# 0.9 Quantitative aspects of chemical change

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## Quantitative aspects of chemical change

An equation for a chemical reaction can provide us with a lot of useful information. It tells us what the reactants and the products are in the reaction, and it also tells us the ratio in which the reactants combine to form products. Look at the equation below:

$\mathrm{Fe}+\mathrm{S}\to \mathrm{FeS}$

In this reaction, every atom of iron ( $\mathrm{Fe}$ ) will react with a single atom of sulphur ( $\mathrm{S}$ ) to form one molecule of iron sulphide ( $\mathrm{FeS}$ ). However, what the equation doesn't tell us, is the quantities or the amount of each substance that is involved. You may for example be given a small sample of iron for the reaction. How will you know how many atoms of iron are in this sample? And how many atoms of sulphur will you need for the reaction to use up all the iron you have? Is there a way of knowing what mass of iron sulphide will be produced at the end of the reaction? These are all very important questions, especially when the reaction is an industrial one, where it is important to know the quantities of reactants that are needed, and the quantity of product that will be formed. This chapter will look at how to quantify the changes that take place in chemical reactions.

## The mole

Sometimes it is important to know exactly how many particles (e.g. atoms or molecules) are in a sample of a substance, or what quantity of a substance is needed for a chemical reaction to take place.

You will remember from Relative atomic mass that the relative atomic mass of an element, describes the mass of an atom of that element relative to the mass of an atom of carbon-12. So the mass of an atom of carbon (relative atomic mass is $12\phantom{\rule{2pt}{0ex}}\mathrm{u}$ ) for example, is twelve times greater than the mass of an atom of hydrogen, which has a relative atomic mass of $1\phantom{\rule{2pt}{0ex}}\mathrm{u}$ . How can this information be used to help us to know what mass of each element will be needed if we want to end up with the same number of atoms of carbon and hydrogen?

Let's say for example, that we have a sample of $12\phantom{\rule{2pt}{0ex}}\mathrm{g}$ carbon. What mass of hydrogen will contain the same number of atoms as $12\phantom{\rule{2pt}{0ex}}\mathrm{g}$ carbon? We know that each atom of carbon weighs twelve times more than an atom of hydrogen. Surely then, we will only need $1\phantom{\rule{2pt}{0ex}}\mathrm{g}$ of hydrogen for the number of atoms in the two samples to be the same? You will notice that the number of particles (in this case, atoms ) in the two substances is the same when the ratio of their sample masses (12 g carbon: 1g hydrogen = 12:1) is the same as the ratio of their relative atomic masses (12 u: 1 u = 12:1).

To take this a step further, if you were to weigh out samples of a number of elements so that the mass of the sample was the same as the relative atomic mass of that element, you would find that the number of particles in each sample is $6,022×{10}^{23}$ . These results are shown in [link] below for a number of different elements. So, $24,31\phantom{\rule{2pt}{0ex}}\mathrm{g}$ of magnesium ( $\mathrm{relative atomic mass}=24,31\phantom{\rule{2pt}{0ex}}\mathrm{u}$ ) for example, has the same number of atoms as $40,08\phantom{\rule{2pt}{0ex}}\mathrm{g}$ of calcium ( $\mathrm{relative atomic mass}=40,08\phantom{\rule{2pt}{0ex}}\mathrm{u}$ ).

 Element Relative atomic mass (u) Sample mass (g) Atoms in sample Hydrogen ( $\mathrm{H}$ ) 1 1 $6,022×{10}^{23}$ Carbon ( $\mathrm{C}$ ) 12 12 $6,022×{10}^{22}$ Magnesium ( $\mathrm{Mg}$ ) 24.31 24.31 $6,022×{10}^{23}$ Sulphur ( $\mathrm{S}$ ) 32.07 32.07 $6,022×{10}^{23}$ Calcium ( $\mathrm{Ca}$ ) 40.08 40.08 $6,022×{10}^{23}$

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There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
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I think
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research.net
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