Jordan+&+Darian

 = ....................................................... Stoichiometry =



__Stoichiometry-__ the calculation of quantitative (measurable) relationships of the reactants and products in a balanced chemical reaction.
** Cookie Example: **
 * **You have:**
 * 2 cups flour,
 * 3 cups sugar
 * 1/2 tsp vanilla
 * 2 sticks butter
 * 3 cups chocolate chips
 * **These ingredients are for a total of 4 dozen (48 cookies)**

//**What if you want 12 dozen cookies?**//
 * **You have:**
 * 6 cups four
 * 9 cups sugar
 * 1 1/2 tsp vanilla
 * 6 sticks butter
 * 9 cups chocolate chips
 * **These ingredients are for a total of 12 dozen (114 cookies)**

This example shows that because each total differs, the amount of ingredients needed also differs. This is similar to stoichiometry in the way that to get the total amount needed you need to change the number of each element.

Steps To Stoichiometry:__ **
 * __[[image:stairs.png width="160" height="127" align="right"]]
 * 1) Step up an equation.
 * 2) Balance the equation.
 * 3) Apply the multipilication factor.
 * 4) Determine the number of grams using molar mass

To download, select save, unzip the file, and double click on set up.
This program will not solve your stoichiometry calculations unless you know what you are doing. It does not use any form of generic A → B equation, instead A = what you have and B = what you want.

//**Example:**//

 * Use this equation: 2H 2 O → 2H 2 + O 2
 * You have 400 grams of water. How many grams of Oxygen is left over?
 * To solve this equation, you must use stoichiometry. The grams to grams portion of stoichiometry is the most complex, but if you can do the grams to grams conversion, all of the other types of conversions will follow.
 * You find out what the molar mass is per mole, for both water and oxygen.
 * **Answer:**
 * **400 grams H 2 O (1 mol H 2 O / 18 grams H 2 O) (1 mol O 2 / 2 mol H 2 O) (32 grams O 2 / 1 mol O 2 ) = 355.56 grams**

// Example: //

 * Use this equation 2H 2 O → 2H 2 + O 2
 * You have 20 liters of water. How many grams of oxygen is left over?
 * To solve this equation, you must use stoichiometry. To convert anything involving liters, instead of using molar mass such as that in a grams related equation, you use 22.4.
 * You only need to find the molar mass of Oxygen.
 * **Answer**
 * **20 liters (1 mol / 22.4 liters) (1 mol / 2 mol) (32 grams / 1 mol) = 14 grams**



__Percent Composition__ - tells the percent, by mass of a particular compound.

 * 1) Find the molar mass of the compound.
 * 2) Find the mass of the part of the compound you need.
 * 3) Determine the percent.

// Example: //

 * CO 2= 44 g/mol total
 * C= 12 grams O 2= 32 grams
 * 12/44= .27 * 100 = **27%**
 * 32/44= .73 * 100 = **73%**

// Example: //

 * C 6 H 12 O 6 = 180 g/mol total
 * C 6 = 72 grams H 12 = 12 grams O 6 =96 grams
 * 72/180= .40 * 100 = **40%**
 * 12/180 = .07 * 100= **7%**
 * 96/180 = .53 * 100 = **53 %**



__Empirical Formulas-__ smallest ratio of atoms in a compound.

 * 1) Find the number of moles of each element in the compound.
 * 2) Set the elements in a small whole number ratio.

// Example: //

 * A compound is 2.8 grams N and 6.4 grams O by mass:

...................... 14 g N ...................... 16 g O
 * ................................................................ __Reduces to:__**
 * 2.8 g N __1 mole__ = .2 mol ...................→. **.**1 mol
 * 6.4 g O __1 mole__ = .4 mol ....................→ . **.**2 mol

.....................................** 1:2 ratio ** **N + O 2 = NO 2 **

// Example: //

 * A compound is 65.4 grams As and 34.6 grams O by mass

.................................................... ....... __**Rounds to:**__ ........................ 75 g As ....................... 16 g O ................................ **1: 2.5 ratio** **As + O = As 2 O** **5**
 * 65.4 g As __1 mole__ = .87 ......................... 1 mol
 * 34.6 g O __1 mole__ = 2.7 ........................ 2 2.5 mol



__Molecular Formula-__ a chemical formula based on analysis and molecular weight

 * 1) Find the emperical formula.
 * 2) Find the mass of the emperical formula.
 * 3) Divide the molar mass of the molecular formula by the molat mass of the emperical formula.
 * 4) Multiply the emperical formula by the answer from step 3.

//Example://

 * A compound has a empericial formula of CH and a molar mass of 78 grams.mol. Find the molecular formula.


 * CH= 13 grams/mol
 * 78/13= 6
 * CH * 6 = **C** **6** **H** **6**

**__Citations:__** All content on this page is from notes taken in class or the book. All grahpics on this page are from the search engine google, then their respectable website.

--- Page made by: Jordan Schaer and Darian Gist