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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:

F e + S F e S

In this reaction, every atom of iron (Fe) will react with a single atom of sulfur (S) to form one molecule of iron sulfide (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 sulfur will you need for the reaction to use up all the iron you have? Is there a way of knowing what mass of iron sulfide 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 Grade 10 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 u) for example, is twelve times greater than the mass of an atom of hydrogen, which has a relative atomic mass of 1 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 12g carbon. What mass of hydrogen will contain the same number of atoms as 12 g carbon? We know that each atom of carbon weighs twelve times more than an atom of hydrogen. Surely then, we will only need 1g 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 (12g 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 6.023 x 10 23 . These results are shown in [link] below for a number of different elements. So, 24.31 g of magnesium (relative atomic mass = 24.31 u) for example, has the same number of atoms as 40.08 g of calcium (relative atomic mass = 40.08 u).

Table showing the relationship between the sample mass, the relative atomic mass and the number of atoms in a sample, for a number of elements.
Element Relative atomic mass (u) Sample mass (g) Atoms in sample
Hydrogen (H) 1 1 6.023 x 10 23
Carbon (C) 12 12 6.023 x 10 23
Magnesium (Mg) 24.31 24.31 6.023 x 10 23
Sulfur (S) 32.07 32.07 6.023 x 10 23
Calcium (Ca) 40.08 40.08 6.023 x 10 23

Questions & Answers

What is a vector
Mercii Reply
vector is anything that has both a direction and a magnitude .they are usually drawn as pointed arrows ,the length of which represents a vector's magnitude
what is electronics?
how to calculate the reading on voltmeter or ammeter
mosima Reply
why is HCl considered a strong acid
Sphiwe Reply
it dissociate almost completely
what is metal displacement
Andile Reply
what is an electric field?
Noluthando Reply
is the charge of an electron always 1,6 ×10^-19? and the mass is always 9,1×10^-13?
Neo Reply
how to calculate a distance between charges
Celucolo Reply
juss apply the formula of the Electrostatic force
please what's the simplest way to derived Schrodinger's wavelength equation
how to show polarity
Sandiso Reply
what can ii do to pass physics
Slindile Reply
Use previous years question papers to understand how questions are answered and asked
but it hard ii am a slow learner
Then the best thing to do is that immediately you are done reading through a certain topic, and you think you understood everything in that topic that's when you can use previous question papers and answer questions related to the topic. I think that's not difficult
mm ii will try
Here's a tip in reading textbooks, don't read it like a novel. First, flip through the pages—scan the chapter that you wanted to read. Second, go to the end of the chapter. Usually, there's a quiz at the end, so if will give you the important information that you need to know.
Third, go to the beginning of the chapter and read through the words that were printed in bold: Titles, subtitles, headings, important words —because it helps to break down information.
Fourth, read the first sentence of the chapter—if it is written by a good author; therefore, it will also have a good introduction. Check also the last sentence of the chapter to sum it up. Finally, read the whole chapter. You won't read it twice anymore.
It looks hard, cause there are so much to do but read it thoroughly, it's easy and it will help you to save time and comprehend better. If you don't really have interest on reading—there are various of videos in youtube😊
Oh, so em dashes turn into question marks :' (( nvm. Goodluck to all of us!
How many ways can we calculate the empirical formula
Joseph Reply
How is current divided between resistors in parallel
ii know 2 ways
what's the other way
What is the meaning of Coulomb's law
Nozipho Reply
Electronic magnetic field
Pride Reply
what is an emf ?
Nobuhle Reply
electromotive force
what are moments of a force
Mercy Reply
the moment of a force" is a measure of a tendency to cause a body to rotate about a specific point or axis....a moment is due to force not having equal and opposite force directly along its line of action.
how to calculate the magnitude of the force of repulsion
Vicky Reply

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Source:  OpenStax, Siyavula textbooks: grade 11 physical science. OpenStax CNX. Jul 29, 2011 Download for free at http://cnx.org/content/col11241/1.2
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