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elements of life 2 (el6-9) (tests for ions (halides (add dilute nitric…
elements of life 2 (el6-9)
empirical and molecular formula
example
mass of mg= 0.84g
amount of Mg atoms: 0.84 / 24= 0.035 mol
mass of oxygen= 0.56g
amount of O atoms: 0.56 / 16= 0.035 mol
ratio= 1:1 (IF RATIO IS NOT 1:1 THEN DIVIDE ALL NUMBERS BY THE SMALLEST NUMBER OF MOLES)
EF= MgO
molecular formulae involve waters of crystalization
hydrated compounds contain waters of crystallization but anhydrous dont
ONE mole of a particular hydrated compound always has the SAME NUMBER of moles of water of crystallization
example: heating 3.210g of MgSO4.XH2O forms 1.567g of anhydrous magnesium sulfate. find value of X
mass of water lost: 3.210-1.567= 1.643g
number of moles of water lost: (mass/molar mass) 1.643 / 18= 0.09127 moles
molar mass of MgSO4: 24.3+ 32.1+(4x16)=120.4 g/mol
number of moles in 1.567g: 1.567 / 120.4= 0.01301 moles
1 mole of salt: 0.09127/ 0.01301 = 7.015
= MgSO4.7H2O
% yield is NEVER 100%
percentage yield= actual yield/ theoretical x 100
theoretical= the mass of the product that should be made if no chemicals are lost in the process
actual= always less than theoretical.
example: 9.2g of ethanol was reacted with excess oxidising agent and 2.1g of ethanol was produced. calculate theoretical and % yeild
M of C2H50H= (2x12)+5+16+1= 46 g/mol
moles of C2H5OH: 9.2/ 46= 0.2 moles
M of CH3CHO: (2x12)+4+16= 44g/mol
theoretical (mass of CH3HO: 0.2 x 44= 8.8g
2.1/ 8.8 x100= 24%
the mole and equations
a mole is the number of particles
amount of substance is measured in moles
ONE MOLE= 6.02 x 1023 (Avogadro's constant)
number of moles= number of particles you have/ number of particles in a mole (Avogadro's constant)
e.g. 1.5 x 1024 / 6.02 x 1023
molar mass = mass of ONE MOLE
molar mass= relative formula mass
e.g. mr of CaCO3= 40.1+ 12+ 3(16)=100.1g/mol
number of moles= mass of substance/ molar mass
in a solution concentration is measured in mol/dm3
number of moles= concentration x volume/ 1000
ionic equations
any reaction involving ions that happen in solution
only reacting particles and the products they form are included
titrations
how to make a standard solution (has a known conc)
1) work out how many moles of solute you need by using: moles= concentration x volume / 1000
2) work out the mass in grams of solute needed using: mass= moles x molar mass
3) carefully weigh out this mass of solute- first weigh beaker and record mass and then add correct mass
4) add a small amount of distilled water to the beaker and stir until all solute is dissolved
5) tip solution into volumetric flask using a funnel
6) rinse the beaker and stirring rod with distilled water and add this to the flask to (the washings)
7) top up flask to correct volume with distilled water (making sure bottom of meniscus reaches line)
8) stopper the flask and turn it upside down 6 times to mix solution
you can make a standard solution from a more concentrated solution
vol to use= final conc / initial conc x vol. required
accurately measure out this volume and add it to volumetric flask
use a pipette/ burette and then fill rest of volumetric flask with distilled water
accuracy
measure alkali using pipette and then add to conical flask with indicator (methyl orange or phenolphthalein)
first do rough titration to see where endpoint is roughly
add acid to burette
add acid to alkali whilst swirling stop when there is a permanent colour change
when doing an accurate titration 2cm3 before rough endpoint add acid drop by drop to alkali
work out titre (final reading - initial reading)
repeat until the results are concordant then calc mean
indicators
methyl orange= turns from YELLOW to RED
phenolphthalein= turns from PINK to COLOURLESS
titration calcs
concentrations from titrations
example: 25 cm3 of 0.5 mol/dm3 HCl used to neutralize 35cm3 of NaOH calc conc of NaOH
balanced equation: HCl + NaOH ----> NaCl + H2O
number of moles of HCl: 0.5 x 25 / 1000 = 0.0125moles
ratio= 1:1
conc of NaOH: 0.0125 x1000 / 35 = 0.36 mol/dm3
volumes from titrations
example: 20.4cm3 of a 0.5 mol/dm3 solution of sodium carbonate reacts with 1.5 mol/dm3 nitric acid. calc volume of nitric acid
balanced equation: Na2CO3 + 2HNO3 --> 2NaNO3 + H2O +CO2
number of moles of Na2CO3: 0.5 x 20.4/ 1000= 0.0102mol
ratio=1:2
0.0102 x 2= 0.0204 mol
vol of HNO3: 0.0204 x 1000 / 1.5 = 13.6 cm3
volumes needed to neutralize masses
example: calc vol of 2 mol/dm3 HCl required to neutralize 3.86g of Ca(OH)2
balanced equation: Ca(OH)2 + 2HCl --> CaCl2 + 2H2O
molar mass of Ca(OH)2: 40.1 + (2x16) + 2 =74.6 gmol-1
number of moles in 3.86g of Ca(OH)2: 3.86 / 74.1= 0.0521 mol
vol of HCl: 0.104 x 1000/ 2= 52 cm3
periodic trends
arranged by atomic number
all elements within a period have the same number of electron shells
all elements within a group have the same number of electrons in their outer shell
elements in a group have similar physical and chemical properties
period 2 and 3
for METALS the melting points increase across period - metal bonds get stronger due to decreasing ionic radius and an increasing number of delocalised electrons. leads to a HIGHER CHARGE DENSITY (attracts the ions more strongly)
simple molecules have low melting points due to weak intermolecular forces
elements with giant covalent structures (c and si) have highest melting point in periods
larger molecules have highest melting points
noble gases have lowest melting points as they exist as single atoms so have weak intermolecular forces
ionisation
first ionisation enthalpy= the energy needed to remove 1 electron from each atom in 1 mole of gaseous atoms to form 1 mole of 1+ gaseous ions
general equation
X(g)---> X+(g) + e-
the lower the ionization enthalpy the easier it is to remove an outer electron and form an ion
things that affect ionization enthalpies
atomic radius
the further outer shell is from nucleus the less the electrons will be attracted to nucleus - ie with be lower
nuclear charge
the more protons in the nucleus the more it will attract electrons - ie will be higher
electron shielding
inner electrons shield the outer shell electrons from attractive force of nucleus- ie will be lower
first ionization enthalpies DECREASE down group
shells are further from nucleus so electrons are less attracted as you go down group
amount of shielding increases
number of protons increases down group
first ionization enthalpies INCREASE across periods
number of protons increase across group
s block have low ionization enthalpies
low nuclear charges means they lose outer electrons easily - makes them reactive
p block have higher charges dur to increase in protons making it difficult for them to lose a proton
group 2
they react with water to produce HYDROXIDES
they react with oxygen to produce OXIDES
oxides and hydroxides are bases
metal hydroxides are soluble in water and hydroxide OH- ions are alkaline
more strongly alkaline solution as you go down group as hydroxides get more soluble
as you go down group hydroxides get MORE SOLUBLE
as you go down group carbonates get LESS SOLUBLE
carbonates decompose to form CO2 and metal oxides
thermal stability
INCREASES DOWN GROUP
carbonate ions are large anions and are unstable if a group 2 element (cation) is present
the cation draws electrons of carbonate towards it polarizing it and distorts it making it less stable
LARGE CATIONS CAUSE LESS DISTORTION THAN SMALL CATIONS- due to lower charge density (charge of ion relative to its volume)
making salts
acid + base----> salt + water
salts are formed from positive cations and negative anions so they are neutral
solubility
lithium, sodium, potassium and ammonium salts are soluble
nitrates are soluble
most chlorides, bromides, iodides are soluble except silver halides, lead and copper iodide
most sulfates are soluble
most hydroxides are insoluble
most carbonates are insoluble
to make a insoluble salt use precipitation reaction
to make soluble salts use a metal or an insoluble base and then react with an acid
to make a soluble salt you can use an alkali which are soluble with acid
tests for ions
flame test
1) dip nichrome loop in HCl and then dip into sample
2) hold loop in blue Bunsen burner flame and observe colour change
colours:
Li= crimson, Na= orange/yellow, K= lilac, Ca= brick red, Ba= green, Cu=blue green
sodium hydroxide (test for metal ions)
add a few drops of sodium hydroxide to solution and coloured precip can form
colours:
Ag= brown, Ca= white, Cu= blue, Pb= white, Fe2+= green, Fe3+=reddish brown, Zn= white but dissolves colourless, Al= white then dissolves colourless
carbonates
add HCl and if carbonate is present limewater turns cloudy
sulfates
add HCl and barium chloride
if white precip forms sulfate is present
ammonium compounds
ammonia gas is alkaline so will turn red litmus paper blue
or add sodium hydroxide and heat mixture if ammonia is given off then ammonia ions are present
hydroxides
dip red litmus paper into solution, if present will turn blue
halides
add dilute nitric acid then add silver nitrate solution
chlorine produces white precip
bromine produces cream precip
iodine produces yellow precip
nitrates
heat solution with sodium hydroxide and aluminium foil
if ammonia is produced nitrate ions are present - can be tested for with litmus paper