Types of Reaction

There's a lot of types of reactions such as:

1. Synthesis
2. Decomposition
3. Single Replacement
4. Double Replacement
5. Combustion
6. Neutralization

• SYNTHESIS:

In the Synthesis Reaction, 2 or more chemical will combine with each other to form another new chemical with the form:
          A+B --> AB
Ex: 2 Na + 1 Cl2 -->2 NaCl ( don't forget your balance man )

• DECOMPOSITION:

In the Decompostion Reaction, one reactant will be broken down to 2 or more product in the fo22rm :
          AB --> A + B
Ex: CaCO3 --> CaO + CO2

• SINGLE REPLACEMENT:

In the Single Replacement Reaction, the element replace an ion in an ionic compound. Metal replaces positive ion ( Cations ) and non-metal replaces  negative ion ( Anion ) in the form :
         A + BC --> AC + B or
         A + BC --> BA + C
Ex: 2 Na + BaCl2 --> 2 NaCl + Ba

• DOUBLE REPLACEMENT: 

In the Double Replacement Reaction, 2 ionic reactant compound exchange ions to form 2 new ionic products in the form:
         AB + CD --> AD + CB
WARNING!!: Unless there's a solid formed, the reaction is not occur. And you must write a Net Ionic Equation if it occurs including the states.
Ex: 2 Na3PO4 + 3 Ca(NO3)2 --> 6 NaNO3 + Ca3(PO4)2
Net Ionic Equation: 3Ca (s) + 2 (PO4) (aq)  --> Ca3(PO4)2 (s)

• COMBUSTION: 

Combustion Reaction is where burning involved in the form:
          AB + O2 --> AO + BO
Ex: C3H8 + 5O2 = 3CO2 + 4H2O

•  NEUTRALIZATION:

Neutralization is basically same as double replacement but it's a reaction between an Acid and a Base in the form :
          Acid + Base --> Salt + H2O
Ex: HCl + NaOH --> NaCl + H2O

 

Activity series chart

Solubilites Chart

Molar Volume

Molar Volume
-Molar volume of a gas at STP gases expand and contract with changes in temperature and pressure.
-We have a standard condition to compare volume of gases called STP(Standard Temperature Pressure)

•            STP = 1 atmosphere at pressure and a temp at 0 degrees celsius or 273.15K
•            At STP 1 mole of gas occupies 22.4L, thus we can create the conversion factors 
• 22.4L/mole of gas  OR mole of gas/22.4L

EXAMPLE

Calculate the volume occupied by 3.4g of ammonia at STP.

SOLUTION

Molar mass of ammonia (NH^3) = (1x14) + (3x1) = 173 mol - - - - - - moles of ammonia = 3.4 g x 1 mole / 17g

Molar volume = 22.4L
                       = 0.2 moles x 22.4L/mole
                       = 4.5L
Volume of occupied by 3.4 g of ammonia at STP = 4.5L

Solution Preparation and Dilution

      Chemicals are shipped around the world in their most concentrated forms. If they were not, we would be shipping lots of water along with the chemicals, which is less cost effective. 

      Sometimes we need to add some water to a solution to form a lower molarity solution.(or add more solute, to form a larger molarity solution.)

      The key idea is that the moles of solute is constant.

      General Form

                             M1L1=M2L2

                              moles solute before=moles solute after

       E.g

                    Minh has 3.00 L of 18.0 H2SO4. But teacher tells he to make 0.900L of 2.00M H2SO4. What should he do?

                    

                      Here is Minh's solution.

                       M1L1=M2L2

                     18.00M x L1= 0.900L X 2.00M

                                      L1=0.100L

                        0.900-0.100=0.800L of water

                       Minh just need to add 0.800L of water to the original solution and take out 0.900L of it.

Lab 4C Formular of a Hydrate

Objectives
1.  To determine the percentage of water in an unknown hydrate.
2.  To determine the moles of water present in each mole of this unknown hydrate when given the molar mass of the anhydrous salt.
3. To write an empirical formula of the hydrate.

Supplies
Equipment:  lab burner, crucible and lid, crucible tongs, pipe stem triangle

Chemical reagents:  approximately 5 g of a hydrate, water.

Procedures
1. Try to dry the crucible by heating it up, make sure the flame is blue

2. Remove the crucible and cool it for 3 mins

3. Determine mass of the crucible

4. Put in the hydrate, record the mass of crucible

5. Heat up for 5 mins

6. Cool it down for 3 mins then record the mass of crucible

Some more infomations

Hydrate is the compound containing water

Anhydrous is a substance contains no water.

Experiment’s errors:

1.     Hydrate decomposed before heating finish

2.    Water in air can affect the crucible’s mass

All types of Mole Conversion ( summary )

Mole x 6,022 x 10^23 = Molecules/Particles/Formula Units

Molecules/Particles/Formula Units x #atoms = atoms

Mole x molar mass ( g ) = mass

Mole x 1/Molarity = Volume

Mole x 22,4 = Volume at STP

Molar Concentration/ "Molarity" of solutions

Molar concentration is the number of moles of solute in one litre of solution. "M" is used to denote molar concentration with units of moles/litre.

Formula: Molarity = moles of solute (mol)/ volume of solution (L)

Examples:

If there is 1.5 moles of NaCl in 1.0L of soltuion, what is the molar concentration?

M= 1.5mol NaCl/1.0L

= 1.5M NaCl or 1.5mol/L NaCl

Calculating the Empirical and Molecular Formula and Percent Composition

There are two kinds of formulas that we can express in chemistry "11".   The empirical and molecular formula.

The Empirical Formula


-Gives the lowest terms of atoms OR moles.
-All formulas of ionic components are empirical formulas.

The Molecular Formula


-Gives all atoms which make up a molecule.
-This can be ionic or covalent compounds.


EG: C6H12O6

To put it into empirical formula, reduce to lowest terms.  Therefore C6H12O6 = CH2O(empirical).

Remember: 1) molecular  = empirical x whole #
2) molecular formula mass = empirical x whole #
3) mass(mole) = EFM(g) x whole #

How to determine empirical formula given mass


EX: Determine the empirical formula of Fe and O given 10.87g of Fe and 4.66g of O.

1) Convert grams ------> moles

10.87 x 1/55.8g = 0.1948 mol
 4.66 x 1/16.0g = 0.0291 mol

2) Divide each molar amount by the smallest molar amount.

Fe 0.1948/0.1948 = 1
O 0.291/0.1948 = 1.49 ~ 1.5

3) Scale ratios to whole #'s by multiplying.

1.5 x "2" = 3(whole number)
1 x "2" = 2(whole number)


Therefore, the empirical formula is Fe2O3

PERCENT COMPOSITION


- The % by mass of the elements in a compound.
                -calculate molar mass.
                -calculate each element's % of that total(ONE DECIMAL PLACE)
This tells what part of a compound each component element makes up.

Remember: %composition = mass of element/mass of compound x 100%


The ratio of moles can be determined from percent composition.  How?

1)Assume 100.0g of all material
2)Convert all %'s to grams.
3)Follow the steps to solve empirical formula.