Element Symbols

Not all elements are labelled simply. You have the standard C for carbon, or H for hydrogen, which are nice and straightforward but its no fun being easy to understand. Science likes to try and confuse you sometimes. This means that not all chemical symbols make sense. I mean, look at sodium. Its chemical symbol is Na. How on earth do you get Na from sodium?

As it turns out, scientists back in the Middle Ages, liked to speak Latin. And so, all our elements that we’ve known about forever, were originally given a Latin name. So sodium, before it was known as sodium, was actually called natrium. Makes sense now.

Here is a list of all the elements with funny symbols and their original names.

Antimony (Sb) – Stibium

Copper (Cu) – Cuprum

Gold (Au) – Aurum

Iron (Fe) – Ferrum

Lead (Pb) – Plumbum

Mercury (Hg) – Hydragyrum

Potassium (K) – Kalium

Silver (Ag) – Argentum

Sodium (Na) – Natrium

Tin (Sn) – Stannum

Tungsten – Wolfram

Evolution: The Finale!

The last point in our series on evidence for evolution is actually quite simple. It is the age of the Earth. We know, having studied our planet and space, that the Earth is approximately four and a half billion years old. Life has been around for about 3500 million years. This information begs the question – how could life not have changed in that time? There is no way that all life would have remained stagnant and exactly the same for all that time. Having gone from volcanoes and lava covering the Earth, through to forests covering the world to different environments in different places (but at the same time) – how could nothing have changed?

And now I release you, my little scientists, to go and educate the world about the evidence for evolution. And for the little genius out there reading this that finds the evidence to make it a law – you win the Nobel Prize. Congratulations.

Evolution: Comparative Embryology.

So you know by now that when a Mummy and a Daddy love each other very much, they have sex. Well, I’d hope you do by now anyway. And as you (again, hopefully) know, having sex can create a baby. And a baby starts out life as just a few small cells and then grows from there. By the end of about 8 weeks in humans, the child has basic limb shapes and is well on their way to being an incredible annoyance to everyone in the family.

What scientists have observed in a multitude of different animals is that during the early stages of development in the womb, many organisms develop in the same manner before becoming differentiated (fancy word for different).

Again, this points towards the suggestion of a common ancestor.

The next (and final) card in our complex house of cards is: A Surprise! You’ll have to wait and see.

Evolution: The Pentadactyl Limb

Comparison of the anatomy of various plants and animals provide indirect evidence of their evolution from common ancestors. The front flipper of a seal, a cat’s paw, a horse’s front leg, a bat’s wing and your own hand all look different and perform different functions. However, they all consist of the same number of bones, muscles, nerves and blood vessels arranged in a similar basic pattern. The basic pentadactyl limb (a limb with 5 digits) can be traced back to the fins of a certain fish from which the first amphibians are thought to have evolved. These fundamentally similar structures are called homologous structures.

 

 

 

 

 

 

 

 

 

 

 

 

Basically, there are a lot of animals (mostly mammals) that have a very similar arm structure. They have 1 upper bone, 2 lower bones and lots of bones that make up “fingers”. All of this is pointing towards the idea of them having a common ancestor.

Next in our line up: the reclusive Comparative Embryology.

Evolution: Biogeography or Geographical Distribution

Yes, I know I said a dirty word; geography, but believe it or not, geography is actually kind of useful in Science. And all I’m really asking for you guys to know is the name of most of the continents (there are 7) and the general shape.

Biogeography is the study of the distribution of plants and animals, both now and in the past (bio/biology = living things, geography = places). What you start to find (and this is evidence for both the theory of evolution and the theory of plate tectonics) is that along the borders of the continents, you find some very similar, if not the exact same, fossils.

Scientists used these similarities and came up with the idea that at some point in the Earth’s history, all the continents were joined together. Then they started to separate and become the landmasses that we know and love today.

This is all in the lead up to help explain this next idea. Scientists propose that, when the continents started to drift apart, there were living organisms! Shocking, I know. As the continents moved and the environments changed, it is suggested that the organisms adapted to better suit the environment. The example most commonly given is that of the flightless bird. We all know that the emu can’t fly. We all know the ostrich can’t fly. Now, hands up any of you who mixed these two birds up occasionally when you were younger?

Understandable, really. They’re both creepy, angry looking birds. I’m getting a bit paranoid one of them is watching me, just thinking about them.

Anyway, other flightless birds also include cassowaries, kiwis, rhea and penguins. Think about all of the countries that you would find these birds in. You can start to see why we might say that they all evolved  from a common ancestor, right?

In summary, the next point in our evolution theory evidence stack is the similarity between species in vastly different environments.

Next topic up for discussion: The Amazing, Stupendous, PENTADACTYL LIMB!!

General Info

Okay, so for those of you who happen to be reading this (but aren’t part of my assignment), this is a “technologically mediated learning tool/resource”. It has been designed/written for University of Wollongong, Chem101 students for an education assignment. I am studying to become a high school science teacher.

Girls, you are welcome to leave comments or email me with any questions or suggestions you may have. Start from the bottom (periodic table) and work your way up (redox).

Please, feel free to offer suggestions for improvement because chances are I’ll keep adjusting it once I have some spare time.

Hope its useful and interesting!

Redox Reactions

Having covered oxidation numbers, we can now look at redox reactions (remember that reduction and oxidation always occur together).

When we deal with reduction and oxidation we usually use half equations. We usually do this when we set up an electrochemical cell. This is when the sheet that is usually on the back of the Periodic Table comes in very handy. It is a list of reduction potentials. This is the list we use to determine what is reduced and oxidised in reactions.

Handy hint: if you ever get confused as to which way it goes, just remember that fluorine likes gaining an electron sooooooo much more than potassium.

The higher the number (or the biggest positive number) is the species that is going to be reduced.

For the half equations, remember to write the electrons on the right side of the equation.

Fe²⁺ + 2e^– → Fe

This is the reduction half equation for iron (II). Oxidation half equations have the electrons on the other side (all you have to do is reverse the equation given on the sheet).

If you have 2 half equations and need to add them together, you need to make sure that the electrons cancel each other out.

We’ll work through a copper and zinc example.

Cu²⁺ + 2e^– → Cu                            E= 0.16V

Zn²⁺ + 2e^– → Zn                            E= -0.76V

Clearly, the copper half equation is going to be reduced. This means we now flip the zinc half equation and because we reverse it, we also change the reduction potential sign.

Zn → Zn²⁺ + 2e^–                            E = 0.76V

Its nice and easy because the electrons already equal each other but sometime you might need to times one of the equations by something to make them equal (think lowest common multiple stuff from maths).

We add them together now and get our final equation and the voltage generated.

Zn + Cu²⁺ → Zn²⁺+ Cu              E = 0.92V

Easy as, right?

So they are the basics of redox reactions. The redox potential isn’t really that important unless you’re trying to calculate the voltage running through an electrochemical cell.

We’ll try something a little bit harder now. These aren’t so much constructing the redox equations, as balancing them out.

Mn 2+  +  BiO3 –  ===>  MnO4 –  +  Bi 3+

First, we split them up. Our two main species/elements Mn and Bi.

Mn²⁺  →   MnO₄^–

So, now we balance the Mn, which is fine. Next, the oxygen gets balanced. There are 4 oxygen atoms on one side, there needs to be 4 on the other side. This is done by adding water to the other side.

4H₂O + Mn²⁺  →   MnO₄^–

We balance the hydrogen now (quick note – this is only for acidic solutions, basic solutions have a slightly different process). Add hydrogen to the other side, so that it is balanced.

4H₂O + Mn²⁺  →   MnO₄^– + 8H⁺

So, as long as all the species present are balanced, we get to balance the charge now. Each side always need to be equal.

We have a +2 charge on one side and an overall charge of +7 on the other side. So we need 5 electrons to balance it out completely. After you’ve done that, you should have a very pretty equation that looks like this:

4H₂O + Mn²⁺ →    MnO₄^– + 8H⁺ + 5e^–

Have a go at the other half equation with bismuth.

You should end up with something that looks like this:

6H⁺ + BiO₃^– + 2e^–  →  Bi³⁺ + 3H₂O

Now, we add both equations together. No, wait! The electrons need to be balanced first. To do that we have to get the lowest common multiple for the electrons. Two electrons and five electrons; the lowest common multiple is 10.

2x (4H₂O + Mn²⁺ →    MnO₄^– + 8H⁺ + 5e^–)                                  8H₂O + 2Mn²⁺ →    2MnO₄^– + 16H⁺ + 10e^–

5x (6H⁺ + BiO₃^– + 2e^–  →  Bi³⁺ + 3H₂O)                                        30H⁺ + 5BiO₃^– + 10e^–  →  5Bi³⁺ + 15H₂O

 

Cancel out any species that appear on both sides of the equation (like the electrons) and voila!

2Mn²⁺ 14H⁺ + 5BiO₃^–   →  5Bi³⁺ + 7H₂O + 2MnO₄^–

Our equation is now balanced. Well, I’m 95% sure that its balanced.

For basic solutions, its exactly the same with just one extra step. After you’ve added hydrogen but before you add any electrons, add an equal number of hydroxide ions to both sides. The hydrogen and hydroxide (on one side) form water.

That’s all there is to it. Simple, huh? Okay, maybe not at first but after practise (see the website below), it gets pretty easy.

http://wc.pima.edu/~skolchens/C152OL/Ch21/REDOX.htm

http://kurtniedenzu.cmswiki.wikispaces.net/file/view/Redox_Equations_WorkSheet.pdf   – I recommend this one

Click to access redox.pdf

Click to access redox_chem220.pdf

Enjoy!

Reduction and Oxidation

Moving in a slightly different direction, we’re going to start reduction and oxidation. They tend to go together in reactions and so are known as redox reactions in general.

A few quick definitions:

Reduction: a gain in electrons, a addition of hydrogen and/or removal of oxygen.

Oxidation: a loss of electrons, a removal of hydrogen and/or addition of oxygen.

Handy hint to remember which one is which – OIL RIG

Oxidation is Loss, Reduction is Gain. Its how I still remember them even after 5 years.

These next two terms are a little bit harder because they seem to be backwards to commonsense.

Reductant: The species being oxidised, or the species that causes reduction.

Oxidant: The species being reduced, or the species that causes oxidation.

In working with redox reactions, we have to understand the concept of oxidation numbers. We assign oxidation numbers to the various elements in a chemical equation. Some of them change, some of them don’t. My favourites tend to be things like hydrogen and oxygen because their oxidation numbers don’t change (+1 and -2 respectively).

The way we assign oxidation numbers is very similar to the way we balance equations, both sides have to equal the same thing.

Oh, and before I completely forget to mention it, an atom by itself has an oxidation number of zero.

I’ll work through an example or two and then its your turn *grins evilly*.

Lets start with aluminium hydroxide. Write down the chemical formula (we’ll assume it has a valency of 3).

Al(OH)₃

We know that oxygen has an oxidation number of -2 and hydrogen has an oxidation number of +1. The overall molecules charge is zero. This is where we throw in a bit of algebra.

Al + (-2×3) + (1×3) = 0

Al + -6 +3 = 0

Al – 3 = 0

Al = +3

As we can see, aluminium has an oxidation number of +3.

If we have the compound SO (just for fun, name it for me), what is the oxidation number of sulfur?

Remember that oxygen has an oxidation number of -2.

S + (-2) = 0

S = +2

Sulfur has an oxidation number of +2.

Now that we know how to assign oxidation numbers to different elements, we can tell if a redox reaction occurs. If the oxidation number changes, it is a redox reaction. If it doesn’t change, it isn’t.

N.B: A heads up, occasionally, very occasionally, the oxidation numbers of our standards (oxygen, hydrogen and some others) do change but not often.

Here are some worksheets to have  go at to practise assigning oxidation numbers but there are plenty more out there on the net.

Click to access oxidationNumbersWkst.pdf

Click to access Oxidation%20Numbers%20Worksheet.pdf

Moles and Theoretical Yields

Checklist time:

• Write the chemical formulas for compounds
• Write chemical equations for different chemical reactions
• Calculate moles and molarity

Knowing all these things, we can now calculate theoretical yields and limiting reagents. Well, to be honest, you kinda have to calculate both if you want to know one.

Down to business. This gets a little bit more tricky than what we’ve been through so far. It isn’t anymore difficult, the process is just a bit longer.

Let’s work through an example so that it makes a teensy bit of a smidge more sense.

First things first, write down the chemical reaction. What are you waiting for? Oh right, you haven’t been given it yet. Okay, lets see, hmm…I’ve got it! Its a fun reaction that Year 7 absolutely adore! We’re going to add sulfuric acid to zinc metal. Go ahead and write it down. Done?

H₂SO₄ + Zn →  ZnSO₄ + H₂

We’ll add an excess of sulfuric acid, just to make sure that all the zinc gets used up *claps hands over mouth*, whoops, I’ve just given you the answer to my next question about the limiting reagent. I’ll go through it anyways, just to be thorough.

Limiting reagent. In simple terms – the limiting reagent is just what we have the least of, it means that it limits the production of the products to a certain amount. Since we have added an excess of acid, it means that the zinc has to be our limiting reagent, its not like its self-renewing or anything (and now that just reminds me of that Tim Tam ad, mmm…chocolate). Having an excess actually makes it easier because we can now ignore the acid.

Alright, we’ve done our scientist thing and added the acid to a 5 gram strip of zinc metal. It has made pretty bubbles but now we want to know how much zinc sulfate we made.

First step?

That’s right, figure out how much zinc we had at the start (remembering that any chemical type calculations deal with moles).

Moles of zinc =  5 (g)/65.38 (g/mol)

Moles of zinc = 0.0764 mol

So, do you remember when I said that the stoichiometry of a chemical equation was important later? Well, its later now.

Basically, stoichiometry is a ratio. This example is nice and simple because it is 1:1. What that means is that 1 mole of sulfuric acid reacts with 1 mol of zinc to form 1 mol of zinc (II) sulfate and 1 mol of hydrogen gas. Except, that we don’t have 1 mol, we have 0.0764 moles. Which is fine.

To simplify, for a 1:1 ratio, its just moles of zinc = moles of zinc sulfate.

With that in mind, we can say that the moles of zinc sulfate formed = o.0764 moles.

And now all that’s left is to convert moles back into grams. We re-arrange our equation (moles=mass/molar mass) to calculate mass.

mass = moles x molar mass                                                     but remember to use the molar mass of zinc sulfate now.

mass = 0.0764 (mol) x 161.45 (g/mol)

mass = 12.33 g

Cools! We made nearly 13 grams of zinc sulfate. Well, theoretically we did. A theoretical yield is just how much you calculate that you should make. In labs, you usually calculate a % yield, which is just however much you actually made, divided by the theoretical amount.

So, if we only made 11.51 grams, our percentage yield is about 93%. Not bad!!

Just a quick note about stoichiometry. 1:1 ratios are easy-peasy. Throw in some different numbers and sometimes, its a bit hard to wrap your heads around. I won’t go through this next one completely but hopefully, you guys can.

We’ll add hydrochoric acid to lead (II) nitrate (not that they let us play with lead all that much anymore. Sad face).

2HCl + Pb(NO₃)₂ → PbCl₂ +2HNO₃

Assuming the equation is balanced (which it is but you should always check, you never know when we teachers will pull a fast one), we can see that we need twice as many moles of hydrochloric acid than lead (II) nitrate for it to react completely. So if I gave you a problem that said we had 3 moles of lead (II) nitrate and 5 moles of hydrochloric acid, which one would you say is the limiting reagent?

Yep, you got it, hydrochloric acid. So we’d only make 5 moles of lead (II) chloride at the very most.

Make sense?

Chemical Reactions and Moles

We can tick off writing chemical equations now, or at least we should be able to do that. That being the case, let’s move on to calculating how much of something we actually have and how much we can make.

Chemists have a few magic numbers but one of our favourites is Avogadro’s number (it’s one of our favourites because it is always given to us in exams). Avogadro was a smart cookie that decided that we needed a new measurement for our fun-fun-fun chemicals. He reckoned that we should measure elements and compounds in moles. One mole of a substance is equal to 6.022 x 10²³ atoms/molecules. This is actually how we get our atomic weights. The weights given on the Periodic Table is actually the weight of 1 mole of the element, so we call it the molar weight or molar mass. Some even refer to it as the molecular weight/mass, but any of those terms are acceptable.

Very important: molar mass is measured in grams per mole (g/mol or gmol^-1).

Okay now, calculating moles is actually pretty simple. All you do is divide the mass of the compound (in grams) by the compound’s molar mass.

Moles =  mass (g)/molar mass (g/mol)

Just quickly, we’ll go through an example. Let’s pretend that we have, say, 7.9 grams of magnesium bromide. First things first, write down the chemical formula for magnesium bromide. 3, 2, 1…pens down please. *Drum roll* Magnesium bromide is MgBr_2. 

Next, go calculate the molar mass of magnesium bromide. All you have to do is to add the molar weights together.

Molar mass = (24.305 +79.904 + 79.904) g/mol

Molar mass =  184.113 g/mol

We now have the mass and the molar mass, which means we can now calculate the moles.

Moles = 7.9 (g)/184.113 (g/mol)

Moles = 0.0429 mol              (I know, I know, my significant figures are horrible, but we’ll worry about that later)

So, we can now calculate moles. We can do much the same to calculate the molarity of a solution. We measure molarity is moles per litre (mol/L), meaning we divide the moles by the volume of the solution.

Molarity (M) = moles (mol)/volume (L).

Again, a quick example. Lets pretend that we dissolve the magnesium bromide into 500mL of water. What is the molarity of the solution?

Molarity (M) = 0.0429 (mol)/0.5 (L)

Molarity (M) = 0.858 mol/L

And that’s pretty much how it works in the basics. Check out the next post for theoretical yields of reactions.