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Welcome into the first part of International Baccalaureate program in Chemistry ! If you were ever worried by Chemistry, let me tell you this simply : there is no need. You will be introduced to all the necessary knowlegde in the chapters to come from the mole concept to the chemical equations and gases... And I am sure you will, at the end, even find the dreadful equations — the ones from which you used to run away before IB — easy.

The basics

Source of the periodic table

Above, you can see the symbols of all the elements discovered by chemists. Elements cannot be broken down by chemical processes, only by physical processes (which is — almost — none of our concern).

Elements differ from each others in many ways. In fact, they boil and vaporize at different temperatures, form different types of bonds, can act as bases, acids or both,... So that we must (and we will !) have to study their properties and differences before you can predict how they will react — and eventually make cool experiments in your garden (although for safety reasons you should not try too soon ).

But lets get back to theory first. Before we are ready to move on and know more about the different elements, there are a few key concepts we should understand : isotopes, atoms, molecules, compounds and moles :

• One element has different "versions" of itself in nature. The different versions don't have the same weights but have the same chemical properties and are called the isotopes of that element. (We will talk about isotopes in more detail in the next chapter.)

• An atom is the smallest part of an element.
• Molecules are made of more than one atom, of the same element or not. Molecules made of two atoms of the same substance are diatomic molecules. Except the elements in the right end column of the periodic table, most elements don't exist in nature as single atoms but as molecules or compounds. 

• Coumpounds are molecules made of more than one element. Molecules or compounds can be broken down by chemical processes into their elements. All compounds are molecules, but not all molecules are compounds as they can be made of only one element.

Example : H₂O or H₂ are molecules. Only H₂O is a compound.

• One mole is the amount of an element or a molecule containing 6.02 x 10²³ atoms of an element or 6.02 x 10²³ molecules. 6.02 x 10²³ correspond to a constant which is Avogadro's number (N_A or L). 12 grammes of carbon-12 contain 6.02 x 10²³ atoms of carbon-12. The mass of one mole of any substance is noted M and called the molar mass (units : g mol^-1).

However, be careful when you play around with moles. One mole of oxygen gas means we have 6.02 x 10²³ molecules of oxygen (O₂) but 1.2 x 10²⁴ moles of the element oxygen (O) !

Fun facts about moles: It might be hard for you to imagine what 6.02 x 10²³ things amount to. Well, the size of the visible universe is of 10²⁵ and the age of the universe is in the 10¹⁸ ! So the number of particules is somewhere between these two values and is, as you know can see, very, very high. 

For those who want to lenghten the break, you might be interested to know that... It was Albert Einstein who, among other things, discovered Avogadro's number. When a lump of suger disolves in water, the liquid becomes more visquous. "It is possible he got interested in the problem while drinking tea" would later say one of his first co-workers. Either way, Albert Einstein described it in mathematical terms and eventually measured the viscosity of different solutions... and deduced the size of the sugar molecules and from it Avogadro's number !

We can now go back to the periodic table. From it, we can derive that elements are caracterized by at least two numbers :

• Their atomic number. For now, we will say it is the number above them, which also gives their position in the periodic table. But do not relax yet, there is more to come on this notion in the next course !  
• Their relative atomic mass, which is situated below the symbol. The relative atomic mass is defined as the weighted mean of all naturally occuring isotopes of the element relative to `1/12` the mass of a carbon-12 atom. Because it is a ratio, it has no unit. The relative molecular mass of molecules is found by adding the relative atomic masses of all the atoms present. If the relative atomic mass of hydrogen is 1.01, its molar mass is 1.01 g mol^-1. So in 1.01 g of hydrogen, you also have one mole of it !

Finally, to end this first chapter, we will look at the different formulas associated with compounds.There are three fromulas you should know and be able to use in exercises :

• The empirical formula. It is defined to be the simplest whole number ratio of atoms of each elements in a particule of a substance. If you know the mass or the percentage composition by mass of each element in a compound, it can be found the following way :

In the compound, there are 74.8 % of carbon, 25.2 % of hydrogen. Find the empirical formula.

Check the percentages add up to 100% : 74.8+25.2 = 100. You can now divide the percentage composition by the atomic mass of the elements (it is as if you consider you have 74.8 grams of carbon for the total mass of the molecule being of 100 grams):

`74.8/12=6.23`

`25.2/1.01=24.95`

The next step is to divide the bigger number by the smaller one : `24.95/6.23=4,01`.

NOTE: If there are more than two numbers, divide all the numbers by the smaller one !

We take the closest whole number: 4. From it, we can now say the empirical formula of the compound is CH₄ (and if you have any doubts, there is only one C because `6.23/6.23 =1` ).

NOTE:  Be careful ! : Sometimes, the percentage composition of oxygen is omitted... Always check the percentage composition of all elements add up to 100% !

• The molecular formula. Two different compounds can have the same empirical formula. For example, CH₂O and C₃H₆O₃ have the same empirical formula because their elements have the same ratios to one another. The molecular formula helps you to differenciate them. It is defined as the number of atoms of each element in the compound. If you know the empirical formula and the molar mass of the compound, the molecular formula can be derived. Because example sometimes speak better than words, here is a little example for you :

You are given an empirical formula : CH₂O. The molar mass of the compound is of 90 grams.

Calculate first the molar mass of CH₂O. Mr₍CH₂O₎= 12 + 2x1 + 16 =30 grams. Then, divide the molar mass of the compound by the molar mass of the compound given by the empirical formula : `90/30=3`. The compound in its molecular formula is simply : 3 x CH₂O = C₃H₆O₃.

• The structural formula. Once you have the molecular formula, you can derive the structural formula. The structural formula shows the bonds and arragement of atoms in a compound. It will be discussed in more depth and you will learn to use it later in the year.

Putting it into equations

You are now ready to make your first (fictive) experiments... and to learn how to put them down into equations. For us to say that a chemical reaction has happened, the reactants must have been TRANSFORMED into new chemical substances, the products.

In molecules, atoms are held together by chemical bonds. Chemical bonds need energy to be broken and release energy when they are formed. However, all bonds don't need/release the same energy... and it may well be that your chemical reaction takes energy from the surrounding (or release energy to the surroundings) if the bonds broken have an higher energy than the bonds formed (or bonds formed have an higher energy than bonds broken). We say there is an energy change between the system and the surroundings.

But all of this doesn't really help us in writing down an equation does it ? Well, not yet, but it is part of the basics you should know. Let's take a (relatively) simple equation you will meet again later in the course :

3H₂ (g) + N₂ ` `(g)   2NH₃ (g)

From it, we will know derive what you should do :

• step 1 : if both reactants and products are present in the final mixture put the arrow illustrated above, otherwise (meaning there are only products in the final solution) put the following arrow: `->` .
• step 2 : put the reactants on the left side of the arrow the products on the right side. As you write down the different compounds also put down in brakets what states they are in : (s) ~ solid ; (l) ~ liquid ; (g) ~ gas ; (aq) ~ aqueous solution.
• sept 3 : if there are specific conditions for the reaction to take place, write them above the arrow.
• step 4 : Look closely at the reactants and products and make out how many atoms you have of each element. At this stage the above equation will look like this : H₂ (g) + N₂ ` `(g)   NH₃ (g) . 

As you can see, there isn't the same number of atoms in the reactants and the products for any given element... and your equation is FALSE. Because you have forgotten (or didn't know yet) that the number of atoms of an element MUST BE THE SAME in the reactants and the products and that the overall mass of the system REMAINS CONSTANT. If you decide to sort out your chemistry notes, you will re-arrange them in a different order and you (hopefully) can't throw any away (because you cry of guilt each time to try to )... So you will have the same number of notes at the beginning and at the end. Well, chemistry follows the same logic. Atoms are re-arranged but NOT lost...

• Step 5 : As a result, you must balance your equation. Usually you should treat H atoms and O atoms last, if you want to make things easier. So we will start with the N atoms. There are 2 N atoms on the reactants' side, and only 1 on the products' side. As a result, you must multiply the right product by 2 (here it is easy as there is only one product). You have a new equation : H₂ (g) + N₂ ` `(g)   2 NH₃ (g)... And you can count down the atoms of each elements again. You have 2 N atoms on each sides and 2 H atoms of the reactants' side and 6 H atoms on the products' side. The N-atoms are balanced but the number of H-atoms must still be multiplied by 3 on the reactants' side to be balanced. So you finally have : 3 H₂ (g) + N₂ ` `(g)  2 NH₃ (g). 3 and 2 are called stoichiometry coefficients and the equation is balanced. The stoichiometry (stoicheion = "element" and metron = "measure" in greek) is the quantitative relationship between the elements. 

NOTE : You must use common sense to guess where to add the coefficients. You should add them so as:

• to change as little as possible the number of atoms of other element than the one considered
• to NOT change the number of atoms of the elements you have already worked on.

Finally, it is worth knowing that the molar ratio of two molecules in a chemical equation is equal to the ratio between their coefficients. We have not treated the case of ionic equations in this chapter as they will be thouroughly discussed later in the course (chapter Bonding).

Calculating the limiting reactant (or reagent).

Now comes a very important part of this course... And you first real calculations. No need to be afraid ! I said real, I didn't say hard  ! Here is a little problem I made up for you :

Determine the limiting reageant and the amount in moles and gramms of Al₂O₃ produced. We are given 31g of Zn and 300 cm³ of 1.25 mol dm^-3  HCl.The following equation is also given to us :

Zn (s) + 2 HCl (aq) `->` ZnCl₂ (s) + H₂ (g)

We first have to calculate the number of moles of both Zn and HCl present at the start of the reaction. If you have molecules, it implies calculating their relative molecular mass.

Relative atomic mass of Zn =  65.39

As we mentionned above, the number of moles is given by dividing the mass of Zn present by its relative atomic mass. For a molecule, the number of moles would be given by dividing the mass present by the relative molecular mass.

Number of moles of Zn  =  `31/(65.39) = 0.47` moles.

The number of moles of H₂O can be calculated directly from its concentration in the solution. Be carefull, it's important to convert it into dm³ and not leave it in cm³ (hence we divide the 150 cm³ by 1 000 to get dm³) ! Then you just have to multiply the volume of the solution by the concentration (number of moles present per dm³).

Number of moles of HCl = `1.25 xx (300/1000) = 0.38` moles.

NOTE : 1 Liter = 1 dm³

Then, there are two possible ways to calculate the limiting reageant. I personnaly prefer the first one because once it makes things way simpler if the reaction doesn't go to completion...

Method 1

Place your equation in a table as shown below. To find the number of moles present at the end FOR THE REACTANTS, you consider an unknown number of moles : x (where x is the number of moles used). For a given reactant, its number of moles will be given by the equation = (number of moles present at the start) - [(coefficient)-x].

Why do you look depressed  ?...

It is logical : the number of moles present are going to become lower as the reaction goes on for the reactants, hence the substraction. Then, the bigger coefficient of the reactant, the more we need of that substance. So, x must be multiplied by the coefficient.

FOR THE PRODUCTS, as there are none present at the start of the reaction, the number of moles present at the end will just be equal to x multiplied by the coefficient.

For the limiting reagent, we consider the reaction goes to completion. So one of the reactants must have zero moles left.

NOTE : If the reaction doesn't go to completion and you have the number of moles left of one of the reagants, you can easily find the number of moles of the other reagants and the products... Just makes the small equation for the given reagant equals its number of moles left and calculate x ! And plug x in the other equations to find the number of moles present for all the other molecules.

Setting the 2 equations equal to zero, we have :

`0,47-x = 0`  hence, `x = 0,47` for Zn

`0.38-2x = 0` hence `x =(0.38)/2 = 0.19` for HCl

It appears HCl gives the smallest value of x. Therefore, it is the limiting reageant and the x it gives us will be used in the following calculations.

And at the end of the equation we are left with (don't forget to use `x=0,19` ONLY not the other value !) :

`0.47-0.19 = 0.28` moles of Zn are left

`1xx0,19=0,19` moles of ZnCl₂ have been produced and the mass of ZnCl₂ produced is : `{(65.39)+(35.5xx2)} xx0,19 = 25.9` g. And you are forbidden to look surprised as it is only the relative molecular mass multiplied by the number of moles.

`1 xx 0,19 = 0,19` moles of H₂ are left

Method 2

It is a method very similar to the first one. In fact, I would consider it the "trying to guess" version of the second one. Because it doesn't use the mathematical tools, as the equations grow more complicated, you are likely to make more mistakes than with the first one (but I cannot be hold responsible, it is useless to sue me for your calculations mistakes ). However, it is sufficient in IB... so just chose the one who feel confortable with !

We see from the equation there should be twice as many moles of HCl than of Zn. As a result, knowing there are 0.47 moles of Zn, there should be :

`0.47xx2=0.94` moles of

HCl. However, we know from the calculations above, we have only 0.19 moles of HCl. Therefore HCl is the limiting reagant and will be entirely used up in the equation.

NOTE : If the coefficient in front of Zn was 3 and the coefficient before HCl was still 2, there should be `0,47xx(2/3)` moles of HCl ! Be very careful for the coefficients !

Now we know the limiting reagant, we can derive from its number of moles the number of moles left of Zn and the number of moles produced of ZnCl₂ and H₂.

To find the number of moles left of Zn, we substract the number of moles used from the number of moles present at the beginning : `0,47-0,19=0,28` moles - And I am sure you have noticed it is just the same as the equation given in the method 1 ! How surprising  !

NOTE : If there is a coefficient in front of Zn, for exemple, 3, then you should multiply the number of moles used by three before substracting them from the number of moles that were present at the start.

The number of moles for a given product is given by 0,19 multiplied by the coefficient of that product. The mass is again the number of moles multiplied by the relative molecular mass.

You have finished your exercise !

...

...

(almost)

Once you have calculated all of this, there is still one thing the IB might ask you : to find the percentage yield of an experiment. They will give you the necessary data. Then, just use the following :

% yield =

Gases and volumes made easy

The distinction between pure liquids and solutions should be made clear before we move on to gases. Pure liquids are made of only one substance whereas solutions are made of more than one substance. The density of a liquid is equal to its mass divided by its volume and its units are in g.cm^-3. Because we have seen - in the precedent part (in an exercise) - that the number of moles present is equal to the concentration of the solution in mol.dm^-3 multiplied by the volume of solution in ` `dm³, you should already know how to handle questions concerning concentration of solutions.

Now, we can move on to gases, or, more accurately to ideal gases. Ideal gases have specific caracteristics :

• there are no attractive forces acting between their particules
• their particules occupy no space

If a gas had the ideal gases' caracteristics, it would not be able to become a liquid !  In fact, one of the reasons is that as a gas changes to a liquid, the attractive forces between its particules grow stronger and tighter... which is obviously not possible if these attractions are inexistant.

The ideal gas equation is the following (it is normally in the data booklet) :

`PV=nRT`

where P is the pression in Pa (pascals), T is the temperature is Kelvin, V is the volume in m³, n is the number of moles and R is the gas constant  (= 8.314 JK^-1mol^-1).

NOTE : To convert celsius into Kelvins you should add 273 to the Celsius value. At - 273,15°C, 0 Kelvins or in other words at the absolute zero... there are no thermic agitation, meaning the atoms/molecules cannot move ! And this temperature has almost been achieved inside the LHC (Large Hadron Collider) ! 

If you want to know a bit more about how this equation came to life click here. Of course, we are talking more about history of science here, so for those who can't go to history class without taking a nap, please... just don't clic !

Sadly, there are still a few things you should know... The ones at the back, don't fall asleep, I am watching you  ! Basically, you can find back all those formulas by considering one of the three variables (Pressure or Volume or Temperature) is held constant, for a fixed number of moles.

`V/T =` a constant (`(nR)/P` = a constant, where P is held constant)

The equation meaning the two variables are proportional, the graph should not surprise you.

`PxxV =` a constant (`nRT =` a constant, where T is held constant)

`P/T =` a constant (`(nR)/V =` a constant, where V is a constant)

In the next chapter, we will see how to differenciate one isotope from another by going at an even smaller scale and analyzing the atoms, study the mass spectrometer and eventually learn the electron arrangement in elements after having studied the emission spectra of different elements. Some fancy words you don't know?... I will wait for you in the next lesson!