diluting solutions to prepare workable solution

-chemicals are shipped around the world in their most concentrated forms
-need to be able to make solutions of any concentration from a more concentrated source.

Ex: 2L 16.0 HCl I need 0.800L of 2.00M HCl

the key idea is that the moles of solute is constant (more water in less concentrated solute)
Formula: M1L1=M2L2
16M HCl x L1= 2.00M HCl x 0.800L=1.60moles

Molar Concentration or Molarity of solution

-Molar Concentraion
   - A homogeneous mixture one substance is dissolved in another---solution
   - The chenmical that is in the smaller quantity is called solute
                                               larger quantity is called solvent
   - Molar Concentration or Molarity is the number of moles of solute in one L of a lolution, we use M to denote molar concentration and it has the units of "mole/L"


FOMULA:  - Molarity = moles of solute(mol)/Volume of soution(L)  or  M = mol/L
       

  Ex: Calculate the molar concentration of a soluion that has 0.510 moles of NaOH in 1.400L of solution.

                               0.510mol/1.400L= 0.364M or  0.364mol/L

  Ex: Calculate the number of grams of calciumhyfroxide in 1.30L of a 0.75M Ca(OH)2 solution.
                               moles Ca(OH)2 = 0.75M  x 1.30L = 0.975mol
                               0.975mol x 74.1 = 72g

Percent composition

- percentage by mass of a "species" in a chemical formula

Ex: what percentage composition of CO2
*assume you have one mole
  CO2 = 44g/mole
  C     = 12g/mole----12g/mole/(44g/mole) x 100%=27.3%
  O    = 32g/mole-----32g/mole/(44g/mole) x 100%=72.3%

*percent composition of a compound tells you which elements are in the compound how much of each here is
Ex: calculate % composition of
-Cu(NO3)2
 Cu(NO3)2 =  63.5g+28g+96g=187.5g/mole
 Cu   63.5g/ 187.5g=33.9%
 N     28g/187.5g=14.9%
 O     96g/187.5g=51.2%

Empirical+Molecular Formula

-Empirical formula --gives the largest term ratio of atoms (or moles) in the formula
*all ionic compounds are empirical formula
Ex: C4H10---molecular formula
       C2H5----Empirical formula

Ex: -consider that we have 10.87g of Fe and  4.66g of O, what is the empirical formula?
      MM of Fe---55.8g----- 10.87g x (1mole/55.8g) =0.195mol
      MM of O------16g---------4.66 x (1mole/16g) = 0.291mol


      -divide both by 0.195(the smallest molar amount)
      Fe    1(0.195/0.195) x2      O    1.5(0.291/0.195)x2
      Fe    2     O    3


      - scale ratio to whole number----Fe2O3




Ex: A compound contains 31.9%K , 28.9%Cl2, 39.2%O, what is the empirical formula
       *assume you have a hundred gram


         K--31.9g x (1mole/39.1g)=0.816mole/0.814-------------1 
         Cl2--28.9g x (1mole/ 35.5g)=0.814mole/0.814----------1
         O--39.2g x (1mole/16g)=2.45mole/0.814-----------------3


          *KClO3---empirical formula     
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-Molecular formula: is a multiple of the empirical formula and shows the actual number of atoms that combine to form a molecule.
**To multiple
        N= molar mass of the compound/molar mass of the empirical formula**
  

   Ex : A molecule has an empirical formula of C2H5 and a molar mass of 58g /mole, what is the    molecular formula?
               MM of C2H5= 29g/mole
               N= (58g/mole)/(29g/mole)=2
               2 x C2H5 = C4H10---molecular formula








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