CALCULATING THE EMPIRICAL FORMULA OF ORGANIC COMPOUNDS!!!

The empirical formula of an organic compound (any substance that contains carbon) can be found by:
-burning the compound (react with O2)
-collecting and weighing the products
-from the mass of the products, the moles of each element in the original organic compound can be calculated


Ex. 5g of sample→15.0g of CO2 & 8.18g of H2O
     
      CxHy+zO2→xCO2+y/2H2O
     
      this is the balanced chemical equation for the burning of CxHy
      
      the moles of C and H in CxHy=moles of C in CO2 and the moles of H in H2O
     
     15g*(1mole/44g)=0.341mole C      8.18g*(1mole/18.02g)=0.454mole water*2=0.908mole H
      
     0.908 moles of H and 0.341 moles of C must have been in the original substance
CxHy=C0.341H0.908
    
    Carbon: 0.341/0.341=1           Hydrogen=0.908/0.341=2.66*3=8
     
    Empirical Formula=C3H8
      
    *always check the masses

Empirical+Molecular Formula

Empirical formula definition: gives the lowest term ratio of atoms (or moles) in the formula

*All ionic compounds are empirical formulas

Reference from wiki:

http://en.wikipedia.org/wiki/Empirical_formula

Ex. C₄H₁₀ (butane) à molecular formula

àC₂H₅

Ex. consider that we have 10.87g of Fe and 4.66g of O, what is the empirical formula

#1 convert g à mole

Fe 10.87g×1mol/55.8g = 0.195mol

O  4.66g×1mol/16.0g = 0.291mol

#2 divide both by the smallest molar amount

Fe 1    O 1.5

#3 scale ratios to whole numbers

Fe 1×2=2

O 1.5×2=3

Therefore empirical formula = Fe₂O₃

Molecular formula: is a multiple of the empirical formula and shows the actual number of atoms that combine to form a molecule.

To calculate multiple:

      Molar mass of the compound

N=--------------------------------------------------------------------

  Molar mass of the empirical formula

Reference from wiki:

http://en.wikipedia.org/wiki/Molecular_formula

Ex. A molecule has an empirical formula of C₂H₅ and a molar mass of 58g/mol, what is the molecular formula?

MM C₂H₅=12×2+1×5=29g/mol

N=(58g/mol)/(29g/mol)=2

Mf=2(C₂H₅)=C4H10

Ex. A compound contains 7.44g C, 1.24g H, and 9.92g O. the molar mass of the compound is 180g. What is the molecular formula?

C 7.74×(1mol/12g)=0.62  à1

H 1.24×(1mol/29g)=1.24  à2

O 9.92×(1mol/16g)=0.62  à1

EF=CH₂O àMM=30g

N=180g/30g=6 C₆H₁₂O₆(Glucose)

Little Quiz -Nov.29

1. Solve for the MM and % of each element.
CH3COOH
NH4NO3
C11H22O11

2. what would be the mass of 15.00×10^21 atoms of Cobalt?

3.The empirical formula of a compound is SiH3. If 0.0275 mol of compund has a mass of 1.71g, what is the compound's molecular formula?

4.What's the persentage of water in CuSO4·5H2O?

5. A compound contains 58.5% C, 7.3%H and 34.1% N. What is the formula of the compound?

Percentage Composition --Nov.29

Definition: Percentage by mass of “species” in a chemical formula.

Ex. What is the percentage composition of CO₂?

 *assume you have 1 mole

Total MM=44.0g/mol

Percentage of C=(12g/mol)/(44g/mol)×100% = 27.3%

Percentage of O=(16g/mol×2)/(44g/mol)×100% = 72.7%

*(27.3%+72.3%=100%)

Percentage composition tells you which elements are in the compound and how much of each there is.

Fomula(useless):

                                              Mass of element

Percentage composition=—————————— ×100%

                                             Mass of compound

Mole Conversions

Mole Conversions

1. Coversion from Particles Particles↔mass ( 1 moles= 6.022×10^23 particles)
a)particles →moles

b)moles→particles/molecules/formula units atoms


2.Concersions between moles↔molecules
a)moles→grams

b) grams→ mole


Harder Mole Conversions

Grams↔Moles↔Particle↔# Atom in a Practice

a)Conversions between particles to mass.

Ex.  What is the mass of 2.78×10^22 Fe atoms?

2.78×10^22 ÷1mole／6.022×10^23 molecules × 80.1g／1mole=1.1×10^-3


b)Conversions between mass to particles.

How many atoms of Iron in 20.0g of Iron?

20.0g×1mole／55.8g×6.022×10^23atoms／1 mole=2×10^23

The Mole

-equal volume of different gases has a constant ratio
ex oxygen:hydrogen=16:1
   carbon dioxide:hydrogen=22:1
-expressed by comparing it to the mass of another object


Avogadro's Hypothesis
Equal volume of different gases at the same temperature and pressure have the same number of particles.
If they have the same number of particles, the mass ratio is due to the mass of the particles.


Atomic Mass
Mass of 1 atom of the element in atomic mass units(amu or u or Dalton)
ex fluorine=19.0 amu


Formula Mass
All the atoms of a formula of an ionic compound(in amu)
ex potassium fluoride (KF)   39.1+19.0=581amu


Molecular Mass
All the atoms of a formula in a covalent compound (in amu)
ex carbon dioxide(CO2) = 12.0+16.0×2=44.0 amu


Molar Mass
Atomic/Molecular/Formula mass of any pure substance(in grams per mole)
ex 1mole of oxygen=16.0 g/mol
   1mole of carbon =12.0 g/mol
(same number of particles)


Avogadro's Number
The number of particles in 1 mole of any amount of substance is
6.022×10^23 particles/mol


The mole is extremely important to chemists because it allows them to count atoms and molecules.

Density Nov.11

1cm^3 of watere(H2O)   =   1mL

Density of water   = 1.0g/mL
                      or  =1000g/L


For a solid      →g/cm^3

For a liquid    →g/mL

V=M/D

D object > D liquid      sink
D object < D liquid   float