Cheatography
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Conventions and Rules
* Molar mass is the mass of one mole of atoms in a substance.
* 1 mole of any gas is 24dm3 at r.t.p.
* Avogadro's Constant is the number of particles in one mole of any substance. The number is 6 x 1023.
* Relative Atomic Mass is the mass of one atom of an element compared to 1/12 of the mass of one carbon-12 atom.
* Relative Molecular Mass is the mass of one molecule of an element compared to 1/12 of the mass of one carbon-12 atom.
* Relative Formula Mass is the mass of one formula unit of an ionic compound. It is the sum of the Ar of all the ions in the formula unit.
* 1000cm3 = 1 dm3 |
Formula Triangles
Top |
Bottom 1 |
Bottom 2 |
No. of Particles |
No. of Moles |
Avogadro's No. |
Mass in Grams |
No. of Moles |
Molar Mass in g/mol |
Volume of Gas in dm3 |
No. of Moles |
24dm3 |
Mass in g |
Volume in dm3 |
Mass Concentration in g/dm3 |
No. of Moles |
Volume in dm3 |
Molar Concentration in mol/dm3 |
* To find top, multiply bottom values. To find bottom, take top divided by other bottom value.
1. No. of Particles // Moles // Avogadro's No.
2. Mass // Moles // Molar Mass
3. Volume // Moles // 24dm3
4. Mass // Volume // Mass Concentration
5. Moles // Volume // Molar Concentration
Concentration
* The amount of solute dissolved in a unit volume of the solution.
* Usually in g/dm3 or mol/dm3.
* Mol/dm3 is also called Molarity.
* 1M = 1 mol/dm3 |
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|
Molecular Formula & Empirical Formula
Value |
Element A |
Element B |
Mass in Grams |
x |
x |
Molar Mass |
x |
x |
Moles |
x |
x |
Mole Ratio |
x |
x |
1. Given % Composition, find masses of both substances in 100g. If 70% is A and 30% is B, then there is 70g of A and 30g of B in 100g of AB.
2. Find molar mass using periodic table.
3. Find no. of moles by multiplying mass and molar mass.
4. Divide all sides by the smallest number and round off to the nearest whole number to get mole ratio.
5. Molecular Formula is always a multiple of the Empirical Formula.
% Purity and % Yield
% Purity = Mass of Pure Substance / Total Mass x 100%
% Yield = Actual Mass / Theoretical Mass x 100% |
Limiting Reagents & Reactants in Excess
x |
No. of Available Moles |
vs |
No. of Moles Needed |
Limiting Reagents |
No. of Available Moles |
< |
No. of Moles Needed |
Reactants in Excess |
No. of Available Moles |
> |
No. of Moles Needed |
|
|
How to Find Limiting Reagents
A + 2B 2AB
1. Find no. of available moles for A.
2. Find no. of available moles for B.
3. Find no. of moles needed for A/B.
E.g. Moles needed for A = Available moles for B / 2
(Refer to mole ratio)
If moles available is < moles needed, then that reactant is the limiting reagent. |
The concept of limiting reagents is the available moles for reaction vs the needed moles for reaction.
Mole Calculations
Given |
|
|
Find |
Mass of A |
Moles of A |
Moles of B |
Mass of B |
Vol. of A |
Moles of A |
Moles of B |
Vol. of B |
Mass of A |
Moles of A |
Moles of B |
Vol. of B |
Vol. of A |
Moles of A |
Moles of B |
Mass of B |
Refer to mole ratio when converting Moles of A to Moles of B.
1 mole of any gas is 24dm3 at R.T.P.
Acids & Bases
Metal + Acid Salt + Hydrogen Gas
Metal Carbonate + Acid Salt + Water + Carbon Dioxide
Metal Oxide + Acid Salt + Water
Metal Hydroxide + Acid Salt + Water
Base + Acid Salt + Water (Neutralisation)
Alkali + Acid Salt + Water
Alkali + Ammonium Salt Salt + Water + Ammonia Gas
Alkali + Salt Metal Hydroxide + Salt |
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