Making a Cake- Part I

This is not a lesson dedicated to anything in particular, but several useful ingredients you will need when you make your cake. 



States
To describe the state of a compound, scientists use subscripts.
gaseous = (g)
liquid = (l)
solid = (s)
aqueous = (aq) -dissolved in water

Diatomic Molecules
Some elements occur naturally in the environment as a pair(H2, O2, Br2, I2, F2, N2, and Cl2)

You know how 7 is supposed to be such a lucky number. Well there are 7 diatomic elements, arranged almost like the number 7!  Hope this will help you memorize them!

Other molecules just exist in the environment in large groups.
Some include S8 and P4.

Naming Compounds Review

Although it hasn't been that long, let's review how to name covalent and ionic compounds.

Covalent Compounds
-2 non-metals
-use prefixes


Carbon dioxide = CO2
Carbon tetrafluoride = CF4
HF- hydrogen monofluoride
*you don't need to put 'mono' in front of the first particle


Dry ice- solid CO2














Here's a link to some practise problems!
http://www.pafaculty.net/biology/keith/KR_Graph_site/Nomenclature.html

Ionic Compounds
-metal and non-metal
-name metal first, then non-metal
-suffix ending on the second particle(ide)

NaCl = Sodium chloride
Fe2O3 = Iron(iii) oxide
*multivalent (more than 1 charge)
Magnesium Sulphide = MgS

Here's a link to some practise problems!
http://www.pafaculty.net/biology/keith/KR_Graph_site/Nomenclature.html

Now some more fun stuff! Acids and bases.
Acids
-have H in them
-usually start with H
-dissolved in water (aqueous)

-HCl = hydrochloric acid (Cl is an element)
-H2SO4 = sulphuric acid (sulphate- ends ate -> ic)
-HClO2 = chlorous acid (chlorite - ends ite -> ous)


Citric Acid found in lemons











Bases always have the compound hydroxide in them.

Molarity

Molarity is a concentration unit that can be measured by the number of moles of solute per liter of solution. 

The molarity formula is: 

                               moles of solute (mol)

                               ___________________

                                

                               volume of solution (L)

it can also be written as:

 

                             M= N/V

where N is the number of Moles and V is the volume

the unit symbol for mol/L is "M", which can be written as "molar". 

Here are some examples:

If 3.0 L of solution contained 6.0 Mol of FeCI, what is the molarity of the FeCI?

molar concentration =  6.0 mol

                                ________  = 2.0 M

                                   

                                  3.0 L

 

                          Therefore, the molarity of the FeCI is 2.0 M.

Percent Composition

Percent composition is used to discover what percentage of the compound a certain element makes up, and it is actually quite easy to find. You just need to follow easy steps:

1. calculate the molar mass of the compound
2. find the molar mass of each element. Then divide the compound's molar mass by the element's molar mass
3. multiply by 100 to get the percentage (round to the nearest tenth!)

                               mass of element    
                             mass of compound    x 100  = % composition

Ex. What's the percent comp. of the compound Na₂O?

Molar mass of Na -> 23.0 g x 2 atoms = 46 g
Molar mass of O -> 16 g
Molar mass of Na₂O -> 46 g + 16g = 62 g

Next step is to divide each element mass by the mass of the compound:

46 g
62 g  = 0.741                        0.741 x 100 = 74.1 %

The percentage of Na 74.1%

16 g
62 g  = 0.258                        0.258 x 100 = 25.8 %

The percentage of O is 25.8 %

74.1% + 25.8% = 99.9%

**When you add the final percentages together, they should equal ~100 %. That's how you know your answers right!**

Conversions for Empirical & Molecular Formulas!

We've already learned about the Mole, Avogadro's number etc... but now is the time to apply what we have learned previously into more complex situations!

It is necessary to know the 2 different formulas: Empirical and Molecular

Empirical formula - the smallest ratio of moles or moles in a compound.
Molecular formula - all atoms that form the compound.
For example...
                                               

                                                 N₆O₉<-- molecular formula  

                                                 N₂O₃<-- empirical formula

                                    

Can you tell how the two relate? The molecular formula still has the same ratio - N₂O₃- except everything is multiplied by 3.

Converting for Empirical and Molecular formulas!

How to find the Empirical Formula:
To find the E.F, simple conversions must be made. For example, if the number of grams is given, we must convert grams -> moles.

Ex. A compound contains 28 g of carbon and 72 g of oxygen. Find the E.F.

     28 g C    1 mol C                                                       72 g O    1 mol O
                x   12 g C  = 2.333 mol                                               x  16 g O  = 4.5 mol

Once we've converted the grams into moles, we want to divide each mole value by the smallest mole value (this is to get one of our mole values to equal 1, making the conversion to whole numbers simpler)

                    2.333 mol C                                           4.5 mol O
                     2.333 mol = 1 mol                              2.333 mol  = 1.95 mol

You see how there are 1.95 moles of oxygen? Well, we want to leave our answers in whole numbers, so we can round 1.95 to 2. Since the moles of carbon is 1, we can just leave it as a one. Therefore, the E.F is CO₂.

How to find Molecular Formula
To find the M.F, we calculate the molar mass of the compound.

Ex. A gas has the E.F of N₂O₃. What's the molecular formula if there are 228 g per one mole?

N₂ --> 28 g
O₃ --> 48 g
48 g + 28 g = 76 g

The molar mass of N₂O₃ is 76 g.

Next step is to divide the number of grams per mole by the molar mass!

228 g
76 g   = 3

Now that we know that there's 3 times the amount of grams in the original mass than in the molar amount, we can just multiply the E.F N₂O₃ by 3

3(N₂O₃) = N₆O₉

Therefore, the M.F. is N₆O₉.

We can double check to see if it's right:

N --> 6 atoms x 14 g = 84 g
O --> 9 atoms x 16 g = 144 g
144 g + 84 g = 228 g

YES! The answer matches the original mass, so we know our answer is correct!     

Want to learn more about Empirical and Molecular Formulas? Check out these great links!
Empirical Formulas  http://www.youtube.com/watch?v=r2Log6-voWo
Molecular Formulas http://www.youtube.com/watch?v=nslC7lOSc7Y

Diluting Solutions

Chemical products come in different forms, but the most convenient way to get these products around is by their most concentrated form. These diluting solutions can tell us the amount of water that is used and can tell us what the concentrated form was before water is added.

Please note that you must know what solvent and solute is. If you do not know those terms then you are in trouble. You can learn them easily, yes, but know that teachers and I would be very disappointed.

Things to keep in mind:

• the moles of the solute is always constant
• Moles Solute before = Moles Solute after
• the "before" is the concentrated form and the "after" is when water is added
• these are in terms of Molarity and Volume

Equation: M1 L1 = M2 L2

Make sense? Molarity->Moles & Volume->Litres. Pretty easy.

It's all now in the matter of substituting the information given in the question, then using basic algebra to find the missing molarity or volume. Sometimes it will ask to set up the equation because not enough

 information is given, so you just have to determine what you will need. If you are asked for the amount of water was used, you simply subtract the two volumes.

Sometimes it gets tricky by involving only in grams and litres or g/mol. Remember that the grams have to be converted into moles first in order for the equation to work and for g/mol you will have to divide it by what  M1 x L1 equal to, to get your moles; simple algebra again to figure volume.

Molar Volume of a Gas a STP

As we all know, gas changes in volume. Gas expands and contracts from different changes in temperature and pressure. STP stands for Standard Temperature and Pressure and it is to compare the volume of the a gas. STP equals to 1 atmosphere with a temperature of zero degrees celcius ( also know as 273.15K ).

Mole Map

The mole map above shows that 1 mole occupies 22.4L. In this case, we can make the new conversion of a gas, being:

22.4L of Gas                1 mol of Gas

1 mol of gas      (or)    22.4L of gas

Example:
Let's try to calculate the volume of 3.4g of Carbon Dioxide at STP...
First you would want to find the molar mass of the compound:
(1 x 12) + (2 x 16)= 48g/mol

Next, find the moles of carbon dioxide:
3.4g x 1 mol = 0.07 moles
             48g

Then you find the molar volume:
 22.4L  x 0.07 moles = 1.6g
1 mole

which means, approximetly 1.6g of carbon dioxide is volume occupied at STP.

Mole Conversions Part 2

Now it's time for a little quiz!

Short Mole Conversion Quiz

The Magic of Hydrates!


 

In lab 4C we were taught how to find the percentage of water in an unknown hydrate! Cool, eh? And it was simple too. First, you may be wondering what a hydrate is... 

-A hydrate is a compound/substance containing water molecules.



The hydrate

CuSO₄ • 5H₂O

When the hydrate is heated and the water is evaporated, it becomes an anhydrous salt.

-An anhydrous salt is a salt containing no water.

 

The anhydrous salt

CuSO₄ • 5H₂O

Alas, let's start the lab!!

Materials:

-lab burner                                                

-crucible

-crucible tongs

-pipestem triangle

-ring stand & ring

-centigram

-safety goggles

Chemical Reagents:

-approx. 5 g of a hydrate

Process:

1.  Put on safety goggles and tie any loose hair back. Remember; safety first!
2. Set up your ring stand (screw on the iron ring to the stand, then put the pipestem triangle on top of the ring.) Also, plug in your Bunsen burner into the gas valve. The iron ring should be about 5 cm above the Bunsen burner.
3.  Once you have your Bunsen burner and ring stand set up, place the crucible on the pipestem triangle. Heat the crucible for 3 minutes to ensure it is dry. *your fire should be blue, NOT orange! If its orange, it'll release carbon bits on to your crucible!*
4. Now, remove the crucible off the burner with your crucible tongs and let it cool. To check if it's cool, bring the back of your hand close to the crucible. If you can still feel heat, let it cool more!
5. Measure the mass of the crucible and record it
6. Put some of the hydrate into the crucible until it's about 1/4 full. Then, weigh it again. Record the mass of the hydrate.
7. Place the crucible back on the pipestem triangle and heat it for about 5 minutes.
8. Turn off the burner off and allow the crucible to cool. (Remember to use the backk-of-the-hand trick to determine whether it's cool or not). Once its cooled, record the mass of the crucible and the hydrate.
9. Do another heating. This is to ensure ALL the water is evaporated. Heat the crucible again for 5 minutes, let it cool, then record your masses. (The two masses should be 0.03 g apart at maximum.)
10. After having recorded all your masses, cool the crucible and add a drop of water back into the crucible. Record any changes the anhydrous salt undergoes.

Record Table:

Mass of empty crucible
Mass of crucible and hydrate
Mass of hydrate
Mass of crucible and anhydrous salt (first heating)
Mass of crucible and anhydrous salt (second heating)
Mass of anhydrous salt
Mass of water evaporated
Mass of one mole of anhydrous salt
Describe any changes after adding a drop of water

How do we find the % of water given off?
Well, that's simple! Subtract the mass of the anhydrous salt from the mass of the hydrate. Then divide the mass of water given off by the mass of the hydrate. Multiply by 100 to get the percentage of water burned off. To get the mass of the salt, divide the mass of the salt by the mass of the hydrate. *A good way to see if your calculation is right is to add the percentages together-they should add up tp approximately 100.*

Example:

If you have a hydrate that weighs 3.20 g and an anhydrous salt that weighs 1.95 g, what is the percentage of water given off?

3.20 g   <--mass of hydrate
1.95 g   <--mass of anhydrous salt
1.25 g   <--mass of water

1.25 g  
3.20 g  = 0.399625

0.399625 x 100 = 40.0% - Don't forget sig figs!

The percentage of water given off is 40.0%

1.95 g   <--anhydrous salt
3.20 g = 0.609375

0.609375 x 100 = 60.9 %

The percentage of the anhydrous salt is 60.9%

40.0 + 60.9 = 100.9  <-- the mass of the anhydrous salt and water is ~100, therefore the answer is correct!

Conversions

Here's the fun part, converting between grams, moles, and atoms! Yay! 

Atomic mass: the average mass of the subatomic particles (protons, neutrons, [electrons don't have enough mass]) that make up an atom
-abreviations: amu, daltons, and u


Formula Mass: the sum of atomic masses of all the atoms in a formula of an ionic compound
Eg.) BeS= 9.0amu + 32.1amu = 41amu (s.f.)

Try the following questions: Calculate the formula mass.
1) NaCl
2) Magnesium Fluoride
3) Li2SO4
4) Copper(ii) nitrate

Molecular Mass: the sum of the atomic masses of all the atoms in a formula of a covalent compound
-think molecular = molecule = covalent compound
Eg.) CN= 12.0amu + 14.0amu = 26.0amu

Try the following questions: Calculate the molecular mass.
1)CO2
2)Chlorine gas
3)ClO3
4)Carbon tetrafluoride

Molar Mass: the mass of 1 mole of an element
-every element has a different molar mass due to the mass of particles in the nucleus
-molar mass is always measuring the mass of 6.022 x 10^23 atoms
-measured in grams/mole (g/mol)
-atomic mass of an element = molar mass of the same element in g/mol
The molar mass is 9.01g/mol.

Try the following problems: State the molar mass.
1)Vanadium
2)Bromine
Determine the element that has a molar mass of:
1)27.0g/mol
2)83.8g/mol

Relative Mass & Avogadro's Hypothesis

Back when humans knew hardly anything about elements and their properties, scientists found the masses of several sustances: H, O, and CO2.  They found a very curious thing; that the same amounts of different gases had a constant mass ratio.
O:H always had a 16:1 ratio
CO2:O = 11:8

Relative mass: a ratio of the mass of an atom relative to 1/12th the mass of carbon (assigned a mass of 12 amu)

note: amu = atomic mass unit

First they used oxgen(16 amu) but later switched to carbon. 

This all works because of a great man called Avogadro.  He was an Italian scientist who discovered the mole(also called Avogadro's constant) and developed his hypothesis (Avogadro's hypothesis). 

Avogadro's Law- at the same temperature and pressure, equal volumes of different gases have the same number of particles. 


Check out this link for a more detailed explanation!
Avogadro's Law explanation-youtube clip

Moles

So, what is a mole?  If you haven't been listening, you might think it's a brown furry creature.

However, in chemistry, we have another definition. A mole represents a specific quantity, like 12 somethings equal a dozen.  But it represents a much larger number. 

6.022 x 10^23

1 mole always represents the same number. 
If you have 1 mole of donuts, you will have 6.022 x 10^23 donuts. 
If you have 1 mole of eggs, you will have 6.022 x 10^23 eggs. 

How much is this? Here's a couple examples to help you visualize.
This is the Sahara desert.

This cup of water(aprox. 10 moles) contains more moles of H2O than grains of sand in the Sahara.   

Eg.2) If you took 1 mole of marbles, and covered the Earth's surface with them, the layer would be around 80 km thick.
 

Moles are very important, especially in chemistry because they make counting particles possible.