Empirical Formulae and Limiting Reagents and Some Other Stuff

Okay, the final post for unit 2A is just adding on to that Stoichiometry stuff we started earlier.

I'm going to start off with the "some other stuff" part. First up on this course outline is mass to volume and volume to moles calculations... wait, haven't we already done that? I guess maybe it's time for a recap... First you need to remember this handy little formula:

n = (V/22.71) where n = number of moles and V = volume of gas at STP

When doing volume to moles calculations, just whack the volume into the equation and BAM! Conversion done.

Mass to volume is slightly harder. Convert the mass to moles, and then rearrange the above equation to convert moles to volume. Look, I'll even rearrange the equation for you so you don't have to think about it.

V = 22.71n

Limiting reagents is just the next step to the stuff covered in Simple Calculations Involving Moles. Basically, in limiting reagent questions, you get told how much you have of more than one of the reactants. You have to figure out which one is the limiting reagent (i.e. which one will run out first). To do this, first you convert the masses or volumes or whatever of all of the reagents given to moles. Then you look at the ratio between the number of moles and the coefficients to see if they line up.

Let's just say that one substance (let's call it Substance A) in the equation doesn't have a coefficient before its formula, while another substance (let's call it Substance B) has a coefficient of 2 before its formula. Since coefficients tell you the ratio between number of moles of substances in the equation, these coefficients are telling you that the ratio between substances A and B is 1:2 (i.e. Substance B has twice the number of moles of Substance A). If you have the number of moles of both substances, you can then check to see if they fit the ratio.

For example, if you have 3 moles of Substance A, and 7 moles of Substance B, the ratio doesn't quite fit the 1:2 ratio as above. You have more Substance B than is required; therefore, Substance A is the limiting reagent. Similarly, if you have 3 moles of Substance A but only 4 moles of Substance B, you don't have enough Substance B (you need 6 moles), so Substance B is the limiting reagent. One easy way to work out which substance is the limiting reagent is by dividing the number of moles of each substance by their respective coefficients. Unfortunately Chem markers nowadays prefer that you write "if/then" statements rather than pure mathematical equations (which is a bummer because I reckon mathematical equations explain stuff better). For example, you could write something along the lines of "If I have 3 moles of Substance A, then I need 6 moles of Substance B, but I have 7 moles, which is more than I need, so Substance A must be the limiting reagent" or "If I have 3 moles of Substance A, then I need 6 moles of Substance B, but I only have 4 moles; therefore, Substance B is the limiting reagent."

Once you find your limiting reagent, you can do all of the other calculations that the question requires, basing your calculations off the number of moles of the limiting reagent.

The last thing I need to cover is Empirical Formulae. I've left this until last because I think I really do need to use an example to explain how to do these (mind you, some of the topics I've covered before probably also need examples but I'm too lazy to provide them). An empirical formula is basically a formula showing all of the atoms in the simplest ratio. For example, propane's molecular formula is C3H6, but its empirical formula is CH2 (just divide both subscripts by the lowest common denominator). They're pretty easy to work out though.

1. Work out the mass of each element in the sample. This is the hardest step and one day I'll have to provide you with a more diverse range of examples to show you all the different ways you might have to work out the mass of each element in the sample. However, sometimes you might get lucky and have the masses spoonfed to you- always nice!
2. Work out the number of moles of each element.
3. Divide the number of moles of each element by the number of moles of the element with the smallest number (sounds confusing, but it might make more sense when I provide an example).
4. If each number of moles is within 0.1 of a whole number (like 1.05 or 2.01), then round each to the nearest whole number to get your simplest ratio. If not, you might have to multiply each number of moles by 2 or 3 or whatever in order to make each number of moles within 0.1 of a whole number and get a simplest ratio.
5. Write down the elements with the ratio numbers as subscripts. This is your empirical formula.

Now here's an example!

An oxide of iron was reduced to metallic iron by reaction with hydrogen gas. 11.2g of iron oxide formed 7.83g of metallic iron. Calculate the empirical formula of the iron oxide.

Now, an oxide of iron is basically a compound containing iron and oxygen.

This question doesn't provide the numbers straight away, so I guess we'll have to work them out.

Assuming none of the other products of the reaction contained iron, the 7.83g must be the mass of all of the iron in the original compound (remember, mass doesn't increase or decrease as the result of a reaction). To work out the mass of the oxygen, therefore, just subtract 7.83 from 11.2 to get 3.37g.

Now we can work out the empirical formula! I've put my working in a table because I find that it makes my working out neater and easier to follow.

Yay, pretty colours.

The next step is working out the molecular formula, if the question asks you to do so. You can only work out the molecular formula if you're given the molar mass of the molecular formula.

Basically, there's only two steps involved. Work out the molar mass of the empirical formula by adding together the mass numbers of all of the elements, then see how many times the molar mass of the empirical formula of the mass number fits into the empirical formula of the molecular formula (i.e. divide the molecular formula's molar mass by the empirical formula's molar mass). Multiply the result by the coefficients of the empirical formula.

The above question has a part b) involving molecular formulae. Part b) is "[The oxide of iron's] molar mass was 159.6g/mol. Determine its molecular formula."

First we get the molar mass of the empirical formula:

(55.85 x 2) + (16 x 3) = 159.7g/mol

And then you divide the molecular formula molar mass by the empirical formula molar mass, rounding if necessary:

(159.6/159.7) = close to 1

Well, that makes things easy. 2 x 1 = 2, and 3 x 1 = 3. Therefore the molecular formula is the same as the empirical formula- Fe2O3 which is iron (III) oxide. (Yeah, I can't be bothered with subscripts any more, I'm just typing in the subscripts using a smaller front size.)

One last thing. Sometimes, instead of given the mass of each element, you're given the percentage by mass of the compound. This makes things easy. Percentage by mass is basically how many grams of each element you have in 100g of the compound (per cent = out of 100), so just pretend you have 100g of the compound and go from there.

So that's all from 2A Chemistry, unless you're desperate for me to explain the Applied Chemistry bits (describe and explain examples where rates of reaction have been altered in and around the home). 2B just has acids and bases, organic chemistry and oxidation and reduction (a.k.a. redox), but I hate redox, not because I'm terrible at it, but because I found it a rather annoying topic for some reason. ANYWAY, that's it from me for this post. I'll be back soon to talk about acids and bases!

Reactions and Equations part deux

There's just a little bit more that I need to cover on Reactions, Equations and Stoichiometry before we're done with Unit 2A.(If you ask me, all this semester-long unit stuff is a bit off. Having year-long chunks is bad enough. When will people learn that learning is a continuous process, not a fragmented one? Yeah, sure, you sometimes learn stuff in leaps and bounds, but not according to someone else's schedule.. okay, I'll shut up now.)

First up is a few standard equations for you to remember. Remember how I said that when mixing two ionic substances together, the cations bond with different anions? As in, if you have NaCl and KF as reactants, you'd get NaF and KCl as products? Well, we're going to expand on that a bit more.

Acid and Base make a Salt and Water

This is your standard neutralisation reaction. When you have an acid, for example HCl, mixing with a base such as KOH, the K+ ion bonds with the Cl- ion to form the salt KCl. Now you have the H+ and OH- ions left. Two hydrogens and an oxygen make water.

HCl + KOH -> KCl + H₂O

Acid and Carbonate make a Salt, Carbon Dioxide and Water

This is slightly more complex, but not really. When you have an acid, such as HCl (it's just about the easiest acid to type on Blogger...) and a carbonate, such as Na₂CO_{3 (which is not so easy to type on Blogger and requires some copy-pasting from Word), what you first get is that the ions swap around, just like the other reactions we've covered so far. The Na+ and Cl- ions would bond to make the salt, NaCl. The carbonate and hydrogen ions would originally bond together to make carbonic acid,} H₂CO_{3, but because carbonic acid is unstable, it decomposes to form carbon dioxide and water. (At least, that's my understanding of it- please correct me if I'm wrong.)}

_{2HCl +} Na₂CO_{3 -> 2NaCl +} H₂CO_{3 +} H₂O

Note that I added coefficients to balance the above equation.

Acid and Metal make a Salt and Hydrogen Gas

When you have a reactive metal and an acid, the metal would first give up electrons to the hydrogen ions in the acid, becoming a positive ion. Then these positive ions would bond with the negative ions in the acid, forming a salt. (As for which metals are reactive- I'll talk about that when I talk about oxidation and reduction.) At the same time, the H+ ions receive the electrons from the metal, becoming H_{2 molecules, i.e. hydrogen gas. It probably all sounds a bit complex for now, but I promise it'll make sense when you learn about oxidation and reduction.}

_{Fe + 2HCl ->} FeCl_{2 +} H₂

The next important thing to talk about regarding chemical reactions is the writing of ionic equations. So far, we've been doing molecular equations. To convert molecular equations into ionic equations, just follow a few simple steps.

1. When you write out the molecular equations, make sure to write the state symbols as a subscript next to each substance. (I haven't been doing that so far, but hey, it's a pain writing subscripts on Blogger.) l for liquid, g for gas, s for solid and aq for ions in aqueous solution (i.e. dissolved in water).
2. Rewrite all of the aqueous substances out as individual ions. For example, write aqueous NaCl as Na+ and Cl-. Do this on both sides of the equation, but ONLY for aqueous substances (NOT liquids, solids or gases).
3. If an ion appears on both sides of the equation, cross it out on both sides.
4. The remaining substances make up your ionic equation. Yayyyyyyyyy.

Now I need to cover the last little bit of Stoichiometry of Unit 2A but I really can't be bothered doing that today, especially because it's stuff that I'll have to find examples for to explain it well (actually, a few of my other blog posts are in dire need of examples too). Maybe I'll get around to writing about Stoichiometry tomorrow. Maybe.

Reaction Rates

In my last post I said that I was going to talk about reaction rates next. Well, here's a post about reaction rates for you!

When we talk reaction rates, we're talking how fast a reaction occurs. There are different ways to measure reaction rates, for example, measuring how fast the products are used up (by measuring mass or volume) or measuring how fast the products are being produced (by measuring mass or volume).

The main concept to know here is the collision theory. The collision theory states that there are 3 conditions that have to take place for a reaction to occur.

1. The reacting particles must collide.
2. The colliding particles must have enough energy to react. (They must have more energy than the activation energy required- see Energy Effects.
3. The particles must collide with a favourable orientation.

We can then extrapolate on this to explain why certain factors affect the rate of reaction.

Concentration- If you have more particles in the same amount of space, then there's a higher chance that the particles will collide. More collisions in the same amount of time = faster reactions. This is a bit of a silly example, but if you squish more people onto a 6-car train, there's a higher chance that people are going to be squashed into one another. More squashiness in the same amount of time = faster spread of annoyance among passengers, particularly when the idiots at the next station try and squish onto an obviously already overcrowded train.

Temperature- Increasing temperature increases the average energy of the particles. Because of this, more particles will have sufficient energy to react. The increased energy also makes the particles move faster, making collisions slightly more likely.

State of sub-division and surface area- Dividing up substances increases the overall surface area. This means that more particles are exposed to the reaction, making the number of collisions greater.

Catalysts- Catalysts affect reaction rates through a range of different means. They might make the particles more likely to collide with a favourable orientation, for instance. The main effect of catalysts is a reduction in the activation energy required for a reaction to occur. Here's the same diagram that I used in the last post, but with an extra line showing a possible catalysed pathway.

As you can see, if a catalyst is present, the activation energy is not as high, but the overall enthalpy change remains the same.

There's an extra dot point here about explaining the relationship between collision theory, kinetic energy distribution graphs and the rate of a reaction. I feel that I've alluded to it enough when I talked about temperature, but if you want me to elaborate some more, just ask!

Energy Effects

First things first: there's a Law of Conservation of Energy, just like how there's a Law of Conservation of Mass (which states that mass cannot be lost or gained, only changed to another form, which I alluded to in my post about chemical equations). The Law of Conservation of Energy essentially states that energy cannot be gained or lost, it can only be converted to another form or transferred elsewhere. For example, a chemical reaction (a.k.a. the "system") might absorb energy from the surroundings, or release it to the surroundings. In fact, each of these types of reactions has a special name- exothermic reactions release energy from the system to the surroundings (I think that "ex" or "exo" means "outer," like in "exoskeleton" and "export") while endothermic reactions absorb energy from the surroundings into the system.

Now there's this other term called "enthalpy," but I'm not entirely sure what it means because, apparently, the true meaning of this term is beyond the Chemistry 2AB course. So far it seems that enthalpy is sort of like potential energy. It's represented by the letter H.

The next thing I need to tell you is how bond breaking and bond making, the basis of pretty much all chemical reactions, relate to exothermic and endothermic reactions. Well, you see, most bonds need at least some energy to break. Energy must be absorbed into the system to break the bonds. The next step in the reaction is the formation of the bonds. Stronger bonds release more energy into the surroundings than weaker bonds. If the overall net effect is the absorption of energy into the system, then the reaction is endothermic. If the overall net effect is the release of energy into the surroundings, then the reaction is exothermic.

By the way, because the formation of stronger bonds release more energy into the surroundings, breaking stronger bonds requires more energy from the surroundings. The inverse is true for weaker bonds: forming weaker bonds releases less energy, while breaking weaker bonds doesn't require as much energy from the surroundings.

All of this can be summed up in some nice little enthalpy diagrams. Here's an enthalpy diagram for an endothermic reaction:

The delta H is the net change of enthalpy in the system.If it's positive, then the reaction is endothermic. If it's negative, the reaction is exothermic. When including enthalpy in a chemical equation, you can either add a "delta H = x kJ" at the end of the equation, or you can add it in the actual equation. If it's endothermic, you need to add the enthalpy change in the reactants side, as energy is being absorbed into the system; if it's exothermic, you need to add it in the products side as energy is being released. Just think about the nature of the reaction to help you remember which side to put the enthalpy change on.

The Ea, or activation energy, is the amount of energy required to break the bonds. The transition state or activated complex is where the original bonds are being broken and new bonds are being formed. In the above diagram, it's easier to see that the products have a higher enthalpy (potential energy?) than the reactants.

And that's pretty much it about energy diagrams, unless you can find anything else that I need to explain. Next up on the agenda is reaction rates! Yayyyyyyyyyyy.

Solutions

We're on to Term 2 Chemistry now, and the first topic is... solutions! No, not solutions as in answers, but solutions as in homogeneous mixtures. (If you can't remember what a homogeneous mixture is, go to my post about Classification of Matter.)

This is another topic I'm going to go through dot point at a time since the dot points don't follow on from each other that well.

Identify, explain and give examples of saturated, unsaturated and supersaturated solutions

To make things a bit less wordy for me to explain, here's some terminology for you. The solute is the substance that dissolves. The solvent is the substance that the solute dissolves in. For example, when you dissolve salt in water, salt is the solute and water is the solvent.

A saturated solution has the maximum amount of solute that the solvent can hold at that temperature. An unsaturated solution has less than the maximum amount while a supersaturated solution has more than the maximum amount.

But wait, you may ask. How can you have a solution containing more than it can dissolve? Well, you see, generally solubility increases with temperature (if I remember correctly, the exceptions are mainly gases which have decreasing solubility with increasing temperature). If you heat up the solvent and dissolve the solute, sometimes most of the solute will stay dissolved in the mixture even as it cools down. This is one way you can get a supersaturated solution. I'm sure there are other ways, but I don't know what they are.

By the way, I was curious as to why solubility increases with temperature, so I Googled it. It seems that increased temperature makes the particles of the solvent spread apart more, allowing more room for other substances to be mixed in to the solution.

Apply solubility rules to predict when two dilute ionic solutions are mixed

When two ionic solutions are mixed, generally what happens is that anions (negative ions) of the two ions swap places (or the cations/positive ions, if you prefer). For example, when NaCl and AgBr are mixed, the resulting products would probably be NaBr and AgCl. To check their solubilities, just look at the solubility table on your data sheet. I don't think I have to explain how to read a table... I hope I don't have to explain, anyway.

Use the colour of ions to identify reactants and the products in chemical processes

The main thing here is just to be logical and use process of elimination or whatever to figure out what the ions are. I don't have an example to show you though.

Explain colligative properties including the effect of concentration on vapour pressure, melting point and boiling point of a solution

This is something I need to revise. Stay tuned...

Describe the characteristics of strong/weak/non-electrolytes and give examples

Strong electrolytes completely dissociate or ionise in water. Weak electrolytes partially dissociate or ionise in water. Non-electrolytes don't ionise in water.

The difference between dissociation and ionisation is as follows: dissociation is when ions break up in water, while ionisation is when a neutral particle breaks up in water and becomes ions. Just remember that you can't ionise something that's already an ion.

Now for examples. Strong acids and ionic compounds are all strong electrolytes. Even insoluble ionic compounds are considered to be strong electrolytes, if I remember correctly. Weak acids and bases are examples of weak electrolytes. Hydrocarbons (substances containing mainly hydrogen and carbon) are examples of non-electrolytes.

Explain the difference between concentrated and dilute solutions of strong and weak electrolytes

In chemistry, there's a difference between concentration and strength. Concentration is how much of the solute you have in that amount of solvent. Strength is how much the solute dissociates or ionises. A 0.1M solution of HCl is a dilute solution of a strong acid- dilute because there isn't much HCl in the solution, but strong because HCl completely ionises in water. Meanwhile, a 6M solution of acetic acid is a concentrated solution of a weak acid- concentrated because there's lots of acetic acid, but weak because acetic acid doesn't completely ionise in water.

When I feel like it, I'll go back and find out more about how concentration affects properties of solutions. But until then, TTFN!

Term 1- Applied Chemistry

Whew, I've been through all of Term 1's work! Yay!

We're going to finish off Term 1 with the "Applied Chemistry" section on my course outline. Basically the "Applied Chemistry" section is there to show you that chemistry has real life applications outside of passing Chem tests and looking good on your WACE.

For Term 1, my course outline just says that we need to know how different types of bonding can contribute to the properties and uses of different household objects. It basically takes what we know about bonding to the next level.

For example, why are wires made of metal coated in plastic? Well, the metal conducts electricity due to all of those delocalised electrons, while plastic, being covalently bonded, has no freely-moving charged particles to speak of, so it's a non-conductor- great for preventing you from getting electrocuted!

There's not really too much else I can explain here, unless you really want more examples. Just think about what the object in question is made of, how it's bonded, and the properties of that kind of bond, and you should be fine.

Heating Curves and Cooling Curves

Okay, I'm going to make this rather short.

Heating curves are reasonably simple to understand. Here's a general graph of a substance being heated.

A (diagonal line): The substance is solid. Its KE is increasing while its PE remains constant. Therefore, the temperature increases but there is no significant change in the distance between the particles.

B (straight line): The substance is melting. Its PE is increasing while its KE remains constant. The particles begin to move further away. The temperature doesn't increase because the substance is too busy melting. I think this part is also called the latent heat of fusion.

C (diagonal line): The substance is a liquid. Its KE is increasing while its PE remains constant.

D (straight line): The substance is boiling. Its PE is increasing while its KE remains constant. I think this part is also called the latent heat of vaporisation.

E (diagonal line): The substance is a gas. Its KE is increasing while its PE remains constant.

Oh yeah, there's also cooling curves. They're basically heating curves backwards (though the axes obviously stay in the same place). I really can't be bothered drawing one, but I will if you want me to.