- Coefficients in balanced equations tell us the number of moles reacted or produced
- They can also be used as conversion factors
- What you need over what you have
Chemistry 11 Mr. Doktor Block G Run by: Mark W. Guarin Julius O. Francia Timothy Samson
Monday, January 24, 2011
Mole to Mole Conversion
Stoichiometry
- Stoichiometry is a branch of chemistry that deals with the quantitative analysis of chemical reactions.
- It is a generalization of a mole conversion to chemical reactions.
- understanding the 6 types of chemical reactions is the foundation of stoichiometry
Synthesis
- A+B-->AB
- Usually elements ---> compounds
Decomposition
- AB ---> A + B
- Reverse of synthesis
Single Replacement
- A + BC ---> B + AC
Double Replacement
- AB + CD ---> AD + BC
- Metals always go first
Neutralization
- Reaction Between an acid and a base
Combustion
- Reaction of something (usually hydrocarbon) with air
- Hydrocarbon combustion always produces CO2 and H2O
Sunday, January 16, 2011
Empirical Formulas and Molecular Formulas
Empirical formulas are the simplest formula of a compound that show only the simplest ratios.
-Not the actual atoms
Ex: - The empirical formula for Hydrogen gas is H
- Dinitrogen TetraOxide is not N2O4 in an empirical formula format. It is NO2.
Molecular formulas give the actual number of atoms.
We need to know the ratio of each element to determine the empirical formula.
This table will help to determine the ratio for an element.
Atom Mass Molar Mass Moles Smallest Mole Ratio
C 8.4g 12.0g 0.7/0.35 = 2 2
H 2.1g 1.0g 2.1/0.35 = 6 6
O 5.6g 16.0g 0.35/0.35 = 1 1
The simplest ratio could be decimals.
For certain decimals you may need to multiply everything by a common number.
Decimal Multiplying Coefficient
0.5 2
0.33/0.66 3
0.25/0.75 4
0.2, 0.4, 0.6, 0.8 5
Atom Mass Molar Mass Moles Smallest Mole Ratio
C 50.5g 12.0g 4.21/3.16 = 1.32 x 3 4
H 5.26g 1.0g 5.26/3.16 = 1.66 x 3 5
N 44.2g 14.0g 3.16/3.16 = 1 x 3 3
To find the molecular formula you need the molar mass. If you know the empirical formula.
Empirical Molecular
C2H6O ?
? 138g/mol
12.0(2)+1.0(6)+16.0= 46.0g/mol
Empirical Molecular
C2H6O ?
46.0g/mol 138g/mol
138.0/46.0= 3 (C2H6O)3
Empirical Molecular
C2H6O C6H18O3
46.0g/mol 138.0g/mol
-Not the actual atoms
Ex: - The empirical formula for Hydrogen gas is H
- Dinitrogen TetraOxide is not N2O4 in an empirical formula format. It is NO2.
Molecular formulas give the actual number of atoms.
We need to know the ratio of each element to determine the empirical formula.
This table will help to determine the ratio for an element.
Atom Mass Molar Mass Moles Smallest Mole Ratio
C 8.4g 12.0g 0.7/0.35 = 2 2
H 2.1g 1.0g 2.1/0.35 = 6 6
O 5.6g 16.0g 0.35/0.35 = 1 1
The simplest ratio could be decimals.
For certain decimals you may need to multiply everything by a common number.
Decimal Multiplying Coefficient
0.5 2
0.33/0.66 3
0.25/0.75 4
0.2, 0.4, 0.6, 0.8 5
Atom Mass Molar Mass Moles Smallest Mole Ratio
C 50.5g 12.0g 4.21/3.16 = 1.32 x 3 4
H 5.26g 1.0g 5.26/3.16 = 1.66 x 3 5
N 44.2g 14.0g 3.16/3.16 = 1 x 3 3
To find the molecular formula you need the molar mass. If you know the empirical formula.
Empirical Molecular
C2H6O ?
? 138g/mol
12.0(2)+1.0(6)+16.0= 46.0g/mol
Empirical Molecular
C2H6O ?
46.0g/mol 138g/mol
138.0/46.0= 3 (C2H6O)3
Empirical Molecular
C2H6O C6H18O3
46.0g/mol 138.0g/mol
Percent Composition
The percentage by mass of an element in a compound is always the same.
To find the percent by mass determine the mass of each element present in one mole.
Example: Hydrogen in Water (H20)
H=2.0g O=16.0g 18.0g/mol 2.0 / 18.0 = 0.111
= 11.1%
Sodium in Salt (NaCl)
Na=23.0g Cl=35.5g 58.5g/mol 23.0 / 58.5 = 0.393
= 39.3%
To find the percent by mass determine the mass of each element present in one mole.
Example: Hydrogen in Water (H20)
H=2.0g O=16.0g 18.0g/mol 2.0 / 18.0 = 0.111
= 11.1%
Sodium in Salt (NaCl)
Na=23.0g Cl=35.5g 58.5g/mol 23.0 / 58.5 = 0.393
= 39.3%
Monday, December 13, 2010
Density and Moles!
Density is a measure of mass per volume represented by this formula! The units are interchangeable, so if you need to figure out mass or volume, you can rearrange and solve!
Density is measured in g/l or g/ml.
Today we also learned about the density of Gases. They vary with temperature. The formula for finding the density of a gas is mm(molar mass)/ stp(22.4l)
Density is measured in g/l or g/ml.
Today we also learned about the density of Gases. They vary with temperature. The formula for finding the density of a gas is mm(molar mass)/ stp(22.4l)
Atoms-Molecules-Moles
Moles
^
Molecules(chemical formula of an element)
^
Atoms(what matter is made up of)
Awhile back, we learned how to go from atoms to molecules to moles. It's hard to explain, so let's do an example.
How many atoms are there in 1.5mol of Iron?
So first off, we write down what we know.
1.5Mol
You want to cancel moles, so you put moles on the bottom, and since you want atoms, you put atoms on top. Remember 1atom is equal to avogadro's number which is 6.02x10^23.
1.5mol x 6.02x10^23/1mol which equals 9.03x10^23 atoms, remember significant digits! = 9.0x10^23 atoms!
What's the difference between atoms and Molecules?
Molecules are composed of atoms, for example Lithium's atomic mass is 6.9.If an element is diatomic such as Carbon, it is 12x2.
(HOFBRCl)
^
Molecules(chemical formula of an element)
^
Atoms(what matter is made up of)
Awhile back, we learned how to go from atoms to molecules to moles. It's hard to explain, so let's do an example.
How many atoms are there in 1.5mol of Iron?
So first off, we write down what we know.
1.5Mol
You want to cancel moles, so you put moles on the bottom, and since you want atoms, you put atoms on top. Remember 1atom is equal to avogadro's number which is 6.02x10^23.
1.5mol x 6.02x10^23/1mol which equals 9.03x10^23 atoms, remember significant digits! = 9.0x10^23 atoms!
What's the difference between atoms and Molecules?
Molecules are composed of atoms, for example Lithium's atomic mass is 6.9.If an element is diatomic such as Carbon, it is 12x2.
(HOFBRCl)
Tuesday, November 23, 2010
Moles And Volume of a Gas
Mass <---> Moles <---> Volume
Ex. How many liters will 5.0 mol occupy at STP.
5.0Mol * 22.4L
-------- = 112 L
1 Mol
Ex. At STP a sample of Nitrogen gas contains 15.5. How many liters are there?
15.5 Mol * 22.4L
-------- = 347.2 L
1 Mol
Ex. STP an Hydrogen gas occupies 12 liters. How many moles of gas must be there?
12L * 1 Mol
-------- = 0.54 Mol
22.4L
- At a specific pressure and temperature one mole of gas occupies the same volume
- At 0'C 101.3 KPa 1 Mole = 22.4L
- This temperature and pressure is called STP
- 22.4L/Mol is the molar volume at STP
Ex. How many liters will 5.0 mol occupy at STP.
5.0Mol * 22.4L
-------- = 112 L
1 Mol
Ex. At STP a sample of Nitrogen gas contains 15.5. How many liters are there?
15.5 Mol * 22.4L
-------- = 347.2 L
1 Mol
Ex. STP an Hydrogen gas occupies 12 liters. How many moles of gas must be there?
12L * 1 Mol
-------- = 0.54 Mol
22.4L
Subscribe to:
Posts (Atom)