Sunday, January 30, 2011

LESSON 22

MASS TO MASS CONVERSIONS
  • Mass to mass problems involve one addition conversion
  • (WHAT YOU NEED OVER WHAT YOU HAVE)
  Grams of A                                    Grams of B
         |                                                    |
        to                                                   to
         |             (chemical equation)         |
   Moles of A  ----------------------->  Moles of B
                 (multiply by stoichiometric)

eg. What mass of potassium chloride is produced when chlorine reacts with 0.47 g of potassium fluoride?
Cl2  +  2KF  -->  2KCl  +  F2
                            0.47g  x  mol  x  2  x  74.6g  =  0.60 g of sodium chloride
                                           58.1g    2       mol


eg. When Polonium (IV) nitrate reacts with 7.5g of Rubidium fluoride. How many grams of polonium (IV) nitrate are required for a complete reaction.
Po(NO3)4  +  4RbF  -->  PoF4  +  4RbNO3
                          7.5g  x  mol  x  1  x  457g  =  8.2 g of polonium (IV) nitrate
                                     104.5g   4      mol

eg. What mass of cesium bromide is produced when bromine reacts with 0.35 g of cesium iodide?
Br2  +  2CsI  -->  2CsBr  +  I2
                         0.35g  x  mol  x  2  x  212.8  =  .29 g of cesium bromide
                                      259.8g    2       mol
         
-sherilyn

posted by chemistry11 @ 9:12 PM 0 Comments

LESSON 21

MOLE TO MOLE CONVERSIONS
  • Coefficients in a balanced equation tell us the #  of moles reacted or produced
  • They can also be used as a conversion factor
               3X  +  Y  =   2Z   (WHAT YOU NEED OVER WHAT YOU HAVE!!!)

eg. Carbon will react with cadmium oxide to produce carbon dioxide. How many moles of carbon dioxide will be produced if 0.69 mol of CdO is completely reacted?
C  +  2CdO   -->   2Cd  +  CO2
0.69 mol x  1 / 2   =  0.35 mol of CO2

eg. How many moles of indium oxide are required to produce 2.5 mol of indium (pure)?
2In2O3  -->  4In  +  3O2
2.5 mol (4In) x  2 / 4  =  1.3 mol of indium oxide

eg. Nitrogen can react with hydrogen to produce ammonia. How many moles of nitrogen will be needed to produce 0.54 mol of NH3?
N2  +  3H2  -->  2NH3
0.54 mol (2NH3) x 1 / 2  =  0.27  mol of nitrogen

-sherilyn
                                                       

posted by chemistry11 @ 7:44 PM 0 Comments

Thursday, January 20, 2011

Lesson 20: Stoichiometry

Stoichiometry- Quantitative Chemistry

·         Stoichiometry is a branch of chemistry that deals with the quantitative analysis if chemical reactions
·         It is generalization of mole conversions to chemical reactions
·         Understanding the 6 types of chemical reactions is the foundation of stoichoimetry

6 Types of Reactions


1.      Synthesis
2.      Decomposition
3.      Single Replacement (S.R.)
4.      Double Replacement (D.R.)
5.      Neutralization
6.      Combustion



Synthesis A+B →AB
·         Usually elements→compound
·         Balance the following reactions:
1.      2K + Br2 → 2KBr
2.      4Al + 3O2 → 2Al2O3
3.      2Fe + 3Cl2 → 2FeCl3

Decomposition AB→ A+B
·         Reverse of synthesis
Ø  Always assume the compounds decompose into elements during decomposition
·         Balance the following equations:
1.      2NaCl → 2Na + Cl2
2.      2KI → 2K + I2
3.      Mg3N2 → 3Mg + N2

Single Replacement (S.R.) A+ BC → B + AC
·         Balance the following equations:
1.      Zn + CuCl2 → ZnCl2 +Cu
2.      Cl2 + 2NaBr → 2NaCl + Br2
3.      Mg + 2HCl → MgCl2 + H2

Double Replacement (D.R.) AB+CD → AD + BC
·         Balance the following equations;
1.      AgNO3 + HCl → AgCl + HNO3
2.      Fe2O3 + 6HCl → 2FeCl3 + 3H2O

Neutralization
·         Reaction between an acid and a base
·         Balance the following equations;
1.      HCl + NaOH → NaCl + HOH
2.      Ca(OH)2 + H2CO3 → CaCO3 + 2 HOH

Combustion
·         Reactions of something (usually hydrocarbon) with air.
·         Hydrocarbon combustion always produces CO2 + H2O
·         Balance the following equations:
1.      C3H8 + 5O2 → 3CO2 + 4H2O
2.      C5H12 + 8O2 → 5CO2 + 6H2O
3.      C7H6O3 + 7O2  → 7CO2 + 3H2O


Kelly

posted by chemistry11 @ 8:44 PM 0 Comments

Sunday, January 16, 2011

Lesson 20: EMPIRICAL FORMULAS.

  • Empirical formulae are the simplest formulae of a compound.
  • They only show the simplest ratios, not the actual number of atoms.
Molecular formula: C8H18 ----- Empirical formula: C4H9

Empirical formula for a diatomic atom like Cl2 is Cl.

Dinitrogen tetraoxide = Molecular: N2O4 = Empirical: NO2

  • To determine the empirical formula, we need to know the ratio of each element.
  • A sample of an unkown compound is foundto contain 8.4 g of C, 2.1 g of H and 5.6 g of O. Determine the ratio. Determine the empirical formula. 

    |  Atom   |   Mass   |   Molar Mass   |   Moles   |   Moles/Smallest Mole   |   Ratio   | 
                C            8.4             12.0                 0.7                        2                          2
                H            2.1              1.0                  2.1                        6                          6
                O            5.4             16.0                0.35                       1                          1

Empirical Formula: C2H6O

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

                                       DECIMAL       |       MULTIPLYING COEFFICIENT     
                                    0.5                                      2
                            0.33 or 0.66                               3
                            0.25 or 0.75                               4
                         0.2, 0.4, 0.6, 0.8                            5


This guy is awesome..





.. but Mr. Doktor's still the best!




»Loureal Agustin.

posted by chemistry11 @ 10:09 PM 0 Comments

Lesson 19: PERCENT COMPOSITION

  • The percentage by mass of an element in a compound is always the same.
  • To find the pervent by mass, determine the mass of each element present in one mole.
EXAMPLE:
  • Determine the percent by mass of Carbon in Octane (C8H18).
                   Carbon's molar mass multiplied by 8              96 g       =     84%
                   Total molar mass of Octane (C8H18)             114 g


»Loureal Agustin.

posted by chemistry11 @ 9:52 PM 0 Comments

Lesson 18: DENSITY & MOLES

Density is the measure of mass per volume.

D = mass / volume

> density is commonly measured in g/L or g/mL.

EXAMPLE:
  • Water has a density of 1.0 g/mL. Determine the mass of 11.5 mL of water.
  • How many moles are in 11.5 mL of water?
           1.0 g   x   11.5 mL = 11.5 g = 12 g
            mL

           11.5 g   x    1 mol    = 0.64 mol
                             18.0 g

Density of Gases varies with temperature
  • At STP we can find density by:
                Molar Mass                      MM                         g/mol    
              Molar Volume              22.4 L/mol                  L/mol

EXAMPLE:
  • Determine the density of methane (CH4) @ STP.
               CH4    =    12.0 + 1(4)    =    16.0 g/mol

               16.0 g/mol      =    0.714 g/L
               22.4 L/mol


»Loureal Agustin.

posted by chemistry11 @ 9:48 PM 0 Comments