Atomic/ Formular/ Molecular/ Molar mass

 Atom is the smallest particle of a chemical element that retains its chemical properties.


A compound's empirical formula is the simplest integer ratio of the chemical elements that constitute it. For example, water is always composed of a 2:1 ratio of hydrogen to oxygen atoms, and ethyl alcohol or ethanel is always composed of carbon, hydrogen, and oxygen in a 2:6:1 ratio. However, this does not determine the kind of molecule uniquely – dimethyl ether has the same ratios as ethanol, for instance. Molecules with the same atoms in different arrangements are called isomers. Also carbohydrates, for example, have the same ratio (carbon:hydrogen:oxygen = 1:2:1)  but different total numbers of atoms in the molecule.

The molecular formula reflects the exact number of atoms that compose the molecule and so characterizes different molecules. However different isomers can have the same atomic composition while being different molecules.

The empirical formula is often the same as the molecular formula but not always. For example, the molecule acetylene has molecular formula C2H2, but the simplest integer ratio of elements is CH.

The molecular mass can be calculated from the chemical formula and is expressed in conventional atomic mass units equal to 1/12 of the mass of a neutral carbon-12 atom. For network solids, the term formul unit is used in stoichiometric calculations.



In order to grasp the concept of molar mass calculations it is important to understand the molar unit. The mole also called mol is the basic unit of measurement in chemistry. By definition, in modern chemistry, one mole represents the number of carbon atoms in exactly 12 grams of carbon 12 isotope. Remember that carbon-12 has an atomic mass of 12 (six neutrons and six protons).
One mole of anything, however, contains 6.0221367E²³ of that object. This is known as Avogadro's number.

Examples:

1 mole of carbon = 6.0221367E²³ carbon atoms

1 mole of bananas = 6.0221367E²³ bananas

Avogadro's number and hypohesis

Avogadro's number, also known as Avogadro's constant, is defined as the quantity of atoms in precisely 12 grams of 12C. The designation is a recognition of Amedeo Avogadro, who was the first to state that a gas' volume is proportional to how many atoms it has. Avogadro's number is given as 6.02214179 x 10-23 mol.

Amadeo Avogadro lived in the early 19th century, and was an Italian savant known for his role in many different scientific disciplines. His most famous statement is known as Avogadro's Law, and is a hypothesis that states: “Equal volumes of ideal or perfect gasses, at the same temperature and pressure, contain the same number of particles, or molecules.”

This is an intriguing hypothesis, because it says that quite different elements, such as nitrogen and hydrogen, still have the same number of molecules in the same volume of an ideal gas. While in real world settings this is not strictly true, it is statistically quite close, and so the ideal model still has a great deal of value.

Lab 4C

Objectives:
1. To determine the % of water in a unknown hydrate.
2. To determine the moles of water present in each mole of this unknown hydrate, when give the molar mass of the anhydrous salt.
3. To write the empirical formular of the hydrate.

Supplies:
lab burnet, ring stand and ring, crucible and lid, centigram or digital balance, crucible tongs, pipestem triangle, lab apron, safety goggles.
Chemical Reagents: approximately 5g of hydrate, water.

ProcedureMass a crucible and its cover and record in a data table. Place about 2 grams of the hydrate in the crucible and record the mass of the crucible, cover and hydrated salt. Gently heat the crucible and its contents for about 10 minutes. During the last minute of heating remove the cover so that any moisture which has collected on the underside of the cover can evaporate. When the crucible has cooled determine the mass of the crucible, cover, and contents.
Replace the cover and heat the crucible for five more minutes removing the cover during the last minute of heating as you did above. Cool and determine the mass of the crucible, cover, and contents. This last mass should agree with the previous mass to within plus or minus 0.005 grams. If it does not repeat this heating until a constant mass is reached. This is called heating to a constant mass and is the only way of insuring that the reaction is complete.

Data:
Mass of empty crucible and lid                                                                          13.12g
Mass of crucible, lid and hydrate                                                                       15.34g
Mass of hydrate                                                                                                    2.22g
Mass of crucible, lid and anhydrous salt(first)                                                  14.54g
Mass of crucible, lid and anhydrous salt(second)                                             14.55g
Mass of anhydrous salt                                                                                         1.42g 
Mass of water given off                                                                                        0.82g
Mass of one mole of hydrous salt(from your instructor)                                  159.6g/mol
Describle any changes that you observed when adding water to the crucible    Become blue solid.

Molarity

 Another way of expressing concentration, the way that we will use most in this course, is called molarity. Molarity is the number of moles of solute dissolved in one liter of solution. The units, therefore are moles per liter, specifically it's moles of solute per liter of solution.  


Rather than writing out moles per liter, these units are abbreviated as M or M. We use a capital M with a line under it or a capital M written in italics. So when you see M or M it stands for molarity, and it means moles per liter (not just moles).


You must be very careful to distinguish between moles and molarity. "Moles" measures the amount or quantity of material you have; "molarity" measures the concentration of that material. So when you're given a problem or some information that says the concentration of the solution is 0.1 M that means that it has 0.1 mole for every liter of solution; it does not mean that it is 0.1 moles. Please be sure to make that distinction.

All types of mole conversions

1.Conversions PARTICLE/ATOM/FORMULA UNIT-MOLES

•     Example; How many moles of Carbon atoms are there in 3.01x10^24C atoms?

  3.01x10^24C atoms x 1 mol C/6.022x 10^23C atoms = 5.00 mol C


2.Comversions MOLES-PARTICLES/ATOMS/FORMULA UNITS
   

• Example: How many CO2 molecules are present in 0.75 moles of CO2?

0.75mol CO2 x6.022x10^23molecules CO2/1  molCO2=4.5x10^23molecules CO2

•     Example : How many atoms of Oxygen are present in 0.75 mol CO2?

     0.750 mol CO2 x 6.022x10^23molecules CO2/1mol CO2 x 2atoms/1 molecule CO2= 9.03x10^23atoms O



3.Conversions MOLES-GRAMS
  

•  Example: What is the mass in grams of 2.04 moles of Carbon?

   Atomic mass of C= 12.0 u
    Molar mass of C= 12.0g/mole
  2.04mol C x 12.0g C/1mol C= 24.5gC



4.Conversion GRAMS-MOLES
  

•  Example: How many moles are there in 3.45 grams of Carbon?

   Atomic mass of C =12.0u
   Molar mass of C =12.0g/mol
   3.45g C x 1mol C/ 12.0g C= 0.287mol C

•   Example: How many moles are there in 6.2 grams of Magnesium chloride?

   Formula unit mass of Mgcl2= 1 Mg x 24.3u=24.3u
    2 Cl x 35.5u =71.5u/95.3u
   Molar mass of Mgcl2 = 95.3g/mol
   6.2 g Mgcl2 x 1mol MgCl2/ 95.3gMgCl 2= 0.065 mol MgCl 2

Diluting Solution to Prepare Workable Solution

- chemicals are typically shipped around the world in their most concentrated form
- thus making it easier to transport and cost fairly less money
- but have to make concentrate form from any solution from an even more
  concentrated solution
Example 1: You have 2.00L of 16.0M HCl, but you need 0.800L of 2.00M HCl
How do you make it?
- numbers of moles are always constant=not changing
- only difference between the two would be there is more water in the less
 concentrated solution
Therefore, Equation would be= Moles solute before=Moles solute after
  M1L1=M2L2
                   The subscript 1 denotes “before” and subscript is “after”
         Solving for L1 wouls solve how much volume there is in the solution
Applying the formula, M1L1=M2L2
16.0M HCl×L1=2.00M HCl×0.800L
16.0M×L1=1.60moles HCl
L1=0.100L or 1.00×10ˆ2ml
0.800L-0.100L=0.700L→amount of water
Example 2: Concentrated HCl is 11.6 moles/L/ How would you make up 250ml of 0.500 moles/L HCl?
11.6M HCl×L1=0.500M HCl×250ml
L1=(M2×L2)/M1= (0.500 mole/L MCl×250ml)/(11.6 moles/L HCL)
  = 0.011L
  = 11ml 250-11ml= 239ml→amount of water

How to solve:
Preparing Solutions Sample Problem 1
Preparing Solutions Sample Problem 2
Preparing Solutions Sample Problem 3