Mass+and+Volume+Relations-+Moles,+Empirical+Formula+and+Limiting+Reactants

=Mass= To determine the mass of a compound add up the atomic masses of each individual element represented. If the element has a subscript next to it be sure to multiply only that element before adding it to the rest. =Moles= There are many different ways to measure substances. Whether we are attempting to find mass or volume there are different units that can often be difficult to keep track of. For this reason scientists use **moles** when comparing elementary identities. A mole (mol) is an SI based unit describing the amount of substance in a system which contains many elementary entities. These elementary identities may be: atoms, molecules, ions, electrons, or other particles. Remember that a mole is not a weight itself, rather it is a ratio to weight; a mole corresponds to a certain weight of a substance.

=Determining the moles in a substance= We determine the number of moles there is of a substance based on its mass in grams. The atomic mass of elements that is given to you on the periodic table can also be described as its molar mass. This is because it describes how many grams are in one mole of that substance. For example: Carbon has an average atomic mass of 12.01. This means that there are 12.01 grams/mole

If we are given the total grams of a substance, along with the formula of the substance, we can determine how many moles are in that substance. __74 grams KCl | 1 mole KCl__ = .992 moles KCl .... ............... . | 74.6 grams KCl
 * Example 1**: How many moles of KCl are there in 74 grams of KCl?

This can also be done in reverse. If you are given moles of a substance it can be converted into grams:


 * Example 2**: How many grams of KCl are there in .992 moles of KCl?

__.992 mol KCl | 74.6 grams KCl__ = 74 grams KCl .... ................ | 1 mol KCl

Note : Don't forget that if you have a hydrate you must multiply the mass of the H20 by the coefficient and then add it to the rest of the compound's mass. =Avogadro's Number= Amedeo Avogadro, an Italian scientist first proposed the idea that at a constant volume of any gas, mass is proportional to the number of atoms or molecules in one mole of that gas. In 1909, French Physicist Jean Perrin determined the Avogadro's constant and naming in honor of Amedeo Avogadro. Avogadro's number is the number of things (atoms, molecules, particles) in one mole of that element. This is true of every type of atom. In one mole of any element there are 6.022 x 10^23 elementary entities. Therefore, Avogadro's number is what defines a mole. We use Avogadro's number when being asked to determine how many molecules are in a substance (if given grams or moles), or how many grams or moles there are of a substance (if given molecules).


 * Example 3**: How many molecules are there in 60 grams of KCl?

__60 grams KCl | 1 mole KCl__ __|__ __6.022 x 10^23 molecules KCl__ = 4.84 x10^23 molecules KCl .... ................. | 74.6 g KCl | 1 mol KCl

=Empirical Formula= Essentially, we know how to calculate what a formula is composed of. We use the given formula to calculate it's entire atomic mass, then use the ratio to calculate the number of moles of each substance, and then how much (mass) of each substance there is in a given sample. If we reverse the process, we can easily determine the empirical formula of a compound after having been given information about it's composition. The process is actually quite simple and can be put into 4 steps (for these steps we are starting with the number of grams of each substance given in the problem) > - Divide each mole number by the smallest number of moles calculated > - If the number is too far to be rounded, multiply each number (after having been divided) by a factor that will make that number whole Example 4: **A compound was analyzed and found to contain 9.52 g N, 27.18 g O, 6.16 g H, and 57.14 g C. What is the empirical formula of the compound?** N: __9.52 grams| 1 mole N__ = __.68 moles N__ 1.0 x 2 = **2** .. .................... | 14.0 g N... .68 moles Nitrogen has lowest mole value: divide by it O: __27.18 grams | 1 mole O__ = __1.70 moles O__ = 2.5  x 2 = **5** ...................... | 16 g O<span style="color: rgb(255, 255, 255);">........ .68 moles <span style="color: rgb(0, 0, 255);">2.5 can't be rounded to a whole number: multiply all answers by 2 to obtain whole numbers H: __6.16 grams | 1 mole H__ = __6.10 moles H__ = 8.97 x 2 = **18** <span style="color: rgb(255, 255, 255);">.<span style="color: rgb(255, 255, 255);">... .................. | 1.01 g H<span style="color: rgb(255, 255, 255);"> ...... 68 moles C: __57.14 grams | 1 mole C__ = __4.76 moles C__ = 7 x 2 = **14** <span style="color: rgb(255, 255, 255);">.<span style="color: rgb(255, 255, 255);">. ..................... | 12.01 g C<span style="color: rgb(255, 255, 255);"> .... .68 moles Once the number of moles is calculated for each element, place the number of moles as a subscript next to the symbol of that element. Empirical formula = **C 14 H 18 N 2 O 5 ** =Limiting Reactants= In a chemical reaction, there will usually be one chemical that is used up completely, and one that is left as excess.
 * 1) Write out the number of grams for each element
 * 2) Using molar mass information on the periodic table, convert the mass of each element into moles
 * 3) Once you have found the number of moles in each substance, determine which element has the smallest number of moles
 * 1) Round to the nearest whole number and set those numbers as the subscripts to the formula

For example: **A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is the limiting reactant?** > NH3: N= <span style="color: rgb(255, 0, 0);">14.06 g > <span style="color: rgb(255, 255, 255);">........ __H= 1.01 x 3=__ <span style="color: rgb(0, 0, 0);">__<span style="color: rgb(255, 0, 0);">3.03 g __ > <span style="color: rgb(255, 255, 255);">...... <span style="color: rgb(0, 0, 0);">NH3= <span style="color: rgb(255, 0, 0);">17.09 g > O2: O = <span style="color: rgb(255, 0, 0);">16.0 <span style="color: rgb(0, 0, 0);">x 2 = <span style="color: rgb(255, 0, 0);">32.0 g  > NO: N= <span style="color: rgb(255, 0, 0);">14.06 > <span style="color: rgb(255, 255, 255);">....... __<span style="color: rgb(0, 0, 0);">O __<span style="color: rgb(0, 0, 0);">__= <span style="color: rgb(255, 0, 0);">16.0 __ > .<span style="color: rgb(255, 255, 255);">...... NO = <span style="color: rgb(255, 0, 0);">30 g > __2 grams NH3 | 1 mol NH3 | 4 mol NO | 30 g N__O =3.53 g NO > <span style="color: rgb(255, 255, 255);">....................| |17 g NH3 | 4 mol NH3 | 1 mol NO > > __4 grams O2 | 1 mole O2 | 4 mol NO | 30 g NO__ =3 g NO > <span style="color: rgb(255, 255, 255);">............... .| | 32 g O2 | 5 mole O2 | 1 mol NO Although it is only a .53 difference, O2 produces less NO product than NH3, therefore it is the limiting reactant.
 * Limiting Reactant:** the reactant that is used up completely in a chemical reaction and limits the amount of product being produced. Once the limiting reactant is completely used up, the reaction stops.
 * Excess Reactant:** the reactant that is left over at the end of a chemical reaction because the limiting reactant is used up and there is nothing else for it to react with.
 * To determine the limiting reactant we must use stoichiometry and compare the mole to mole ratio of both reactants to a common product. Whichever calculation results in a smaller value for the common product is the limiting reactant.
 * 4 NH3(g) + 5 O2(g)[[image:http://www.chem.tamu.edu/class/majors/tutorialnotefiles/arrow.gif width="36" height="13"]]4 NO(g) + 6 H2O(g) **<span style="display: block; color: rgb(0, 0, 0); text-align: left;">
 * 1) Determine the molar mass of each NH3 and O2:
 * 1) Choose a common product & determine it's molar mass in order to later calculate how much of the it is being produced by each individual reactant: we will choose NO
 * 1) <span style="color: rgb(0, 0, 0);">Use all the information gathered and stoichiometry for the calculations

One helpful way to review this is to use the sandwich analogy: It is a fact that **two** slices of bread and **one** piece of ham make up **one** ham sandwich and no other combination works. Say you have **six** slices of bread and **ten** pieces of ham, how many sandwiches can you make? But if you take the ham route, you would have used up all of the bread before making all ten sandwiches. This means that the amount of bread **limits** the amount of sandwiches that you can make; therefore making it the **limiting reactant** while the ham is the **excess reactant**.
 * If you go by the slices of bread, then you can make **three** sandwiches.
 * If you go by the pieces of ham, then you can make **ten** sandwiches.

[]