mOle to mole conversions

What is a Mole Ratio?  A mole ration is a ration between the amounts in moles of any two compounds found in a chemical reaction. Mole ratios are used as conversion factors between products and reactants in many chemistry problems.

Key things to remember when doing Mole to Mole conversions:
1. Write the chemical equation
2. Balance the chemical equation
3. Understand what the question is asking
4. Now place what you need over what you have. It may not be easy to grasp so here's an example..
If 0.15 mol of methane are consumed in a combustion reaction. How many moles of CO2 are produced?
CH4 + 02 ---> CO2 + H2O

(If you forgot..refer to the steps, for solving mole to mole conversions)


the answer is....
CH4  + 202---> CO2 + 2H2O

Want more problems??

How many moles of bauxite (Aluminum oxide) are required to produce 1.8 mol of pure Aluminum?


Al2O3 ----> Al + O2           




Answer:
2AlO3------> 4Al+3O2






When 1.5 mol of Copper react with Fe (II) chloride. How many moles of Iron should be produced?
2Cu + FeCl2------> 2CuCl+Fe


Answer: 
nothing is changed

Mole to Mole conversions

What is a Mole Ratio?  A mole ration is a ration between the amounts in moles of any two compounds found in a chemical reaction. Mole ratios are used as conversion factors betweenproducts and reactants in many chemistry problems.

Key things to remember when doing Mole to Mole conversions:
1. Write the chemical equation
2. Balance the chemical equation
3. Understand what the question is asking
4. Now place what you need over what you have. It may not be easy to grasp so here's an example..

If 0.15 mol of methane are consumed in a combustion reaction. How many moles of CO2 are produced?
CH4 + 02 ---> CO2 + H2O

(If you forgot..refer to the steps, for solving mole to mole conversions)


the answer is....
CH4  + 202---> CO2 + 2H2O

Stoichiometry (Quantitative Chemistry)

Stoiciometry is a branch of Chemistry that deals with the quantitative anaysis of chemical reactions

• It is a generalization of mole conversions to chemical reactions
• understanding the 6 types of chemical reactions is the foundation of stoichiometry

There are 6 types of reactions we use in stoichiometry, they are...

1. Synthesis (formation)
2. Decomposition
3. Single Replacement (SR)
4. Double Replacement (DR)
5. Neutralization
6. Combustion

SYNTHESIS

A synthesis reaction is when two or more simple compounds combine to form a more complicated one. The general form for this reaction is:

A + B ---> AB

DECOMPOSITON

A decomposition reaction is the opposite of a synthesis reaction - a complex molecule breaks down to make simpler ones. These general form for this reaction is:
AB ---> A + B

Example: 2 H2O ---> 2 H2 + O

 
SINGLE REPLACEMENT

This is when one element trades places with another element in a compound. The general form for this reaction is:

A + BC ---> AC + B (A is a metal)
OR
A +BC---> C+ BA (A is a non-metal)
Example: Mg + 2 H2O ---> Mg(OH)2 + H2


DOUBLE REPLACEMENT

This is when the anions and cations of two different molecules switch places, forming two entirely different compounds. These general form for this reaction is:
AB + CD ---> AD + CB
Example: Pb(NO3)2 + 2 KI ---> PbI2 + 2 KNO3

NEUTRALIZATION

Also known as Acid-base neutralization: This is a kind of double displacement reaction that takes place when an acid and base react. The H+ ion in the acid reacts with the OH- ion in the base, causing water to form. The product of this reaction is some ionic salt and water:
HA + BOH ---> H2O + BA

Example: HBr + NaOH ---> NaBr + H2O

COMBUSTION

Combustion reactions happen when oxygen combines with another compound to form water and carbon dioxide. These reactions are exothermic.

Example:  C10H8 + 12 O2 ---> 10 CO2 + 4 H2O

TEST YOUR KNOWLEDGE ON REACTIONS! (:

Empirical Formulas & Molecular Formulas

* Are the simplest formulas of compounds


Ex. Empirical Molecular
* C4H9 - C8H18
* NO2 - N2O4
* CH7O2 - C3H21O6


* They show only the simplest rations, not the actual atoms.
* Molecular formulas give the actual # of atoms…
* To determine an empirical formula we need to know the ratio of each element. This is why we use the table below; to gather all the information we may need to solve each problem.

Atoms	Mass	Molar Mass	Moles	Mol/ Smallest Mole	Ratio
C	8.4	12.0	0.7	2
H	2.1	1.0	2.1	6
O	5.6	16.0	0.35	1

Step # 1. Fill in the chart with what’s already given to you (the names and masses of the three atoms, plus their molar masses which you can find on the periodic table).

Step #2. Calculate the number of moles simply by following the conversion chart.
* 8.4g x 1mol / 12g = 0.7 mol
* 2.1g x 1mol / 1g = 2.1mol
* 5.6g x 1mol / 16g = 0.35mol

Step # 3. Find the smallest mole (0.35mol in this case) and divide it into itself and the rest of the mole #’s.


* 0.7 / 0.35 = 2
* 2.1 / 0.35 = 6
* 0.35 / 0.35 = 1
* If your answers are not “whole #’s you must multiply everything by a common #. For example:
* Say you end up with the numbers 1.0, 1.5, and 6.0… you would use the number 3 to change 1.5 into a whole # while still keeping the other two numbers whole as well. Thus you would end up with 2.0, 3.0, and 12.

= C2H6O

Atom	Mass	Molar     Mass	Moles	Mole/ Smallest mole	Ratio
Pd	42.56	106.4	0.3992	1
H	0.80	1.0	0.80	2

Step # 3.  You must be given the compound’s molar mass (in this case it’s 216.8 g/mol) in order to figure out its molecular formula.  Besides the molar mass, you also know from your previous calculations that the empirical formula is PdH2. You then calculate the empirical formula’s molar mass (by looking at the periodic table).  In this case it is 108.4.  You can then divide 216.8 by 108.4 to figure out what number to use for the molecular formula’s subscripts.  It is 2… thus you change Pd into Pd2 and H2 into H4.  And there you have your molecular formula!  


Informational and VERY helpful video:

Density

* Density is a measure of mass / volume
* When calculating density the following chart makes the task a lot easier to understand…


Density
* (D = m / v)
(Molar Mass) (Molar Volume)
Mass ------ Moles ------ STP

*
(Avogadro’s #)

Molecules
*
(Subscripts)
*
Atoms


* Here are some examples / solutions to density problems:


Ex. Compound “X” has a density of 3.0 g/ml. Determine the mass of 12 mL of X. How many moles are in 12 mL of X…?

Step # 1. 12 mL x 3g /1 mL = 36g


Step # 2. H2O ~ 2(1.0) + 16.0 = 18.0g


Step # 3. 36g x 1 mol / 18.0g = 2.0 mol


Ex. An unknown compound has a molar mass of 73.0 g/mol. If 0.15 mol occupies a volume of 50 mL, determine the compound’s density.

Step # 1. 0.15 mol x 73.0g / 1mol = 10.95g

Step # 2. 10.95g / 50mL = 0.219 g/mL ~ 0.22 g/mL

Cool Video :)

Moles to Atoms

To go to atoms from moles you must use Avogrado's number (6.02 x 10^23)

1. How many atoms are in 1.5 mol of Iron:

     1.5 mol x 6.02x 10^23 atoms = 9.0 x 10^23 atoms
                           1 mol

2. How much water molecules are there in 0.65 mol?

    0.65 mol x 6.02 x 10^23 molec  = 3.9 x 10^23 molec
                            1 mol

3. How many H2, Hydrogen atoms are there? How many Oxygen atoms? In H20

   i) 3.9 x 10^23 molec x 2 atoms = 7.8 x 10^23 atoms
                                         1 molec

  ii) 3.9 x 10^23 molec x 1 atoms  = 3.9 x 10^23 atoms
                                         1 molec

4. A cylinder of Helium contains 4.6 x 10^30 atoms. How many moles of Helium?
      4.6 x 10^ 30 atoms x 1mol   = 7.6 x 10^6 mol
                                       6.02 x 10^23 atoms


How many C atoms? How many H atoms? How many O atoms? How many in total? in CH3COOH?

1. 3.0 x 10^24 molec x 2 C atoms  = 6.0 x 10^24 C atoms
                                       1 molec

2. 3.0 X10^24 molec x 4 H atoms  = 1.2 x 10^25 H atoms
                                       1 molec

3. 3.0 x 10^24 molec x 2.0 atoms = 6 x 10^24 O atoms

4. 6.0 x 10^24 + 1.2 x 10^25 + 6 x 10^24 atoms = 2.4 x 10^25 atoms

Mole Conversions

Mole Conversions:
(Converting between Grams and Moles)


* To convert between moles & mass we use molar mass as the conversion factor
* Be sure to cancel the appropriate units!
* Use the chart below to help you with your conversions

Density
* (D = m / v)
(Molar Mass) (Molar Volume)
Mass ------ Moles ------ STP
*
(Avogadro’s #)

Molecules
*
(Subscripts)
*


Atoms

Ex. How many grams are there in 3.0 mol of O2?
* O2 = 2 x 16.0 ~ 36g / mol
* 3.0 mol x 32g / 1mol = 96.0g


Ex. Determine the # of moles of C5H12 that are in 362.8g of the compound.
* 5(12.0) + 12(1.0) = 72g / mol
* 362.8g x 1mol / 72g = 5.039mol


Ex. How many moles of magnesium bromide contain 5.38 x 10^24 formula units?
* 5.38 x 10^24molec x 1mol / 6.02 x 10^23molec = 8.39mol


Ex. Find the mass of 0.89mol of CaCl2.
* 40.1 + 2(35.5) = 111.1g / mol
* 0.89mol x 111.1g / 1mol = 99g


Ex. Find the mass of 0.159mol of SiO2.
* 28.1 + 2(16.0) = 60.1g / mol
* 0.159mol x 60.1g / 1mol = 9.56g

Percent Composition

Don't worry! we have easy instructions on how to plow through percent composition and become a professional at it! (:

Percent Composition - The percent composition of a component in a compound is the percent of the total mass of the compound that is due to that component.

****The percentage by mass of an element in a compound is always the same*****

To find the percentage by mass, determine the mass of each element present in one mole.

The percent composition (percentage composition) of a compound is a relative measure of the mass of each different element present in the compound.




To calculate the percent composition of a component in a compound:


1. Find the molar mass of the compound by adding up the masses of each atom
in the compound using the periodic table.
2.Calculate the mass due to the component in the compound you are for which you are solving by adding up the mass of these atoms.
3.Divide the mass due to the component by the total molar mass of the compound and multiply by 100.

Remember this!

Percent composition = mass due to specific component/ total molar mass                                                    of compound x 100%

here's a video on percent composition