SOOO... what we did in class. Last day, we basically just had to do a quiz on percent composition and empirical and molecular formulas. That took up almost the whole block.. because it was SO SO SO HARD! *ahem* Me, as well as other studentts agree that everything should be easy peasy :) sooooo make it easierrr please ^^!
ok..so anyways, we didn't do much after that.. we had to do a flow chart on Lab 4C, and copy out the table. That was basically all we had to do! In the next few classes, we are going to determine the percentage of water in hydrate, the moles of water present in a hydrate, and how to write the empirical formula of the hydrate!!!
This is a gas hydrate.
But anyways today, we did Lab 4C.
In this lab we worked with a bunsen burner... so us, the students could have been in GREAT DANGER!! WAHHH. No... we weren't... but some saftey equipment we needed included a lab apron and safety googles. The purpose of this lab was to determine the percentage of water as well as the number of moles of water present in an unknown hydrate.
First step: Heat the crucible with the bunsen burner and a ring stand to make sure that it is dry enough.
Second step: Let the crucible cool down, then weigh the empty crucible. After, add some of the hydrate into the crucible and weigh that as well. Begin heating the crucible over the bunsen burner for 5 minutes. To make sure the heating is effective, make sure that the tip of the fire is touching the crucible, as it should glow dull red.
Third step: Wait for it to cool down and weigh it again. Now, do a second heating.
This second heating is to confirm the numbers that were found in the first heating, just so we can be more accurate.
Final step: Add drops of water back into the anhydrous compound (the thing that's left in the crucible) and see what you get.
So in last class we learned how to find both Emperical and Molecular formulas. Today we learned how to found the EF of an organic compound, which is some what harder.
Basically, we can find the EF of any organic compund by burning the compund (collecting/weighing product). We then have a burnt product. We then can calculate the moles of each element from the original organic (unburnt) product.
Now let's try an example to see if we can calculate it right.
EX. What is the EF of a compound that when a 15.00 gram sample is burned it produces 25.0 grams of CO2 and 10.0 grams of H20
Step 1: Calculate all moles of CO2 and H2O
25.0 CO2 x 1mol CO2/ 44.0g CO2 = 0.568 mol CO2
10.0 H2O x 1mol H2O/18.0 H2O = 0.556 mol H2O
Step 2: Find moles of C and moles of H in CO2 and H2O
0.568 mol CO2 x 1molC/1 molCO2 = 0.568 mol
0.556 mol H2O x 2molH/1molH20 = 1.112 mol
Step: 3: Divide both moles by smallest molar amount
C = 0.568/0.568 = 1
H = 1.112/0.568 = 2
There for the Emperical Formula is CH2
Step 4: Check answer.
Convert moles back into grams. The result should be 15.0g. If the answer does not match, that means Oxygen is a component. To find the amount of Oxygen use this formula.
Mass of O = Mass of compound - mass of C - mass of H
This concludes my blog. Have a good night everyone.
So today............we're going to be doing Empirical and Molecular Formula.
What is the empirical formula? It is the expression of the ratio of atoms in a formula in its lowest terms. By this definition, we can realize that all ionic compounds are empirical formulas.
So, for example, let's say we have the formula C10H14N2, which is Nicotine (the thing that makes you addicted to cigarettes).
Now, to get the empirical formula, let's reduce all the subscripts to the lowest terms. We'd get C5H7N. Of course, this is a whole new compound, but we can say this is the empirical formula of nicotine.
Now, find the empirical formula for a compound consisting of 63% Mn and 37% O.
First, we assume that we have 100g of this compound, just to make everything easier.
Now, we take the 63 g(now grams of Mn, because 63% of 100 is 63) and convert it into moles.
Since the molar mass of Mn is 54.9, we do
63 g X 1 mole = 1.1 moles Mn (in sig figs)
54.9g
Then, we take the 37 g of O and do the same.
37 g X 1 mole = 2.3 moles O
16.0g
Now, since we have both parts of the compound in moles, we now divide each by the smallest molar amount. If you're going like HUH? just wait for the next step.
So, we can all agree that the number 1.1 is less than 2.3 right?
That will be what we are dividing by.
We take the 1.1 moles of Mn and divide by 1.1, giving us the whole number 1.
Then, we take 2.3 moles of O and divide by 1.1, giving us 2.1, which is around 2.
The final part is to scale the ratios into whole numbers. But good thing for us, the numbers in this example are already whole numbers, so we don't need to multiply again. (In other words, we don't get 1.5, which means we would have to multiply by 2 to get a whole number)
We will get 1 Mn, and 2 O.
Our final answer for this question is...... MnO2.
OK..Now let's do some molecular formula. This is a multiple of the empirical formula and it can show the actual number of atoms that are necessary to combine into a molecule.
For this, we have a special formula. That is, the molar mass of a compound divided by molar mass of the empirical formula, will give us the molecular formula.
Let's say
The simplest formula for vitamin C is C3H4O3. Experimental data indicates that the molar mass of vitamin C is about 180 g/mol. What is the molecular formula of vitamin C?
First, we take the empirical formula, and add up the molar masses of each element to get the total molar mass. Get your periodic table for this - you'll need it.
The molar mass of C is 12.0g, and we have 3, so 3 X 12 = 36.0 Then, we have 4 Hydrogen, with each hydrogen at 1.0g each, so 4 X 1 = 4.0 Finally, we have 3 Oxygen at 16.0 g each, so 3 X 16 = 48.0 36.0 + 4.0 + 48.0 = 88.0 g/mol
Now we have both molar mass of compound, and the molar mass of the empirical formula. The last step for us is to put it into the formula.
So 180 g/mol = 2.045, which is around 2. 88 g/mol
This means that we need to take that original formula and multiply it entirely by 2. 2 X C3H4O3 This gives us C6H8O6, which means the molecular formula of vitamin C is C6H8O6
This is a gas hydrate. 
