atomic structure, amount of substance, and periodicity (A1 physical and inorganic chemistry) Flashcards
time of flight mass spectrometry
relative atomic mass = (abundance A * isotopic mass A) + (abundance B * isotopic mass B) / total abundance
t = d/v
t = tof
d = distance travelled (m)
v = velocity
KE = 1/2mv²
t = d√m/2KE
- sample is vaporised so it can travel through mass spectrometer
- ionisation of sample via electron impact (firing electrons at sample to knock electrons off) or electrospray (dissolving sample in volatile and polar solvent then passing a high voltage through causing sample to gain a proton from solvent, more gentle method for larger molecules that prevents fragmentation)
- positive ions accelerated through electric field, concentrated into a thin ion beam and passed through successively decreasing electrically charged plates to be provided with same kinetic energy, particles with lower m/z ratios will accelerate quicker
- particles travel with no electrical field through chamber with a constant speed and kinetic energy
- ions are detected as they hit a plate at the end of the spectrometer and an electrical current is made, particles with lower m/z ratios reach the detector faster
must be kept under high vacuum to prevent ions produced from colliding with molecule in the air
ideal gas equation e.g. what volume of H₂ is produced when 12g of potassium reacts with water at 100kPa of pressure and 298K (gas constant is 8.31JK⁻¹mol⁻¹)
pV = nRT
p = pressure (Pa)
V = volume (m³)
n = moles
R = gas constant
T = temperature (K)
standard conditions are 100kPa and 298K
m → x10 → dm → x10 → cm
m → x100 → cm
m² → x100 → dm² → x100 → cm²
m² → x10,000 → cm²
m³ → x1000 → dm³ → x1000 → cm³
m³ → 1,000,000 → cm³
2K (s) + 2H₂O (l) → 2KOH (aq) + H₂ (g)
moles of potassium = 12 / 39 = 0.31mol
2 moles of potassium react with one mole of hydrogen
moles of hydrogen = 0.31 / 2 = 0.155mol
V = nRT / p = 0.155 * 8.31 * 298 / 100,000 = 3.84x10⁻³m³
using balanced equations to work out theoretical mass e.g. how much CaO can be made when 34g of Ca is burned completely in oxygen
- write out balanced equation
2Ca (s) + O₂ (g) →2CaO (s) - find mass (Mr * mol)
40 * 2 = 80g
56 * 2 = 112g - divide masses on either side by mass of leading side then by mass in the question
112 / 80 = 1.4g
1.4 * 34 = 47.6g
give the definition for first ionisation energy and explain successive ionisation energies
first ionisation energy is the minimum amount of energy required to remove one mole of electrons from one mole of atoms in their gaseous state
X(g) -> X⁺(g) + e⁻
one mole of a substance contains the same amount of particles as there are atoms in 12g of carbon-12
ionisation requires energy, so it is always an endothermic process with a positive value
in successive ionisations, there is an increase in ionisation energy as we are removing an electron from an increasingly positive ion
there will be jumps in energy at certain points when we remove an electron from a new shell closer to the nucleus
explain the trend and exceptions in first ionisation energy across periods (using period 3 as an example), and down groups
there is a general increase across periods because increased nuclear charge with the same shielding effects and a marginally smaller atomic radius leads to increased nuclear attraction from valence electrons, therefore requires more energy to lose an electron
aluminium’s first ionisation energy is lower than magnesium’s, this provides evidence for sub-shells - aluminium’s outermost electron occupies the 3p1 orbital which is higher energy than the 3s2 orbital (magnesium’s outermost electron), meaning its electron is slightly further from the nucleus and slightly shielded by the 3s2 orbital, therefore it requires less energy to be removed
sulfur’s first ionisation energy is lower than phosphorous, this provides evidence for electron repulsion within an orbital - sulfur’s outermost electron is taken from a fully occupied orbital, electrons repel eachother so this requires less energy to remove than phosphorous’s outermost electron, which is in a singly-occupied orbital, despite sulfur having a slightly higher nuclear charge and the same shielding effect
going down groups ionisation energy decreases, as there is more shielding and larger atomic radius, so weaker attraction of outermost electron to nucleus (evidence for shells, proves Niels Bohr’s model of the atom correct)
titration calculations
a. 18.3cm³ of 0.25moldm⁻³ HCl was required to neutralise 25cm³ of KOH, calculate the concentration of KOH
b. 15.7cm³ of 0.450moldm⁻³ H₂SO₄ was required to neutralise 0.120moldm⁻³ of NaOH, calculate the volume of NaOH being neutralised in cm³
- write balanced equation
a. HCl + KOH → KCl + H₂O
b. 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O - calculate moles and molar ratio
a. 18.3x10⁻³ * 0.25 = 4.58x10⁻³ mol
1:1 ratio
b. 15.7x10⁻³ * 0.450 = 7.07x10⁻³ mol
2:1 ratio of NaOH:H₂SO₄
7.07x10⁻³ * 2 = 0.0141 mol - calculate unknown value
a. 4.58x10⁻³ / 25x10⁻³ = 0.183moldm⁻³
b. 0.0141 / 0.120 = 0.118dm³ * 1000 = 118cm³
titrations
- acid or alkali of known concentration in burette
- acid or alkali of unknown concentration but known volume in conical flask, with drops of indicator
- add chemical in burette to conical flask until indicator changes colour, add drop-wise toward end point
- read how much chemical from the burette was required to neutralise the chemical in the conical flask (from bottom of meniscus, at eye level) and record results to 2 decimal places, repeat until 2 concordant results are achieved (within 0.10 cm of each other)
phenolphthalein indicator is colourless in acidic solutions and pink in alkali, better when we have a weak acid and strong base as equivalence point matches phenolphthalein’s transition range
methyl orange indicator is red in acidic solutions and yellow in alkali, better to use when we have a strong acid and a weak base as equivalence point matches methyl orange’s transition range
a compound contains 23.3% magnesium, 30.7% sulfur and 46.0% oxygen, what is the empirical formula of this compound?
- write percentages as masses
- divide these by relative atomic mass to get number of moles
- divide all values by the smallest number of moles
- if necessary (e.g. 2.5), multiply by two and decipher empirical formula
23.3g / 24.3 = 0.96 / 0.96 = 1
30.7g / 32.1 = 0.96 / 0.96 = 1
46.0 / 16.0 = 2.88 / 0.96 = 3
MgSO₃
a hydrocarbon combusts completely to make 0.845g of CO₂ and 0.173g of H₂O, what is the empirical formula of the hydrocarbon?
- divide masses by Mr to get number of moles
- multiply this by molar ratio within compound
- divide values by smallest number of moles to decipher empirical formula
0.845g / 44 = 0.0192
0.173 / 18 = 0.0096
1 mole of CO₂ contains 1 mole of C atoms
1 mole of H₂O contains 2 moles of H atoms
0.0096 * 2 = 0.0192
0.0192 / 0.0192 = 1
CH
percentage yield and factors that affect it
percentage yield = actual yield / theoretical yield * 100
reversible reactions that do not go to completion
incomplete reactions due to unfavourable conditions
unwanted side reactions
loss of product during transfer, for example during filtering, evaporating, and pouring
atom economy
% atom economy = (Mr of desired product / Σ Mr of all reactants) * 100
remember to multiply Mr of desired product by its number of moles in the equation
importance of atom economy
industries try to develop chemical processes with the highest atom economy, this is because high atom economies mean that raw materials are used more efficiently meaning more sustainability, produce less waste, benefiting the environment, and produce less by-products so less time and money is spent separating these from the desired product
how to find molecular formula from empirical formula e.g. the empirical formula of a molecule is CH₂O, it has a relative molecular mass of 180, what is its molecular formula?
- calculate the empirical mass
- divide the Mr by the empirical mass
- multiply all the atoms in the empirical formula by this value
12 + (2 * 1) + 16 = 30
180 / 30 = 6
6 * 2 = 12
C₆H₁₂O₆
explain the melting points of period 3 elements
Na, Mg, Al - metals, therefore metallic bonding, increasing melting points as they have increasing charge density (more positive charge paired with smaller ionic radius) and donate more electrons to the sea of delocalised electrons, therefore there are stronger forces of electrostatic attraction which require more energy to overcome
Si - giant covalent structure, has the highest melting point in the period, silicon atoms held together by many strong covalent bonds which require a lot of energy to overcome
P, S, Cl, Ar - simple molecular structures, decreases going across period as they have smaller molecular structures and therefore weaker van der waals forces holding them together which are easier to break, slight increase between phosphorous and sulfur (4, 8, 2, 1)
explain the trends in atomic radius going across periods and down groups
decreases across periods as there is an increased nuclear charge with the same shielding effect, so valence electrons are pulled more strongly towards nucleus
increases down groups as outer electrons experience more shielding, which overrides increase in nuclear charge
