Mass to Mass conversions

• Mass to mass problems involve one addition conversion

Example:

•  Lead (IV) Nitrate reacts with 5.O grams of Potassium Iodide. How many grams of Lead (IV) Nitrate are required for a complet reaction?
• Pb(NO3)4 + 4 KI---> 4 KNO3 + PbI4
• 5.0 g x 1 mol / 166 g x 1 / 4 x 455.2 g / 1 mol = 3.4 g

Example:

• How many grams of O2 are produced from the decomposition of 3.0 grams of Potassium Chlorate?
• 2 KClO3 ---> 2 K + Cl2 + 3 O2
• 3.0 g x 1mol / 106.6 g x 3 / 2 x 32 g / 1 mol = 1.4 g  

Example:

• When solid Zinc reacts with hydrochloric acid what mass of Hydrogen gas is produced when 2.5 grams of Zinc react?
• Zn + 2 HCl ---> H2 + ZnCl2 
• 2.5 g x 1 mol / 65.4 g x 1 / 1 x 2 g / 1 mol = 0.076 g 

Example:

• If 100 grams of Octane are burnt in a car engine what mass of Water is produced?
• C8H8 + 25 / 2 O2 ---> 9 H2O + 8 CO2 
• 100 g x 1 mol / 118 g x 9 / 1 x 18 g / 1 mol = 142 g 

-Eva          

Classifying Chemical Reactions

I. 4 General Types of Reactions

                A. Direct combination, decomposition, single replacement,  double replacement

                B. By knowing the type of chemical reaction you can predict the  products

II. Direct Combination Reactions (AKA Synthesis Reactions)

                A. 2 or more reactants come together to form a single product. The single product is always more complex than the reactants.

A + B = AB

III. Decomposition Reactions

                A. The Reactant breaks down into smaller parts

                B. When a complex, single compound is broken down into two or more smaller/ simpler compounds or elements

AB = A + B

IV. Single Replacement Reactions

                A. A single element replaces another element that is part of a compound

                B . A more active element will usually replace a less active element

                               

AB + C = AC + B

V. Double Replacement Reactions

                A. Atoms or ions from 2 different compounds replace each other

                B. This type of reaction has 2 compounds as reactants and 2 compounds as products

AB + CD = AD + CB

Nicole.

Mole to Mole conversions

• coefficients in balanced equations tell us number of moles reaced or produced 
• they can also be used as conversions factors

           3 X + Y ---> 2 Z

• What you need over what you have

Example: 
 

• If 0.15 moles of methane (CH4) are consumed in a combustion rxn how many moles of CO2 are produced?
• CH4 + 2 O2 ---> CO2 + 2H2O
• 1.5 moles of CH4 x 1 / 1 = 0.15 mol of CO2

Example:

• When 2.1 x 10^-2 mol of Aluminium Hydroxide react with Sulphiric acid. How many moles of Aluminium Sulphate are produced?
• 2 Al(OH)3 + 3 H2SO4 ---> 6 HOH + 1 Al2(SO4)3 
• 2.1 x 10^-2 mol x 1 / 2 = 4.2 x 10^-2 mol of Al2(SO4)3

Example:

• How many moles of Bauxite (Aluminium Oxide) are requred to produce 1.8 mol of pure Aluminium?

2 Al2O3 ---> 4 Al + 3 O2

• 1.8 moles of Aluminium x 2 / 4 = 0.9 mol of Aluminium Oxide

Example:

• When 1.5 mol of copper react with Iron (II) Chloride how many moles of Iron should be prodeced?
• Cu + FeCl2 ---> Fe + CuCl2
• 1.5 mol x 1 / 1 = 1.5 mol of Fe    

-Eva 
 

Molecular Formulas

• if you know an empirical formula to find molecular formula you need the molar mass

Example:

• The empirical formula for a substance is CH2O and its molar mass is 60.0 g / mol. Determine the molecular formula.
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  Empirical Formula	Molecular Formula
  CH2O	C2H4O2
  30.0 g / mol	60.0 g / mol

  
  

Example:

• The empirical formula for a compound is C2H6O and the molar mass is 138 g / mol. Determine the molecular formula.
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  Empirical Formula	Molecular Formula
  C2H6O	C6H18O3
  46 g / mol	138 g / mol

  
  
   
  -Eva

Empirical Formulas

A. Empirical formula: The simplest form of a compound

                a. They show only the simplest ratios, not the actual atoms

                                i. The empirical formula for chlorine gas is Cl, not Cl₂

                                ii. dinitrogen tetraoxide

                                                -It does not equal N₂O₄

                                                -To write it properly you have to only use the ratio;
                                                 for every nitrogen there are two oxygen atoms

                                                -The proper empirical formula is NO₂

B.  Molecular formula: Give the actual number of atoms

Ex.

Molecular Formula	Empirical Formula
P4O10	P2O5
C10H22	C5H11
C6H18O3	C2H6O

C. To determine the empirical formula we need to know the ratio of each element

Atom	Mass	Molar Mass	Moles	Moles/ Smallest Mole	Ratio

Ex. A sample of an unknown  compound is found to contain 8.4 g  of "C", 2.1 g of "H, and 5.6 g of "O". Determine the empirical formula.

Atom	Mass	Molar Mass	Moles (mass) x ( 1 mole/molar mass)	Moles/ Smallest Mole	Ratio
C	8.4 g	12.0	(2.4g) x (1 mol/ 12.0) = 0.7 mol	(0.7 mol/ 0.35 mol) = 1	2
H	2.1 g	1.0	(2.1 g) x (1 mol / 1.0) = 2.1 mol	(2.1 mol/ 0.35) = 6	6
O	5.6 g	16.0	(5.6 g ) x (1 mol/ 16.0) = 0.35 mol	(0.35 mol/ 0.35 mol) = 1	1

Empirical Formula: C₂H₆O

Ex. Find the empirical formula of a compound, given that a 48.5 g sample of the compound contains 1.75g of carbon and 46.75 g of bromine.

Atom	Mass	Molar Mass	Moles (mass) x ( 1 mole/molar mass)	Moles/ Smallest Mole	Ratio
C	1.75 g	12.0	(1.75 g) x (1 mol/ 12.0) = 0.145 mol	(0.145 mol/  0.145 mol) = 1	1
Br	46.75 g	79.9	(46.75 g) x (1 mol/ 79.9) = 0.585 mol	(0.585 mol / 0.145 mol) = 4	4

Empirical Formula: CBr₄

                       

                a. 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

Ex.  Determine the empirical formula of a compound that contains 2.16 g of aluminum, 3.85 g of sulfur, and 7.68 g of oxygen.

Atom	Mass	Molar Mass	Moles (mass) x ( 1 mole/molar mass)	Moles/ Smallest Mole	Ratio
Al	2.16 g	27.0	(2.16 g) x (1 mol/ 27.0) = 0.08 mol	(0.08 mol/ 0.08 mol) = 1	1( x 2) = 2
S	3.85 g	32.1	(3.8 g) x (1 mol/ 32.1) = 0.119 mol	(0.119 mol/ 0.08 mol) = 1.5	1.5 (x 2) = 3
O	7.68 g	16.0	(7.68 g) x (1 mol/ 16.0) = 0.48 mol	(0.48 mol/ 0.08 mol) = 6	6 (x 2) = 12

Empirical Formula: Al₂S₃O₁₂

_-Nicole