amount of substance Flashcards
mole
amount of substance in grams that has same no of particles as there are atoms in 12g of carbon-12
molar mass
mass in grams of 1 mol of a substance given unit g/mol-1
mol =
mass/mr
empirical formula
simplest whole no ratio of atoms of each element in the compound
empirical equation
Step 1 : Divide each mass (or % mass) by the atomic mass of the element
Step 2 : For each of the answers from step 1 divide by the smallest one of
those numbers.
Step 3: sometimes the numbers calculated in step 2 will need to be multiplied
up to give whole numbers.
These whole numbers will be the empirical formula.
molecular formula
actual no of atoms of each element in compound
heating in a crucible method
weigh empty clean dry crucible + lid
add 2g of calcium sulfate to crucible and weigh
heat strongly with bunsen for couple of mins
allow to cool
weigh crucible and contents again
heat crucible again and reweigh until you reach constant mass (ensures reactions complete)
concentration =
moles / volume
units conversions
cm3 - dm3 (/1000)
24000cm3 = 24dm3
standard solution method
- Weigh an empty weighing bottle before adding approximately 3g of the required acid
salt. - Reweigh the weighing bottle and calculate the mass of acid that was added.
- Transfer the solid to a 100 cm3
beaker. - Add approximately 50 cm3
of distilled water to the acid and stir with a glass rod so it
dissolves. - Using a funnel, transfer the solution to a graduated 250 cm3
volumetric flask. Wash
the rod and sides of the beaker into the graduated volumetric flask and fill with
distilled water to the 250 cm3
mark. - Stopper the flask, then mix thoroughly by inverting and shaking vigorously. This is the
standard solution.
ideal gas equation
pv = nRT
p=pressure
v=vol
n=moles
R=8.31Jkmol-1
T=237K
NO OF PARTICLES=
amount of substance (mols) x avagadro constant
percentage yield =
actual/theoretical x100
(q mass is actual and one we work out is theoretical)
atom economy =
( mass of useful products / mass of all reactants ) x100
