1.3- Chemical Calculations Flashcards
(a)
Relative atomic mass (Ar)
the average mass of one atom of the
element compound to one-twelfth the mass of an atom of carbon-12
(a)
Relative isotopic mass
the mass of one atom of an isotope
relative to one-twelfth the mass of one atom of carbon-12
(a)
Relative formula mass (Mr)
the total average mass of all the
atoms in the formula relative to one-twelfth the mass of an atom
of carbon-12
(a)
Molecular formula
The actual number of atoms of each element present in the molecule.
(b)
Mass spectrometer: Ionisation
In the ionisation chamber, vaporised sample particles are bombarded by electrons, resulting in collisions that remove an electron from the particles, creating positive ions.
(b)
Mass spectrometer: Acceleration
The positive ions are accelerated to a high speed by an electric field so they all have the same kinetic energy.
(b)
Mass spectrometer:
Deflection
Ions are deflected by a magnetic field based on their mass and charge: lighter ions are deflected more than heavier ones, and ions with higher positive charges are deflected more. These factors are combined in the mass/charge (m/z) ratio.
(b)
Mass spectrometer:
Detection
Only ions with a specific m/z ratio reach the ion detector, where they gain electrons, producing a current proportional to the isotope’s abundance. The signal is amplified and recorded, and by varying the magnetic field, different ion streams are sequentially directed to the detector.
(b)
RAM
Ar = sum of isotope mass x isotope abundance / 100
(c)
Simple mass spectra: Chlorine
Chlorine has two isotopes – chlorine-35 and chlorine-37.
35Cl—35Cl+ (m/z 70)
35Cl—37Cl+ and 37Cl—35Cl+ (m/z 72)
37Cl —37Cl+ (m/z 74)
Ratio 9:6:1
(d)
Empirical and molecular formulae determined from given data
- Find the amount in moles of each element (divide by the molar mass)
- Find the ratio of the number of atoms present (divide by the smallest value in step 1)
- Convert these numbers into whole numbers (atoms combine together in whole number ratios)
(d)
A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Its relative molecular mass is 180. What are the empirical and molecular formulae?
Empirical formula: CH₂O
Molecular formula: C₆H₁₂O₆
(e)
Relationship between the Avogadro constant, the mole and molar mass
number of moles = mass of substance (in g) ÷ molar mass (Mr)
mass = number of moles x molar mass of atoms
molar mass of atoms = mass of substance (in g) ÷ number of moles
(f)
Relationship between grams and moles
moles= molarmass(g/mol)/
mass(g)
(g)
Concentration in terms of grams formula
Concentration (g/dm3 ) = Mass (g) / Volume (dm3 )
(g)
Concentration in terms of moles per unit volume
Concentration = amount of moles of solute/volume of solution
(h)
The molar gas volume
Number of moles (n) = volume given/ molar gas volume (24.5 dm3)
(i)
Ideal gas equation
pV = nRT
p= pressure (Pa)
V= volume (m3)
n= number of moles (mol)
R = molar gas constant (8.31 J mol–1 K–1) (data booklet)
T = temperature (K)
(i) Conversions
Kpa = Pa
1atm = Pa
m = dm = cm
°C = K
1 kPa = 1 × 1000 Pa
1 atm = 1.01 × 100000 Pa (data booklet)
m = dm = cm (x1000)
°C (+273) = K
(k)
Atom economy formula
(total Mr of required product/ total Mr of recatants) × 100
(k)
Percentage yield
% yield = (mass of product obtained/ maximum therotical yield) × 100
(l)
Percentage error
percentage error = 0.1 / x × 100
Water of crystallisation
- Calculate the moles of anhydrous
- Find moles of water
- Divide both number of moles by the smallest number to find ratio
- Formula
A 2.50 g sample of hydrated magnesium sulfate, MgSO₄·xH₂O, is heated until all the water is removed. The mass of the anhydrous MgSO₄ is 1.20 g. Calculate the value of x in the formula.
x = 7 → Formula = MgSO₄·7H₂O
