Hahaha,,,, today,, we did a experiment,,,
which is lab 4C FORMULA OF A HYDRATE.
The procedure was:
First,,, Wear your safety GOGGLES!!
Then set up the pipestem triangle, iron ring, stsand and bunsen burner....
Heat the crucible with busen burner for around 3 min,, to make sure it is dry,,,,,,
-Let crucible cool down and measure the mass of it.
-put the given hydrate inside the crucible and record the mass.
_heat the crucible(with hydrate) until the buttom of crucible is dull red,,, for about five min
-after that,, let it cool down for approximately five min,,,, and record down the mass of it
-to make sure all water is driven off,,,, reheat it for another five min
-cool it down for another five min again and record the mass
-lastly,,,, drops a few water to the contents of the crucible and see what happen!! XDD
Links:
Another experiment for formula of a hydrate:
http://chem.lapeer.org/Chem1Docs/HydrateLab1.html
A video clip for the expriment:
http://www.youtube.com/watch?v=OuF4hjTFdsg
HAHAHA,, That's all.....
Have A Nice Day! ^^ XDD
Tuesday, December 7, 2010
Friday, December 3, 2010
DECEMBER 3rd
Do you remember the CH compounds or the compounds containing Carbon?
They are referred to as the ORGANIC COMPOUND--
The Question is...
How do you calculate the Empirical formula of Organic Compounds?
For example, given:
They are referred to as the ORGANIC COMPOUND--
The Question is...
How do you calculate the Empirical formula of Organic Compounds?
For example, given:
What's the empirical formula of a compound when a 5.00 gram sample is burned producing 15.0 grams of CO2 and 8.18 g of H2O?
Let the empirical formula of the compound be CxHy.
CxHy + zO2 -> xCO2 + y/2 H2O
1. Find the moles of C and H in the compound.
CO2: 15.0 g x 1mol/44.0 g = 0.341 mol
H2O: 8.18 x 1mol/18.02g = 0.454 mol
2. 0.341 moles of C 0.341/0.341 = 1 x 3 =3
0.454 x 2 = 0.908moles of H 0.908/0.341 = 2.66 x 3 = 8
3. Ratio of C:H = 3:8
therefore, the empirical formula is C3H8.
4. Now, check the masses:
0.341 mol of C x 12g/1mol = 4.092 g of C
0.908 mol of C x 1.0g.1mol = 0.908 g of H
Total mass = 5.00 g
Therefore, there is no Oxygen atom in the compound.
*Here is a challenging question for an extra thinking~
Adipic acid is an organic compound composed of 49.31% C, 43.79% O, and the rest hydrogen. If the molar mass of adipic acid is 146.1g/mol, what are the empirical and molecular formulas for adipic acid?
if there were 100 g of the sample, it would contain
49.31 g C
43.79 g O
6.90 g H
convert everything to moles
49.31 g C------>4.11 mol C
43.79 g O------>2.74 mol O
6.90 g H -------->6.83 mol H
now, we have the molar ratio
C:O:H = 4.11 mol C : 2.74 mol O: 6.83 mol H
reduce the ratio to form in which the smallest term is 1
do this by dividing each term by the smallest value (2.74)
4.11 mol C : 2.74 mol O: 6.83 mol H = 1.5 mol C : 1 mol O :2.5 mol H
to eliminate fractions, multiply by the LCD (2 in this case)
1.5 mol C : 1 mol O :2.5 mol H = 3 mol C : 2 mol O :5 mol H
thus, the empirical formula is C3H5O2
======================================…
to find the molecular formula, we have to know how many times the empirical weight is the molecular weight
we do this by dividing the molecular weight by th empirical weight
Empirical Weight (EW)
=(12 g/mol C)(3 mol C) + (1.01 g/mol H)(5 mol H) + (16 g/mol O)(2 mol O)
= 73.05 g
Molecular Weight (MW) = 146.1 g
MW / EW = 146.1 g / 73.05 g = 2
multiply this by the subscripts of the elements in the empirical formula
===============
thus, the molecular formula is C6H10O4
thus, the empirical formula is C3H5O2
Did you get it??
49.31 g C
43.79 g O
6.90 g H
convert everything to moles
49.31 g C------>4.11 mol C
43.79 g O------>2.74 mol O
6.90 g H -------->6.83 mol H
now, we have the molar ratio
C:O:H = 4.11 mol C : 2.74 mol O: 6.83 mol H
reduce the ratio to form in which the smallest term is 1
do this by dividing each term by the smallest value (2.74)
4.11 mol C : 2.74 mol O: 6.83 mol H = 1.5 mol C : 1 mol O :2.5 mol H
to eliminate fractions, multiply by the LCD (2 in this case)
1.5 mol C : 1 mol O :2.5 mol H = 3 mol C : 2 mol O :5 mol H
thus, the empirical formula is C3H5O2
======================================…
to find the molecular formula, we have to know how many times the empirical weight is the molecular weight
we do this by dividing the molecular weight by th empirical weight
Empirical Weight (EW)
=(12 g/mol C)(3 mol C) + (1.01 g/mol H)(5 mol H) + (16 g/mol O)(2 mol O)
= 73.05 g
Molecular Weight (MW) = 146.1 g
MW / EW = 146.1 g / 73.05 g = 2
multiply this by the subscripts of the elements in the empirical formula
===============
thus, the molecular formula is C6H10O4
thus, the empirical formula is C3H5O2
Did you get it??
Here is a summary of the steps
1. Find # of moles of the compounds containing C and H
2. Find the # of moles of atom of C, H
3. Determine the ratio of C:H based on the # of moles (divide by the smallest number)
4. IF there are decimals, use either rounding or multiply to get a whole number
5. FINALLY, PLEASE check the masses of each atom of C and H to see if it equals the mass (if given) in the question. If it does not, then there must be a O atom in the compound. Find the mass of O atom and convert to moles in order to determine the C:H:O ratio!
Wednesday, December 1, 2010
December 1, 2010
Wow, I almost can't believe it, it is already the start of December!
Today is officially December 1st so that means it is 24 days until Christmas & 17 more days until Winter Vacation! We are almost there, so keep going everybody!!
Well enough about Christmas (which is my favourite holiday) and lets talk Science.
Today we learned a new topic..... are you ready for this?..........
Empirical + Molecular Formaula!!!!!!
Now this topic is very detailed so stay with me here;
Lets first start with Empirical Formula; (EF)
- EF is the lowest term ratio of atoms in a formula. Otherwise known as reducing.
ex. C6H12 = molecular formula
C3H6 = CH2 = empirical formala
ex. consider that we have 10.87g of Fe and 4.66g of O. What is the EF?
1st Step. Convert grams to moles
Fe 10.87g x 1 mol / 55.8g = 0.195 mols of Fe
O 4.66g x 1 mol / 16.0g = 0.291 mols of O
2nd Step. Divide both by the smaller molar amount
Fe 0.195 / 0.195 = 1
O 0.291 / 0.195 = 1.5
3rd Step. Scale ratios to whole numbers
Fe 1 x 2 = 2
O 1.5 x 2 = 3 Therefore the answer is: Fe2O3
ex. A compund contains 31.9 % K, 28.9 % Cl, 39.2 % O. What is the EF?
* Assume you have 100g
1st Step. Convert grams to moles
K 31.9g x 1 mol / 39.1g = 0.816 mol
Cl 28.9g x 1 mol / 35.5g = 0.814 mol
O 39.2g x 1 mol / 16g = 2.45 mol
2nd Step. Divide each by the smaller molar amount
K 0.816 / 0.814 = 1
Cl 0.814 / 0.814 = 1
O 2.45 / 0.814 = 3
3rd Step. Scale ratios to whole numbers
Dont need to do because all ratios are whole numbers already (1, 1, 3)
Therefore the answer is: KClO3
Now I will explain Molecular Formula; (MF)
- MF is the multiple of empirical forms. It shows the number of atoms that combine to form a molecule.
- Formula: n = mm of the compund / mm of the empirical formula. (mm = molar mass)
ex. A molecule has an empirical formula of C2H5 and a molar mass of 58 g/mol.
What is the MF?
MM of C2H5 = 29.0 g/mol
n = 58 g/mol
----------- = 2
29.0 g/mol
MF = 2(C2H5) = C4H10
ex. The empirical formula of a gas is CH2. What is the MF if MM is 42 g/mol?
MM of CH2 = 14 g/mol
n = 42 g/mol
----------- = 3
14 g/mol
MF = 3(CH2) = C3H6
ex. A compound contains 7.44g of C, 1.24g of H, 9.92g of O. The MM is 180 g/mol. What is the MF?
1st Step. Find the empirical formula since it is not shown in the question.
C 7.44 x 1 mol / 12g = 0.62 mols
H 1.24 x 1 mol / 1g = 1.24 mols
O 9.92 x 1 mol / 16g = 0.62 mols
2nd Step. Then divide each by the smallest molar amount which is 0.62
C = 1 H = 2 O = 1 therefore EF = CH2O
3rd Step. Now you can find the MF.
MM of CH2O = 30 g/mol
n = 180 g/mol
----------- = 6
30 g/mol
MF = 6(CH2O) = C6H12O6
That concludes of all what we did today, now I see that in the last blog Percent Composition was not explained so I will explain that right now. Percent Compostion is what we did last class.
Percent Composition: (PC)
- PC is percentage by mass of a "species" in a chemical formula.
ex. What is the PC of CO2?
*Assume you have one mole
Total MM = 44.0 g/mole
% of C = 12.0 divided by 44.0 x 100% = 27.3 %
% of O = 32 divided by 44 x 100% = 72.7 %
Notice that they both add up to a hundred percent, in these equations the two or three answers should always add up to a hundred percent or at least very close to.
ex. Compound contains 5.1g of Cl, 22.0g of C, and has a total mass of 44.1g and contains some amount of O. Calculate the PC.
Mass of Oxygen = 44.1 - 5.1 - 22.0 = 17.0 g
% of Cl = 5.1 divided by 44.1 x 100% = 11.6 = 12%
% of C = 22.0 divided by 44.1 x 100% = 49.9%
% of O = 17.0 divided by 44.1 x 100% = 38.5%
All add up to 100.4 percent. This is a way to check your work too by the way.
ex. Calculate the PC of the underlined components of (NH4)2SO3
Total MM = 116.1
MM of all the NH4 = 36
% of NH4 = 36 divided by 116.1 x 100% = 31.0%
Whew, alot of numbers as you can see :). Now its time to treat you with some video's & practice questions for your own practice!
- Worksheets:
Empirical + Molecular Formula
1) http://eaglepoint.lexingtonchristian.org/hs/science/dhumphreys/Shared%20Documents/Honors%20Chemistry/Empirical%20Formula%20Worksheet%205.pdf
2http://www.ccboe.net/Teachers/nutial_laura/files/CBDD907F339B420CB25588150BD481A9.doc
3) http://www.ezclasssites.com/data/captainkirk/EmpiricalFormulaWS.pdf
4) http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson59.htm
5) http://lhs2.lps.org/staff/sputnam/practice/UnitV_EmpForm.htm
6) http://westwood.sjsd.net/~dshoesmith/FOV1-0003789A/FOV1-000378A1/Percent%20Composition%20and%20Molecular%20Formula%20Worksheet.doc?FCItemID=S01C8C50E
7) http://www.cbv.ns.ca/rv/campbell/Resources/M%20F%20EF.pdf
Percentage Composition
1) http://misterguch.brinkster.net/PRA023.pdf
2) http://www.fordhamprep.org/gcurran/sho/sho/worksheets/worksht58a.htm
3) http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson58.htm
4) http://misterguch.brinkster.net/001_024.doc
5http://cmsweb1.loudoun.k12.va.us/52820831134912597/lib/52820831134912597/Moles/Homework/masspercomp.pdf
6http://www.ccboe.net/Teachers/epperson_steve/files/36F0A3C2EC0648DBB63EF8981578B7A0.doc
7http://asd1.schoolwires.com/1741206142323657/lib/1741206142323657/Percentage_Composition_Worksheet.pdf
- Videos
1) http://www.youtube.com/watch?v=xbEeyT8nK84
2) http://www.youtube.com/watch?v=_H009sTvYE0
3) http://www.youtube.com/watch?v=gfBcM3uvWfs
Ooops almost forgot...... HW:
1) Empirical and Molecular Formula
2) Reminder: Quiz next week regarding this material and the material will be covered on Frida (next class).
Thank You and Have A Great Weekend!!!!!
Today is officially December 1st so that means it is 24 days until Christmas & 17 more days until Winter Vacation! We are almost there, so keep going everybody!!
Well enough about Christmas (which is my favourite holiday) and lets talk Science.
Today we learned a new topic..... are you ready for this?..........
Empirical + Molecular Formaula!!!!!!
Now this topic is very detailed so stay with me here;
Lets first start with Empirical Formula; (EF)
- EF is the lowest term ratio of atoms in a formula. Otherwise known as reducing.
ex. C6H12 = molecular formula
C3H6 = CH2 = empirical formala
ex. consider that we have 10.87g of Fe and 4.66g of O. What is the EF?
1st Step. Convert grams to moles
Fe 10.87g x 1 mol / 55.8g = 0.195 mols of Fe
O 4.66g x 1 mol / 16.0g = 0.291 mols of O
2nd Step. Divide both by the smaller molar amount
Fe 0.195 / 0.195 = 1
O 0.291 / 0.195 = 1.5
3rd Step. Scale ratios to whole numbers
Fe 1 x 2 = 2
O 1.5 x 2 = 3 Therefore the answer is: Fe2O3
ex. A compund contains 31.9 % K, 28.9 % Cl, 39.2 % O. What is the EF?
* Assume you have 100g
1st Step. Convert grams to moles
K 31.9g x 1 mol / 39.1g = 0.816 mol
Cl 28.9g x 1 mol / 35.5g = 0.814 mol
O 39.2g x 1 mol / 16g = 2.45 mol
2nd Step. Divide each by the smaller molar amount
K 0.816 / 0.814 = 1
Cl 0.814 / 0.814 = 1
O 2.45 / 0.814 = 3
3rd Step. Scale ratios to whole numbers
Dont need to do because all ratios are whole numbers already (1, 1, 3)
Therefore the answer is: KClO3
Now I will explain Molecular Formula; (MF)
- MF is the multiple of empirical forms. It shows the number of atoms that combine to form a molecule.
- Formula: n = mm of the compund / mm of the empirical formula. (mm = molar mass)
ex. A molecule has an empirical formula of C2H5 and a molar mass of 58 g/mol.
What is the MF?
MM of C2H5 = 29.0 g/mol
n = 58 g/mol
----------- = 2
29.0 g/mol
MF = 2(C2H5) = C4H10
ex. The empirical formula of a gas is CH2. What is the MF if MM is 42 g/mol?
MM of CH2 = 14 g/mol
n = 42 g/mol
----------- = 3
14 g/mol
MF = 3(CH2) = C3H6
ex. A compound contains 7.44g of C, 1.24g of H, 9.92g of O. The MM is 180 g/mol. What is the MF?
1st Step. Find the empirical formula since it is not shown in the question.
C 7.44 x 1 mol / 12g = 0.62 mols
H 1.24 x 1 mol / 1g = 1.24 mols
O 9.92 x 1 mol / 16g = 0.62 mols
2nd Step. Then divide each by the smallest molar amount which is 0.62
C = 1 H = 2 O = 1 therefore EF = CH2O
3rd Step. Now you can find the MF.
MM of CH2O = 30 g/mol
n = 180 g/mol
----------- = 6
30 g/mol
MF = 6(CH2O) = C6H12O6
That concludes of all what we did today, now I see that in the last blog Percent Composition was not explained so I will explain that right now. Percent Compostion is what we did last class.
Percent Composition: (PC)
- PC is percentage by mass of a "species" in a chemical formula.
ex. What is the PC of CO2?
*Assume you have one mole
Total MM = 44.0 g/mole
% of C = 12.0 divided by 44.0 x 100% = 27.3 %
% of O = 32 divided by 44 x 100% = 72.7 %
Notice that they both add up to a hundred percent, in these equations the two or three answers should always add up to a hundred percent or at least very close to.
ex. Compound contains 5.1g of Cl, 22.0g of C, and has a total mass of 44.1g and contains some amount of O. Calculate the PC.
Mass of Oxygen = 44.1 - 5.1 - 22.0 = 17.0 g
% of Cl = 5.1 divided by 44.1 x 100% = 11.6 = 12%
% of C = 22.0 divided by 44.1 x 100% = 49.9%
% of O = 17.0 divided by 44.1 x 100% = 38.5%
All add up to 100.4 percent. This is a way to check your work too by the way.
ex. Calculate the PC of the underlined components of (NH4)2SO3
Total MM = 116.1
MM of all the NH4 = 36
% of NH4 = 36 divided by 116.1 x 100% = 31.0%
Whew, alot of numbers as you can see :). Now its time to treat you with some video's & practice questions for your own practice!
- Worksheets:
Empirical + Molecular Formula
1) http://eaglepoint.lexingtonchristian.org/hs/science/dhumphreys/Shared%20Documents/Honors%20Chemistry/Empirical%20Formula%20Worksheet%205.pdf
2http://www.ccboe.net/Teachers/nutial_laura/files/CBDD907F339B420CB25588150BD481A9.doc
3) http://www.ezclasssites.com/data/captainkirk/EmpiricalFormulaWS.pdf
4) http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson59.htm
5) http://lhs2.lps.org/staff/sputnam/practice/UnitV_EmpForm.htm
6) http://westwood.sjsd.net/~dshoesmith/FOV1-0003789A/FOV1-000378A1/Percent%20Composition%20and%20Molecular%20Formula%20Worksheet.doc?FCItemID=S01C8C50E
7) http://www.cbv.ns.ca/rv/campbell/Resources/M%20F%20EF.pdf
Percentage Composition
1) http://misterguch.brinkster.net/PRA023.pdf
2) http://www.fordhamprep.org/gcurran/sho/sho/worksheets/worksht58a.htm
3) http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson58.htm
4) http://misterguch.brinkster.net/001_024.doc
5http://cmsweb1.loudoun.k12.va.us/52820831134912597/lib/52820831134912597/Moles/Homework/masspercomp.pdf
6http://www.ccboe.net/Teachers/epperson_steve/files/36F0A3C2EC0648DBB63EF8981578B7A0.doc
7http://asd1.schoolwires.com/1741206142323657/lib/1741206142323657/Percentage_Composition_Worksheet.pdf
- Videos
1) http://www.youtube.com/watch?v=xbEeyT8nK84
2) http://www.youtube.com/watch?v=_H009sTvYE0
3) http://www.youtube.com/watch?v=gfBcM3uvWfs
Ooops almost forgot...... HW:
1) Empirical and Molecular Formula
2) Reminder: Quiz next week regarding this material and the material will be covered on Frida (next class).
Thank You and Have A Great Weekend!!!!!
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