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Topic 1 Quantitative Chemsitry
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Mind Map on Topic 1 Quantitative Chemsitry, created by kevin86604 on 05/21/2014.
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Resource summary
Topic 1 Quantitative Chemsitry
1.1 The mole concept and Avogadro's constant
Measurements and unites
kg, s, K, m3, Pa or Nm -2
g, min, ˚C cm3, atm
Amount of substance
Element
Atoms
Compound
Moles
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
Molecular Formula
Relative molecular mass
Avogadro's constant
6.02 x 10^23 mol -1
The number of particles per mole
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
1.2 Formulas
Emperical formula
It gives the ratio of the atoms of different elements in a compound.
Molecular formula
Shows the actual number of atoms of each element present in a molecule
1.3 Chemical Equations
Balancing equations
Equation: CH3+O2 -> CO2 + H2O
Balanced Equation: CH4+2O2 -> CO2+2H2O
1.4 Mass and gaseous Volume relationships in chemical reactions
Limiting reactant
The limiting reactant is the reactant that determines the theoretical yield of product
Theoretical yield
Theoreticall yield is the mass or amount of product produced according to the chemical equation
Percentage yield
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
1.5 Solutions
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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