Balancing Equation

Today we are gong to talk about how to make the number of atoms of each kind on the relevant side equal to those on the product side

Here are some rules :
1. balance the atoms which only occur in one molecule on each side
2. balance the whole group
3. dont jump all over an equation
4. in an elemental form

Examples~
1. Al + CuCl2 ----> Al2Cl3 + Cu
step 1 : (4)Al + (3)CuCl2 -----> (2)Al2Cl3 + (3)Cu
step 2 : u should check if the molecules on both sides are equal
(4Al , 3Cu, 6Cl = 4Al , 3Cu, 6Cl)

2. KOH + H3PO4 -----> K3PO4 + H2O
step 1 : (3)KOH + H3PO4 ----> K3PO4 + (3)H2O
step 2 : checking
(3K, 7O, 6H, 1P  = 3K, 7O, 6H, 1P)

Diluting Solutions to Prepare Workable Solutions

We should make solutions of any concentrration from a more concentrated source.

We can make it by a simple equation.

M1V1 = M2V2

Example 1 :

Theres a 40.0mL of 0.400M NaOH solution is diluted to a final volume of 200.0mL, calculate the new concentration.

Let x be the new concentration.

0.400 x 40.0/1000.0 = 100/1000x

                             x= 6.25

The new concentration is 6.25M

Example 2 :

A 0.700M solution is concentrated by evaporation to a reduced final volume of 200.0mL and a molarity of 1.65M. Calculate the original volume.

Let V be the original volume.

0.700V = 200/1000 x 1.65

        V = 2.33
The original volume was 2.33L

Molar Concentration/ "Molarity" Conversation

Molarity is a number of moles of solute in 1 L of solution. We use "M" to denote molar concentration and ift has the units of "moles/L"

E.g. A 100 M solution is MORE concentrated than a 5 M solution.

Formula: Molarity = moles of solute(mol)/ volume of solution(L) aka M = mol/L

We also learned how to calculate the volume. We will learn more about this topic next class

Formula of a Hydrate

    So today we did a lab about calculating the hydrate. One common example of the hydrate is called calcium chloride. Another common hydrate is called the carbohydrate. So here is one hydrate which is anhydrous, aka Cobalt (II) chloride.



And after i did some research in the internet, I found out there is this another form of Cobalt (II) chloride which is called Cobalt (II) chloride hexahydrate. Which looks like this

So after doing this lab I think I am much prepared for the quiz which will be next wednesday. 

Calculate The Empirical Formula Of Organic Compound~

An organic compound : any substance that contain CARBON

When we do the calculation, we can write a balanced chemical equation for the burning of CxHy
(CxHy + zO2 -----> xCO2 + y/2H2O)

1. We should calculate the moles of CO2 and H2O produced
2. Find the mole of C and H in the CO2 and H2O
3. Find the ratio of C : H
4. Multiply the ratios to get a whole number

For example...
1.  What is the empirical formula of a compound that burns to produce 8.45g of CO2 and 1.73g of H2O?
mol of CO2 = 8.45/44 = 0.19        mol of C in CO2 = 0.19C
mol of H2O = 1.73/18 = 0.10       mol of H in H2O = 0.2H
Mole ratio = 1:1
Therefore,the empirical formula is CH

2. When 6.28g of an organic compound is burned,10.22g of CO2 and 5.18g of H2O is produced?What is the empirical formula?
mole of CO2 = 10.22/44 = 0.232   mol of C in CO2 = 0.232C
mole of H2O = 5.18/18 = 0.288     mol of H in H2O = 0.576H
Check Mass :
0.232C x 10.22g = 2.371g
0.576H x 5.18g = 2.98g
6.28g - 2.371g - 2.98g = 0.929g ----->   Oxygen
0.929/16 = 0.058mol Oxygen
Mole ratio = 0.232 : 0.576 : 0.058
                  = 4        : 10      : 1
Therefore,the empirical formula is C4H10O

Empirical formulas and Molecular formulas

Empirical formula: it gives the lowest term ratio of atoms (or moles) in the formula. *All ionic compounds are empirical formulas*

Ex: C3H6 (propene) ---> molecular formula
     CH2 ----> empirical formula


Molecular formula: it is a multiple of the empirical formula and shows the actual number of atoms that combine to form a molecule

Ex. A molecule has an empirical formula of C4H10 and a molar mass of 29 g/mol, what is the molecular formula
Ans: C2H5

More Mole Conversion(2)

Today we are going to teach you to solve Two Step Mole Calculartion Problems~

If we were asked to convert 22 grams of copper to atoms of copper, we'd have to go from one end of the map to the other. Instead of doing a simple one step calculation, we'd need to do a two-step calculation, with the first step going from grams to moles and the second step going from moles to atoms.

How can we solve this kind of problem? Well, we start off by doing the same thing that we did in our last example: We had to convert grams to moles before, and we can see from the map that we have to convert grams to moles now, too. To refresh your memory, here's the calculation from last time:

In the next step, we do the same thing over again, except that we need to add another T to the T-chart. When you do this, take the units of the thing at the new top left and put them on the bottom right (in this case, moles). Then take the units of what you want (in this case, atoms) and put it in the top right. Finally, put in your conversion factors, which from the chart above is Avogadro's number, or 6.02E23. Since this number refers to the number of atoms in a mole of a substance, we put this in front of "atoms of copper". Again, put the number "1" in front of moles, because we're saying that there are 6.02E23 atoms in ONE mole of an element.
When we add all these terms in, we can cross out the units that cancel out, as shown. To get the answer, multiply all the numbers on the top together and divide by the numbers on the bottom. Your answer should then be set up like this:

And that's how you do mole problems!