Monday, 13 October 2014

Chemistry: Moles and equations



Chemistry

Moles and equations


A)     Define and use Mole calculations


One Mole: the amount of substance that contains as many elementary entities as there are carbon atoms in 12g of carbon-12


-Moles can be calculated using these equations:
- Amount in moles = mass in grams / molar mass                                      =     n = m / M

-Amount in moles = volume of gas / 24                                                     =     n = Vgas / 24


 -Amount in moles = concentration x volume                                              =     n = c x V

-Amount in moles = number of particles / Avogadro's number                           =     n = # / 6.02 x 10²³




B)      Define Avogadro’s number


Avogadro’s number: the number of carbon-12 atoms in 12g of carbon-12. Is equal to 6.022045 x 10²³




C)      Explain and calculate Empirical and Molecular formula


Empirical formula: gives the simplest whole number ratio of the atoms of each element present in the compound.


Molecular formula: gives the actual number of atoms of the different elements in one molecule of a compound


-e.g

A compound contains 73.47% carbon, 10.2% hydrogen, and 16.33% oxygen. Its relative Molecular Mass in 98. Calculate the empirical and molecular formulas of the compound.

Method:


1) Divide percentage of each element by the elements Atomic mass

2) Divide all figures by the smallest calculated

3) Use this ratio to generate the empirical formula

4) Use the formula show to calculate how many times the empirical formula goes into the molecular formal mass

5) Multiply the empirical formula by this number to calculate the molecular formula



                      C                                             H                                             O


                       73.47/12                                   10.2/1                                 16.33/16


                       =6.1225                                    =10.2                                     =1.02


                    6.1225/1.02                             10.2/1.02                              1.02/1.02


                           =6                                        =10                                           =1



Empirical Formula :              C₆H₁₀O


Molecular Formula Mass      =      Empirical Formula Mass x n

     98         =       [(6 x 12) + (10 x 1) + (16)] x n

     98         =       98 x 1

                                1 x Empirical Formula         =       Molecular formula

                                                   1 x C₆H₁₀O          =        C₆H₁₀O




D)     Be able to calculate percentage errors


Percentage error = (maximum error / measurement taken) x 100

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