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Mind Map on Topic 1 Quantitative Chemsitry, created by kevin86604 on 05/21/2014.
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908086
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2016-02-19T07:15:36Z
Topic 1
Quantitative
Chemsitry
1.1 The mole concept and Avogadro's constant
1.2 Formulas
1.3 Chemical Equations
1.4 Mass and gaseous Volume relationships in chemical reactions
1.5 Solutions
Measurements and unites
kg, s, K, m3, Pa or Nm
-2
g, min, ĖC cm3,
atm
Amount of
substance
Element
Atoms
Compound
Moles
Molecular Formula
Relative molecular mass
Avogadro's constant
6.02 x 10^23 mol
-1
The number of particles per
mole
Number of moles(n) =Mass/Molar mass
A mole is the amount of a
substnace which contains the
same number of chemical species
as there are atoms in exactly 12g
of the isotope carbon-12
Example
Calculate the amount of water, H20, that contains 1.80x 10^24
molecules
Solution
n = N/L n= 1.80x 10-24/6.02x
10^23
n= 2.99 mol
Emperical formula
Molecular formula
Balancing equations
Limiting reactant
Theoretical yield
Percentage yield
The limiting reactant is the
reactant that determines the
theoretical yield of product
Theoreticall yield is the mass or
amount of product produced
according to the chemical equation
experimental
yield/theoretical yield
x 100
STP= 273K and 101.3 kPa (1atm)
For gas: Number of moles
(n)= Volume (V) / Molar
Volume
The ideal gas equation
PV=nRT
Equation: CH3+O2 -> CO2 + H2O
Balanced Equation: CH4+2O2 -> CO2+2H2O
It gives the ratio of the atoms of different elements in a compound.
Shows the actual number of atoms of each element present in a molecule
Concentration
Example: A solution of sodium hydroxide has a
concentration of 8.00g dm-3. What is its
concentration in mol dm-3
n=m/M M=2.99+1.01+16.00 =
40.00g n=8.00/40.00= 0.200
[NaOH] = n/V = 0.200/1.00 = 0.200
mol dm-3
n=n/V
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908086
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2016-02-19T07:15:36Z
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