As stated previously in one of the older blogs, volume at stp can be found using the conversion factor 22.4L/Mol!
Heat can be included as a separate term in chemical reactions. It's called enthalpy.
reactions that release heat are exothermic
reactions that absorb heat are endothermic!
Heres an example!
If 3.0 Grams of Aluminum Oxide decomposes, what volume of oxygen is produced?
So first, write down what you're solving for, cancel units then convert to moles. Once you have your moles, write down what you have over what you need, then convert into liters using STP. You should get your answer then!
Chemistry 11 Mr. Doktor Block G Run by: Mark W. Guarin Julius O. Francia Timothy Samson
Monday, February 7, 2011
Moles to mass to moles to mass?
Wow, that sure sounded confusing. Well, some problems you will encounter in Chemistry 11 will give you an amount of moles and will ask you to determine the mass. Converting moles to mass only requires one additional step!
Heres an example!
How many grams of water are produced if 0.24 moles of phosphoric acid is neutralized by Bariun Hydroxide?
So first write your chemical equation.Balance. Right what you're solving for, which is water! Write what you have over what you need. Then take the grams of all the elements in H20 (Hydrogen and Oxygen) and divide it by a mole. That's how you solve it!
If you're struggling, here's some tips.
1) If you're stuck, always start out by writing down your chemical equation then balance. Double check, if your equation is wrong, your answer will come out wrong.
2) Write down what you need to solve for and always remember what you need to cancel.
3) Significant Digits, always count the number of sig figs, in this question it is 2. You do not wanna lose half a mark!
Heres an example!
How many grams of water are produced if 0.24 moles of phosphoric acid is neutralized by Bariun Hydroxide?
So first write your chemical equation.Balance. Right what you're solving for, which is water! Write what you have over what you need. Then take the grams of all the elements in H20 (Hydrogen and Oxygen) and divide it by a mole. That's how you solve it!
If you're struggling, here's some tips.
1) If you're stuck, always start out by writing down your chemical equation then balance. Double check, if your equation is wrong, your answer will come out wrong.
2) Write down what you need to solve for and always remember what you need to cancel.
3) Significant Digits, always count the number of sig figs, in this question it is 2. You do not wanna lose half a mark!
Monday, January 24, 2011
Mole to Mole Conversion
- 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
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%
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