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
What is percent composition? Well... it is a compound that is a relative measure of the mass, of the different elements that are in the compound. Anyways...you need to be able to figure out the molar mass of each element, so if you are not sure what the molar mass of elements are, please read the previous blogs!!
This is an example of what a element would consist of, if the percent composition was represented by a pie graph
Now... how to actually calculate the percent composition of an element...
Ex 1/ What is the percentage composition of H20
First... you need to find:
Total Molar Mass of H2O.= 18.0g
Molar Mass H = 2.0g/ mol
Molar Mass O = 16.0/ mol
After that you find the % of each element, by dividing it by the total molar mass.
% of H = 2.0g per mol/ 18.0grams x100 = 11.1%
% of O = 16.0 per mol/ 18.0 grams x 100 = 88.9%
11.1%+88.9%= 100%
* Because there were no numbers in the questions, then you don't have to count the significant figures, so just round the number by one decimal place
Ex 2/ What is the percentage composition of Sc2(SO4)3
Total Molar Mass: 333.3 grams
Molar Mass Sc: 45.0 g/mol
Molar Mass S: 96.3 g/mol
Molar Mass O: 192.0 g/mol
% of Sc = 45.0 grams per mol / 333.3 grams x100 = 13.5%
% of S = 96.3 grams per mol / 333.3 grams x 100 = 28.9%
% of O = 192.0 grams per mol / 333.3 grams x 100 = 57.6%
The total percentage totals up to 100%
Ex 3/ What is the percentage composition of Zn(ClO3)2
Total Molar Mass: 232.4 grams
Molar Mass Zn: 65.4 grams/mol
Molar Mass Cl: 71.0 grams/mol
Molar Mass O: 96.0 grams/mol
% of Zn: 65.4 grams per mol / 232.4grams x100 = 28.1%
% of Cl: 71.0 grams per mol / 232.4 grams x100 = 30.6%
% of O: 96.0 grams per mol / 232.4 grams x 100 = 41.3%
The total adds up to 100% YAY!
Ex 4/ If a compound contains 48.0 grams of C, 6.0 grams of H, 64.0 grams of O, and contains some amount of Be and has a total mass of 127.0 grams. Calculate the % composition.
127.0-48.0-6.0-64.0 = 9.0 grams
% of C: 48.0 g per mol / 127.0 grams x 100 = 37.8%
% of H: 6.0 g per mol / 127.0 grams x 100 = 4.7%
% of O: 64.0 g per mol / 127.0 grams x 100 = 50.4%
% of Be: 9.0 g per mol / 127.0 grams x 100 = 7.1%
*Remember, there are numbers in the question, so sig figs count!
FINALLY......
Ex 5/ If a compound contains 137.3 g of Ba, 28.0 g of N, and contains some amount of O and has a total mass of 261.3 grams. Calculate the percent composition.
261.3 - 137.3 - 28.0 = 96.0 grams
% of Ba: 137.3 g per mol/ 261.3 x 100= 52.5%
% of N: 28.0 g per mol / 261.3 x 100 = 10.7%
% of O: 96.0g per mol / 261.3 x 100 = 36.7%
In the case the percentages only add up to 99.9%. Due to the rounding, there may be some cases where the percentages do not add up to 100%. If it is only 0.1% off, it is not a big deal, but to be safe ask Ms. Chen =]
These video's can maybe explain in more detail on how to calculate percent compositions..
After having all that knowledge inputted into your brain... time to use those skills on some other problems!
Whatup? Today we took the mole conversions quiz. Ms.Chen as well as half our class wasn't present today, probably because of the snow storm that hit Vancouver overnight. Anyways, the substitute teacher corrected the mole conversions review in the begining of the class. He then gave us 10 minutes to study. After studying we had the rest of the class to complete the test. I hope everyone did well. See you guys later.
So now that we understand how to convert from grams to moles, moles to grams, particles to moles and moles to particles, let's try converting from particles to grams and grams to particles.
These are now 2 step conversions. Once again we need our periodic table.
Let's start with particles to mass.
If you have 2.78 x 10^22 Fe atoms, how many grams are there?
First, you want to get rid of the atoms, so let's go from atoms to moles.
2.78 x 10^22 atoms of Fe x 1 mole / 6.022 x 10^23 atoms
Now, we also know how to go from moles to grams, so let's get rid of moles.
2.78 x 10^22 atoms of Fe x 1 mole / 6.022 x 10^23 atoms x 55.8g (molar mass of Fe) / 1 mole
Now our final answer is going to be (2.78 x 10^22)(55.8g) / (6.022 x 10^23) in 3 sig figs.
So, we get 2.58g. This means 2.78 x 10^22 Fe atoms weighs 2.58 g.
Now mass to particles...How many atoms of Fe are in 20.0g of Fe?
So, we get rid of grams by multiplying moles/grams.
20.0g of Fe x 1 mole / 55.8g (molar mass of Fe)
Then, since we want particles, we get rid of moles by multiplying particles/moles.
20.0g of Fe x 1 mole / 55.8g (molar mass of Fe) x 6.022 x 10^23 atoms / 1 mole
Our final answer of 20.0g Fe in particles (3 sig figs) is 2.16 x 10^23 atoms.
Well... I'm going to start off with a funny story... My friend VIVIAN CHENG thought the "moles" were the black dots that grow on your face... then she thought it was the ANIMAL mole..... until i explained to her...
Well...we're still learning about moles, and how to convert moles... as much as I dislike moles, we still gotta learn about MOLES!...
Keep in mind Avogadro's Number is 6.022 x 10^23 --> particles/mole
Molar Mass
You can convert molar masses from either:
1. Particles to moles
or
2. Moles to particles
Converting molar masses is somewhat simple. It is just like doing unit conversions.
For example:
1) Convert 4.5 x 10^24 particles to moles
4.5 x 10^24 particles x 1 moles/6.022 x 10^23 particles = 7.5 moles
What I did, was I divided 4.5 x 10^24 by 6.022 x 10^23. The particles cancel out. And because there is only one mole, the answer is simply 7.5 moles.
*Remember that you have to count to the correct number of significant digits.
2) *(2 atoms) Convert 0.82 moles CO₂ to molecules
0.82 moles x 6.022 x 10^23 particles/ 1 moles = 4.9 x 10^23 molecules CO₂
The moles cancel out, so you simply multiply 0.82 x (6.022 x 10^23) particles . Because its 4.9 x 10^23/ 1, it works out to just be 4.9 x 10^23 molecules CO₂
There are two atoms, so there is one more step involved. Well...it's not anything special... but...here goes!
4.9 x 10^23 molecules CO₂ x 2 atoms O/ 1 molecule CO2 = 9.8 x 10^23 atoms of Oxygen
And again... the molecules cancel out, and you just multiply the numbers.
This video may help explain some questions you may have:
Are you tired of converting moles yet? Well we're almost done, so bear with me!
Now.. we are going to learn how to convert moles to grams, and grams to moles. Get your periodic table for this!!! We need it !! To find the molar mass of an element, look for the atomic mass of the element!
Example:
1) Convert 3.06 moles Fluorine to grams
The molar mass of Fluorine is 19.0g/mol
3.06 moles x 19.0g/1 mole = 58.1 grams of F
Here, the moles cancel out, and you just multiply 3.06 x 58.1 grams of F...and.. BAM, you get the answer!
2) To convert 7.65 grams of Fluoride to moles
The molar mass of Fluorine is 19.0g/mol
This is a gas hydrate. 
