chem pwns.

Stiochemistry Investigation: Testing the Stoichiometric Method

Monday, January 31, 2011 /5:53 PM

Today in class we did a lab to determine if stoichiometry can accurately predict the mass of produced chemical reactions. In the experiment we dissolved 2g of Strontium Nitrate in 50ml of water. Then we dissloved 3g of Copper Sulphate in 50ml of water and combined the two solutions together. We created a precipitate (Strontium Suplhate and Copper (II) Nitrate). After mixing the solutions the precipitate will be separated by filtration, dried and weighed.

Next class we will be finishing up our lab by weighing our dried filter paper with precipitate.

~christina :]

Mass to Mass Conversions

Thursday, January 27, 2011 /2:21 PM

- mass to mass porblems involve one addition conversion

Example:

How many grams of H₂are produced from the decomposition of 1.1g of H₂O_2?

2H₂O₂ --> 2H₂ + O₂

1.1g H₂O_{2 x 1 mol}H₂O₂_x2H₂_{x 2g}H₂_{= 0.0647g}H₂

_34gH₂O₂H₂O_{2 1 mol}

~christina :]

Mole to Mass and Mass to Mole

Tuesday, January 25, 2011 /12:54 PM

- some questions will give you an amount of moles and ask you to determine the mass
-converting moles to mass only requires one additional step

Ex: How many grams of Colbalt (II) Suplur are required to produce 1.2mol of Colbalt?

CoS --> Co + S [balanced equation]

1.2mol Co x 1mol CoS = 1.2mol CoS
                    1mol Co
1.2mol CoS x 91g CoS    = 109.2
                     1mol CoS    = 109g

How many grams of Gold (III) Bromide are need to produce 3.7mol of Oxygen?

2AuBr₃ + O_{3 --->}Au₂O₃ + 3Br₂[balanced equation]

3.7mol O x 2mol AuBr₃₌7.4 molAuBr₃

_{1 mol O}

_7.4molAuBr₃x 436.7gAuBr₃_{= 1.6 x}10³

₂AuBr₃

_{How many moles of Silver Iodide are required to produce 9.8g of Silver?}

_{2AgI --> 2Ag + I}₂[balanced equation]

_{9.8g Ag x 2mol Ag =0.1816mol Ag}
_107.9Ag
_{0.1816mol Ag x 2mol AgI = 0.1816mol AgI}
_{2molAg = 0.18mol AgI}

~christina :]

Mole to Mole Conversions

Friday, January 21, 2011 /5:29 PM

- coefficients in balanced equations tell us the number of mole reacted or produced - they can also be used as conversion factors
3X + Y --> 2Z

- what you need over what you have

Examples:
- How many moles of Iron Fluoride are required to produced 2.7mol of pure Iron?
2Fe
F₃ --> 2Fe + 3F₂ _{2.7mol x 2/2 = 2.7mol} -When 3.8mol of Magnesium reacts with Lithium Iodide. How many mole of Magnesium should be produced? Mg + 2LiI --> 2Li + MgI2

3.8mol x 1/2 = 1.9mol

- How many moles of Potassium Nitride are required to produced 8.5mol of pure Potassium?

2K₃N --> 6K + N₂

8.5mol x 2/6 = 2.8 mol

~christina :]

Stoichiometry (Quantitive)

Wednesday, January 19, 2011 /6:20 PM

- stoichiometry is a branch of chemistry that deals with the quantitive analysis of chemical reactions
- it is a generalization of mole conversions to chemical reactions
- understahe 6 types of chemical reactions is the foundation of stoichiometry

6 types of reactions

Synthesis: - A + B --> AB
                - usually elements --> compounds
                - balance the following equations
1. 3Cu + 2P --> Cu₃P₂
2. Sr + S --> SrS
3. 3Mg + N₂ --> Mg₃N₂

Decomposition: - AB --> A + B
   - reverse of synthesis
   - always assume the compounds decompose into elements during decomposition
   - balance the following decomposition reactions
1. 2H₂O --> 2H₂ + O₂
2. C₆H₁₂O₆ --> 6C + 6H₂ + 3O₂
3. 6NaHCO₃ --> 3Na₂CO₃ + 3H₂O + 3CO₂

_{Single Replacement: - A + BC --> B + AC}
_{- balance the following reactions}
1. Mg + 2H₂O --> Mg(OH)₂ + H₂
2. Zn + 2HCl --> ZnCl₂ + H₂
3.2KI + Cl₂ --> 2KCl + I₂

_{Double Replacement:} - AB + CD --> AD + BC
                                 - balance the following : * metals always first*
1. NaCl + AgNO₃ --> NaNO₃ + AgCl
2. Pb(NO₃)₂ + 2KI --> PbI₂ + 2KNO₃
3.AgNO₃ + KCl --> AgCl + KNO₃

_{Neutralization: -reaction with an acid and a base}
_{-balance the following}
_1. HBr + NaOH --> NaBr + H₂O
_2.HCl + NaOH--> H₂O + NaCl
3. HBr + NaOH --> NaBr + H₂O

Combustion: - reactions of something (usually hydrocarbon) w/ air
                    - hydrocarbon combustion always produces CO₂+ H₂O
- balance the following
1. 2Cu + O₂ --> 2CuO
2. C₁₀H₈ + 12O₂ --> 10CO₂ + 4H₂O
3. C₃H₈ + 5O₂--> 3CO₂ + 4H₂O

~christina :]

Chapter Test: The Mole

Monday, January 17, 2011 /5:30 PM

Today in class we did our chapter test on the mole.

~christina :]

UNIT REVIEW

Thursday, January 13, 2011 /6:55 PM

Today we reviewed and Mr. Doktor gave us the block to work on our review package before our big unit test on moles. If we need any help or questions, find him before you write the test.

It will be out of 44 with 15 multiple choice and 29 long answers.

CHLOE KO

Empirical Formula

Monday, January 10, 2011 /7:47 PM

Empiricial Formula

empirical formulas are the simplest formula of a compound
show only the simpliest ratios, not the actual atoms

-the empirical formula for chlorine gas is Cl
-Dinitrogen tetraoxide ≠ N₂0₄= NO₂

example

a sample of an unknown compound is found to contain 8.5g of "N", 2.2g of "H" and 5.5g of "O". Determine the empirical formula.

Molecular Formula

-give the actual number of atoms

to determine the empirical formula we need to know the ratio of each element
to determine the ratio, fill in the table below for each problem:

example:

a sample of an unknown compound is found to contain 8.5g of "N", 2.2g of "H", and 5.5g of "O". Determine the empirical formula.

N₂H₆O

the simplest ratio may be decimals. For certain decimals you need to multiply everything by a common number.

http://www.youtube.com/watch?v=gfBcM3uvWfs

Molecular formulas

If you know an empirical formula to find the molecular formula you need the molar mass.

CHLOE KO

Percent Composition

Thursday, January 6, 2011 /7:37 PM

-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:
Find the percentage composition of a compound that contains 2.88g of gold in a 6.20-g sample of the compound.

2.88g/ 6.20g = 0.4645
                     = 46.4% [multiply by 100 to get percentage]

Determine the mass of Hydrogen present in a 5.1g sample of Water .
2g/ 18g = 0.11%
            = (0.11)5.1g = 56.1g
                              = 56g

Find the percentage composition of a compound that contains 3.36g of Berylium in a 9.99g sample of the compound.
3.36g/ 9.99g = 0.3363
                     = 33.6%