T5: Amount of Substance Flashcards
Hydrogen + Oxygen -> Water balanced equation
2H2(g) + O2(g) -> 2H2O(l)
chemical formulae rules
- elements written as 1 atom: all metals, noble gases, carbon, silicon (Na, Mg, Ca)
- diatomic elements: N2, O2, F2, Cl2, Br2, I2
- covalent compounds: formulae indicated by name (carbon dioxide -> CO2)
- ionic compounds: determine from charges on individual ions
state symbol rules
- solids (s) -> giant ionic substances, metals, large simple molecular substances (I2), macromolecular structures (C, Si, SiO2)
- liquids (l) -> relatively large simple molecular substances (H2O, Br2), Mercury (Hg)
- gases (g) -> relatively small simple molecular substances (O2, HCl, CO2)
Sodium + Water -> Sodium hydroxide + hydrogen balanced equation
2Na(s) + 2H2O(l) -> 2NaOH(s) + H2(g)
Iron(III) oxide + Carbon -> Iron + Carbon dioxide balanced equation
2Fe2O3(s) + 3C(s) -> 4Fe(s) + 3CO2(g)
Aluminium + Nitric acid -> aluminium nitrate + hydrogen balanced equation
Al(s) + 3HNO3(aq) -> Al(NO3)3(aq) + 1.5H2(g)
compare the amounts of substance in both beakers
- twice as many C atoms needed to make up 12g compared to Mg
- each C atom has half the mass of a Mg atom
- same mass doesn’t mean same number of particles
Avogadro constant definition
the number of atoms in 12g of 12C
molar mass unit
g mol-1
number of particles in: 1 mole of Sodium Na+
6.022x10^23 ions of Na
number of particles in: 2 moles of Silicon Si
2 x (6.022x10^23) atoms of Si
number of particles in: 1 mole of hydrogen (H2) molecules
6.022x10^23 molecules of H2
number of particles in: 3 moles of ammonia (NH3)
3 x (6.022x10^23) molecules of NH3
how many atoms of H are in 2 moles of H2O
4 moles
number of particles equation
Number of particles = Moles x Avogadro Number
relative atomic mass (Ar) definition
(average mass of 1 atom of an element) / (1/12th the mass of 12C atom)
relative molecular mass (Mr) definition
(average mass of 1 molecule) / (1/12th the mass of 12C atom)
moles equation
n = M / Mr
mass unit
kg
moles unit
mol
concentration equation
c = n / V
concentration unit
mol dm^-3
volume unit
dm^3
1cm^3 to 1dm^3
divide by 1000
1dm^3 to 1cm^3
multiply by 1000
stage 1 of practical aspects of titrations (steps to make up a standard solution)
Transfers know mass of solid
- weigh the sample bottle containing the solid on a (2dp) balance
- transfer the beaker and reweigh sample bottle
- record the difference in mass
stage 2 of practical aspects of titrations (steps to make up a standard solution)
Dissolves in water
- add distilled / deionised water
- stir (with a glass rod) or swirl
- until all solid has dissolved
stage 3 of practical aspects of titrations (steps to make up a standard solution)
Transfer, washing and agitation
- transfer to volumetric / graduated flask using a funnel
- with washings
- make up to 250cm^3 / mark with water
- shakes / inverts / mixes
why remove any air bubbles from the burette tap
otherwise the volume of the air bubble will be counted in the burette reading (GREATER TITRE)
(remove by opening tap first)
why remove the funnel used to fill the burette before starting the titration
solution drops could fall from the funnel into the burette and affect the burette reading (TITRE DECREASES)
why during the titration, swirl the conical flask and rinse the walls of the conical flask with distilled water
- ensures all reactants mix and no reagent is left unreacted on sides of flask
- doesn’t change the number of moles in conical flask so has no adverse effect on titration
% error (uncertainty) equation
(error of the equipment / amount measured) x 100
how does the error of the equipment change
error * (X -> depends on equipment used)
- burette read twice so 2x error
how would you decrease % error in a titre measurement
increase the measurement made from the burette
- increase the concentration of solution in conical flask
- decrease the concentration of solution in burette
the ideal gas equation
PV = nRT
name P and it’s units in the ideal gas equation
pressure
Pa (pascals)
name V and it’s units in the ideal gas equation
volume
m^3
name n and it’s units in the ideal gas equation
moles
mol
name R and it’s units in the ideal gas equation
gas constant = 8.31
J K^-1 mol^-1
name T and it’s units in the ideal gas equation
temperature
K (kelvin)
rearrange PV = nRT for P
P = nRT / V
rearrange PV = nRT for V
V = nRT / P
rearrange PV = nRT for n
n = PV / RT
rearrange PV = nRT for R
R = PV / nT
rearrange PV = nRT for T
T = PV / nR
dm^3 to m^3
divide by 1000
m^3 to dm^3
multiply by 1000
cm^3 to m^3
divide by 1 x 10^6
m^3 to cm^3
multiply by 1 x 10^6
°C to K
+ 273
K to °C
- 273
0°C is ?K
273
25°C is ?K
298
100°C is ?K
373
empirical formula definition
the empirical formula of a compound is the simplest whole number ratio of atoms of each element
molecular formula definition
the molecular formula of a compound is the actual number of atoms of each element
how do you remove all the water (water of crytalisation)
heat to constant mass
atom economy equation
% atom economy = (mass of desired products) / (total mass of reactants) x 100
why is it desirable to have a high atom economy yield
- increases the mass of atoms in the reactants that are turned into products
- reduces the amount of waste by-products
how can atom economy be improved
atom economy can only be improved by finding a different reaction to produce the desired product which also produces fewer by products
what is the atom economy if only 1 product is produced
100%
percentage yield equation
(actual yield x 100) / theoretical yield
when using a chemical reactant to prepare to prepare a substance it is desirable to have a high % yield because:
- increases the amount of product that is made from the reactants
- % yield can be increased by making improvements in experimental technique
