directly. Ideal gas presents a special situation. The collisions are perfectly elastic and there is no intermolecular force between molecules. The volume occupied by gas molecules is negligible with respect to the volume of ideal gas. It means that the size of molecule has no consequence as far as the volume of ideal has is concerned. These negligibly small molecules, however, exchange momentum and kinetic energy through perfectly elastic collisions. Clearly, volume depends solely on the numbers of molecules present – not on the mass or size of molecules. This understanding leads to Avogadro’s hypothesis :

All ideal gases occupy same volume at a given temperature and pressure. One mole of ideal gas occupies 22.4 litres i.e. 22,400 ml at standard temperature (273 K) and pressure (1 atmosphere) condition.

The gas volume for 1 mole is known as molar volume. The Avogadro’s hypothesis provides an important relation between volume and moles of gas present. This relationship does not hold for real gas. But we are served well in our calculation by approximating real gas as ideal gas. The real gas like hydrogen, oxygen etc. may still be close to the ideal gas molar volume at normal temperature and pressure conditions.

Problem : Air contains 22.4 % oxygen. How many moles of oxygen atoms are there in 1 litre of air at standard condition ?

Solution : The volume of oxygen is 0.224 litre. We know that 22.4 litres contains 1 mole of oxygen molecules. Hence, 0.21 litres contains “n” moles given by :

$n=\frac{0.224}{22.4}=0.01\phantom{\rule{1em}{0ex}}\text{moles of oxygen molecules}$

We know that one molecule of oxygen contains 2 atoms. Therefore, numbers of moles of oxygen atoms is 0.02 moles.

Problem : Calculate the numbers of molecules in 300 cc of oxygen at 27° C and 740 mm pressure.

Solution : We first need to convert the given volume at given temperature and pressure to that at standard temperature and pressure condition. Applying gas law,

$\frac{{P}_{1}{V}_{1}}{{T}_{1}}=\frac{{P}_{2}{V}_{2}}{{T}_{2}}$

$⇒{V}_{2}=\frac{{P}_{1}{V}_{1}{T}_{2}}{{P}_{2}{T}_{1}}=\frac{740X300X273}{760X300}=265.82\phantom{\rule{1em}{0ex}}cc$

Number of molecules of oxygen is :

$⇒N=\frac{265.82X6.022X{10}^{23}}{22400}=7.146X{10}^{21}$

## Analysis techniques

Stoichiometric analysis is about analyzing mass of different chemical species or concentrations of solutions involved in chemical reactions. We employ varieties of different concepts and hypotheses. These principles are primarily based on molecular association of various chemical species in chemical reactions. Most important among these are mole, equivalent weight and gram equivalent weight concepts. Each of these concepts relates amount of different species in chemical reaction in alternative ways. It means that these concepts have contextual superiority over others (suitability in a particular context), but are essentially equivalent.

Stoichiometric analysis is broadly classified as gravimetric or volumetric analysis depending on whether calculations are mass based (gravimetric) or concentration of solution based (volumetric). There is no strict division between two approaches as we encounter reactions which require considerations of both aspects i.e. mass of species and concentration of solution.

In general, gravimetric analysis covers displacement reactions, action of acid on metals, calculations based on balanced equations, decomposition reaction etc. On the other hand, volumetric analysis covers wide ambit of titration including neutralization and redox reactions. Many of the chemical reactions, however, need combination of two analytical approaches. For example, a sequence of chemical process may involve decomposition (gravimetric analysis) and titration (volumetric analysis).

find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!