amount of substance Flashcards
equation for mass with molar mass
molar mass times no.moles
molar mass rearrangements
molar mass = moles/ mass
moles = mass/ molar mass
solutions advantages
safer than pure substances
-easier to store
solution
when one substances dissolved into another
solute
substances that dissolves into solvent
concentration equation
moles of solute/ dm3 of solvent
1m3 in dm3
1m3 = 1000dm3
1dm3 in cm3
1dm3 is 1000cm3
moles in terms of concentration
conc times volume
mol dm3 to g dm3
concentration in mol dm3 times molar mass
equation for conc of 2 solutions
C1V1 = C2V2
conc of c2 equation for 2 solutions
C2 = c1v1/v2
steps of running titrations
add acid in conical flask
-volume
-add indicator
-known conc NaOH to HCl
-stop when solution turns pink
neutralisation reaction equation
acid + base = salt + water
steps for doing titration calculations
-write balanced equation
-moles reacted for known conc solution
-use mole ratio to find amount of other solution reacted
-use amount with initial volume to calculate unknown conc
back titration calculations steps
-write full equations for both reactions
-identify key substance involved in both reactions as reactant
-calculate amount of substance used up in second reaction
-amount of key substance used up in first reaction
-amount of other reactant in first reaction
purity as a percentage equation
mass of solute/ total mass of substance times 100
assumptions about gases made
-made up of identical particles
-particles move randomly
-when particles collide they lose no energy
-particles collide with the container walls their in
-occupy no space
no intermolecular forces between particles
ideal gas equation
pV = nRT
ideal gas equation components
p= pressure in Pa
-V = volume in m3
-n= moles
-R= gas constant
T= temp in K
1Pa into 1 atmosphere
101325= 1 atmosphere
actual yield
amount of desired product created in a reaction
theoretical yield
amount of desired product that could be produced if all reactants fully reacted
explanations for lower yield
-not every single molecule takes part in reaction
-not every single molecule of desired product is collected
calculating theoretical yield
-balanced equation of reaction
-moles of limiting product
-use that to find out moles of desired product
-use moles of desired product to work out mass
-
percentage yield
actual yield as percentage of theoretical yield
percentage yield equation
actual yield/ theoretical yield times 100
percentage atom economy equation
Mr of desired products/ Mr of all reactants times 100
molar mass of salts
Mr added up all together including molar in front of solid and water molecules
decomposition reaction
lattice of solid and water breaking into only solid and water
dissacosiation equations
aq along with separate reactants as products
acid + metal
salt plus hydrogen
acid + metal oxide
salt + water
acid + metal carbonate
salt + water + carbon dioxide
percentage water content formula
molar mass of water/ molar mass of solid + molar mass of water
times 100 to give percentage
solving for x moles of water
molar mass = molar mass of solid plus x(18)
-rearrange and solve for x with given molar mass
solving for x when given water percentage
percentage = molar mass of water/ molar mass of solid + molar mass of water
times 100
finding x using experiment
heat liquid until only crystal salts left
-measure weight of salt
-mass at start and end measured
-
how to make sure all water removed when finding x using experiment
mass will stop decreasing
how to find mass of water in hydrated salt
total mass of hydrated salt and crucible minus anhydrous salt and crucible mass
solving for x using calculations equations
-find moles of solid by dividing mass by molar mass
-use moles of anhydrous solid to work out moles of hydrous salt
-molar mass of hydrous = mass/ moles
-molar mass of salt = molar mass of solid/ molar mass of water
x=molar mass of hydrous salt - anhydrous molar mass/ 18
