Amount of a substance

Question 1

Relative atomic mass

Answer 1

The weighted mean mass of an atom of an element compared with 1/12 the mass of carbon 12

Question 2

Relative molecular mass

Answer 2

The weighted mean mass of a simple covalent molecule compared with 1/12 of the mass of an atom of C-12

Question 3

how to calc relative atomic mass

Answer 3

to calc relative molecular mass you add up all the individual relative atomic masses of the elements contained in the molecule

Question 4

Amount of a substance units and symbol

Answer 4

N, Moles

Question 5

Mole

Answer 5

The amount of substance contained as many particals as there are carbon atoms in exactly 12g of C-12

Question 6

Molar mass

Answer 6

The mass, in g, per mole of a substance.Units are g Mol -¹

Question 7

Amount of a substance equation

Answer 7

Amount, n = Mass/Mr
Moles(mol) g gmol-¹

Question 8

Avagadro’s constant

Answer 8

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

Question 9

Number of particals equation

Answer 9

No of particals= Nª × moles

Question 10

Reacting masses of soilds equation

Answer 10

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

Question 11

empirical formular

Answer 11

the simplest whole no ratio of atoms of each element present

Question 12

molecular formular

Answer 12

the actual no of atoms of each element in a molecule

Question 13

empirical formular equation

Answer 13

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

Question 14

molecular formular equation

Answer 14

molar mass/ empirical formular mass
then x the empirical formular by the answer

Question 15

percentage Yield equation

Answer 15

% 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

Question 16

atom economy equation

Answer 16

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

Question 17

Moles in solution

Answer 17

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

Question 18

Acid titrations method

Answer 18

1. 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
2. The solution B was slowly added to the conical flask until the indicator changed colour at the end point (trial)
3. 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

Question 19

how to prep a standard solution

Answer 19

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

Question 20

identifying an unknown metal by titration

Answer 20

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

Question 21

definition of hydrated, anhydrous and water of crystallisation

Answer 21

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

Question 22

Calc the moles of water of crystallisation

Answer 22

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

Question 23

Moles of a gas

Answer 23

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

Question 24

Moles of a gas not at RTP

Answer 24

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

Question 25

some conversions

Answer 25

1 m³= 1000 dm³
1 dm³= 1000 cm³
1 kPa= 1000 Pa
0⁰c= 273k