Lab 4C 9/12

Lab 4C:
By burning anhydrous, it's a way that find out the relationship between reaction and the change of mass, as the anhydrous salt contains water, we have to burn it by Bunson burner and check out what's the mass eventually. By using empirical formula, we can also find out the name of the compound as it changes its mass.

What I learned!!!!:
-By burning anhydrous salt, water will evaporate, and if the time and the heating temperature remain the same, the percentage of H2O that evaporate will also be the same amount!

-Through the first heating and the second heating as anhydrous salt remain the same mass, which means there don't have any H2O in the ahydrous and the mass will no longer to change by heating.

-Be strong, as the Bunson burner is hot, I got to act like confidence and don't let my partner upset.

Conclusion
I learn how to use the empirical formula to find out the compound and the way of using Bunson burner. In the other hand, I've learned how to take care and teach my partner, it makes me more understanding about this lab.

Calculating the empirical formula of organic compound 3/12

Organic can be found by:
- Burning
- Collecting, weighing product
- Mass, mole that can be calculated
Ex. 5g of CxHy burn, then produce 15g of CO2 + 8.18g H20, Let the Empirical foumula - CxHy
mol CO2 = 15.0g CO2 x ( 1 mol of CO2) / (44.0g CO2) = 0.341 mol

mol H20 = 8.18g H2O x (1 mol H2O) / (18.0 g H2O) = 0.454 mol

- 0.454 water , 0.341 Carbon dioxide
#2

mol C = 0.341 mol CO2 x ( 1 mole C) / ( 1 mol CO2) = 0.341 mol
mol H = 0.454 mol H20 x  ( 2 mole H) /  ( 1 mol H2O) = 0.908 mol

0.908 H, 0.341 C > CxHy > could be C 0.341 H 0.908

ratios of moles = (0.908 mol H) / (0.341 mol C) = (2.66) / (1) (as we can't have any decimal in empirical formula, we have to round it up to the closest whole number)(2.66) x 3 > 7.98 > 8
1        x 3 >     3   > 3

The empirical formula > C3H8
DONE !!

As we learned all those empirical formula and molecular formula from the past, the picture from below is a little summary that conver all we had learned since term 2 has started! It's the relationship between EF, MF and Organic compound!

Empirical + Molecular Formula 1/12

Empirical Formula = gives the lowest term ratio of atoms or molecule in the formula!

All ionic compounds are empirical formulas!C4H10 (Butune) > molecular formula
C2H5                 > empirical formula

Ex.) Consider that we have 10.87g of Fe and 4.66g of O. What is the empirical foumula?
#1 Convert g > mole
Fe: 10.87g/ (55.8) > mole = 0.195g/mol
O : 4.66/ (16) > mole = 0.291g/mol#2 Divide both by the smallest molar amount.
Fe: 0.195 / 0.195 = 1
O : 0.291 / 0.195 = 1.49 > 1.5
(as 1.5 can't be in the empirical formula, we have to multiply it into whole number)  O : 1.5 x 2 > 3 , and the other one also have to multiply the same amount, which is "2" in this situation.
Fe : 1 x 2    > 2#3 Scale ratios to whole numbers
Fe2O3
DONE! : )


Second part:


Molecular Formula:it's a multiple of the empirical formula and shows the actual number of atoms that comine to form a molecule to calculate multiple:

Ex.) A molecule has an empirical formula of C2H5 and a molar mass of 85g/mol. What is the molecula formular?

#1 : molar mass of C2H5 = 12 x 2 + 5 = 29g/mol

N=  (58g/MOL) / (29) = "2"
#2:   "2" x C2H5 = C4H10 
DONE!

Let's take a look at this video ! As Ms.Chen said, there won't have any question hardar than this one, once you can handle this one.  You will be fine I bet : )

Percent Composition 29/11

Percent composition -Percentage by mass of a "species" in a chemical formula
Basicly, it's similar as the way we calculate our mark percentage :)Ex.  Keith has gotten 18 out of 24 mark in Biology 11, so if he wants to calculate what exactly percentage he gets, he can determine it by this formula:   (what you get from an example / what a total number, marks in an example) x 100%
As the formula from below, we can plug in (18/24) x 100% = 75%  ：) Done!




Back to the percent composition,
Ex) What percentage composition of H2O?
＊Assume you have one mole
Total molar mass 18g/mol
Molar mass of H2 = 1+1 = 2g/mol
Molar mass of O   = 16g /mol
Percent of H2 = (2/18) x 100% = 11.1%
Percent of O   = (16/18) x100% = 88.9%

As the blog video insert function doesn't work :(  , Let's go to the link from below and take a look the SHORT VIDEO :)
http://www.youtube.com/watch?v=e11mq1BYlcI

Harder Mole Conversion" November 23

First of all,  Let's take a look the picture from below

It has shown us the relationship between molar mass, molecule, gram and atom.
In the other hand, it's a MOLAR MAP that direct us to convert number into a certain unit that we want

So, eg1: If we want to convert GRAM to ATOM,  what can we do? Let's take a look at this video from bellow =)

So basiclly, "Harder Mole Conversion" is just based on the simple mole conversion, "BUT" it takes more than one STEP to reach the answer

Atomic Mass, Formula Mass and Molar Mass + Mole Conversions - Nov 19

Atomic MassAtomic mass: the relative mass of atoms in ATOMIC MASS UNITS (u).

 i.e.
Atomic Mass of He is 4.00u
Atomic Mass of O is 16.0uAtomic Mass of C is 12.0u
Formula Mass: Total mass of all the atoms in the formula of an ionic compound in Atomic Mass Units
i.e.
NaCl
Atomic mass of Na = 23.0 , Cl = 35.5
= 23.0+ 35.5 = 58.5u


CuCO3
Atomic mass of Cu = 63.5 , C = 12, O3 = 16+16+16 = 63.5 + 12 + 3(16) = 123.5u

Molecular Mass: Total mass of all the atoms that make up a molecule of a covalent compound, organic compound or a polyatomic element in ATOMIC MASS UNIT (u)
i.e
C6H12O6
C6 = 6(12) , H12 = 12(1), O6 = 6(16)= 6(12) + 12(1) + 6(16) = 180u
Molar Mass : mass of mole (6.022 x 10^23 particles) of a substance and is the same numerial value of Atomic mass, formula mass, or molecular mass but expressed IN GRAMS PER MOLEi.e.
the molecular mass of NO is 26.0 u and the molar mass of NO is 26.0g/moleAs the molecular mass and molar mass are both have the same result, but they are just different express =)

 Mole Conversions

As the picture have shown from above, that's the relationship between Mole, Atoms, Gram and ParticlesIf you are still having a lot of WONDERING about THOSE information i 've shown from above.
Please leave a comment, or contact me directly by e-mail =)