Amount of a substance Flashcards
Relative atomic mass
The weighted mean mass of an atom of an element compared with 1/12 the mass of carbon 12
Relative molecular mass
The weighted mean mass of a simple covalent molecule compared with 1/12 of the mass of an atom of C-12
how to calc relative atomic mass
to calc relative molecular mass you add up all the individual relative atomic masses of the elements contained in the molecule
Amount of a substance units and symbol
N, Moles
Mole definition
The amount of substance contained as many particals as there are carbon atoms in exactly 12g of C-12
Molar mass
The mass, in g, per mole of a substance.Units are g Mol -¹
Amount of a substance equation
Amount, n = Mass/Mr
Moles(mol) g gmol-¹
Avagadro’s constant
The number of particals in one mole of that particle is called avogadro constant
.It is given the symbol Nª and has a value of 6.02× 10²³ mol-¹
in a mole of a substance there are 6.02×10²³ particals
Number of particals equation
No of particals= Nª × moles
Reacting masses of soilds equation
To calc the mass of a product formed or reactants used in a specific reaction
1. balance the equation
2. find your known and unknown values
3. Work out the moles of your known value n= mass/mr
4.work out the number of moles of the unknown by using the ratio in the equation e.g 2MgO—-> Mg + O so the ratio of MgO to MG is 1:2
5.then to find mass do, Mr×Moles= mass
empirical formular
the simplest whole no ratio of atoms of each element present
molecular formular
the actual no of atoms of each element in a molecule
empirical formular equation
do percentage mass or mass divided by relative atomic mass to find the moles
divide the moles by the smallest no of moles to find the simplest ratio
molecular formular equation
molar mass/ empirical formular mass
then x the empirical formular by the answer
percentage Yield equation
% yeild= actual mass/mol of product/theoretical mass/mol of product
×100
1. calc amount of reactants used in mol
2.using balanced equation calc amount in mol of product produced
3.convert theoretical moles to theoretical mass
4. compare actual mass of product to theoretical mass to calc % yield
atom economy equation
a reaction with high atom economy has less waste as it is a measure of how much mass of the total mass of reactants is converted to the desired product
molecular mass of desired product/molecular mass of all the products
×100
Moles in solution
Amount, n= C × V
moles = concentration x volume
mol moldm-³ dm³
if volume is given in cm³ ÷ that no by 1000
if conc us given in gdm‐³ then ÷ that no by the Mr
Acid titrations method
- 25.0cm³ of solution A is transfered into a conical flask by a pipette
2.indicator was added to the solution in the conical flask
3.Solution B of unknown conc was added to the burette and the initial burette reading was recorded to the nearest 0.05cm³ - The solution B was slowly added to the conical flask until the indicator changed colour at the end point (trial)
- the final burette reading was recorded to the nearest 0.05cm³
6.The titration was then repeated until 3 concordant results were achieved
Always record reading to 2 dp
to calc mean titre all the results used should be within 0.1cm³ of eachother
how to prep a standard solution
step1) accurately weigh the solid( mass by diff)
step2) dissolve the solid in a beaker using a small amount of distilled water
step3) carefully transfer the solution into volumetric flask. ensure u rinse the beaker with distilled water and add this to a volumetric flask
step4) carefully fill the flask with distilled water until the bottom of the meniscus lines up exactly with the graduation mark
step5) invert the flask several times to mix the solution
identifying an unknown metal by titration
1) calc moles of know reactant
2)use balanced equation to calc mole of unknown substance used in titration
3) calc mole of unknown substance present in the 250cm³ solution
4)calc molar mass of unknown substance using Mass and moles
5) calc reactive atomic mass of unknown element in substance
6) suggest the identity of the element
definition of hydrated, anhydrous and water of crystallisation
hydrated- when water of crystallisation is present in a crystal compound
anhydrous- when all the waters of C have been removed from a compound
Water of C- The water present in a compound giving the compound a crystalltaline appearance
Calc the moles of water of crystallisation
1) weigh empty crucible
2) add hydrated salt to the crucible and reweigh
3) heat the crucible to evpourate the water until the anhydrous salt remains
4) allow the crucible to cool and reweigh
5)( mass of curcible + hydrated salt)- mass of empty= mass of H salt
6) ( mass of crucible+ anhydrous salt)- mass of empty= mass of A salt
7) mass of hydrated salt- A salt= mass of water
8) 18+ mass of water= mol
9) mass of A salt/ Mr of salt and water= mol
10) create a ratio to find moles of water
Moles of a gas
Amount,n=volume,V / molar gas volume, Vm
mol= dm³/ dm³ mol-1
molar gas volume at RTP 24.0dm³ mol‐1
1 Mole of gas= 24.0 dm³
Moles of a gas not at RTP
pV=nRT
p=pressure, Pa
V= volume of gas, m³
n= amount of gas, mol
R=gas constant, 8.314 J mol-1 K-1
T= tempreature, K
some conversions
1 m³= 1000 dm³
1 dm³= 1000 cm³
1 kPa= 1000 Pa
0⁰c= 273k
