1.3- Chemical Calculations

Question 1

(a)

Relative atomic mass (Ar)

Answer 1

the average mass of one atom of the
element compound to one-twelfth the mass of an atom of carbon-12

Question 2

(a)

Relative isotopic mass

Answer 2

the mass of one atom of an isotope
relative to one-twelfth the mass of one atom of carbon-12

Question 3

(a)

Relative formula mass (Mr)

Answer 3

the total average mass of all the
atoms in the formula relative to one-twelfth the mass of an atom
of carbon-12

Question 4

(a)

Molecular formula

Answer 4

The actual number of atoms of each element present in the molecule.

Question 5

(b)

Mass spectrometer: Ionisation

Answer 5

In the ionisation chamber, vaporised sample particles are bombarded by electrons, resulting in collisions that remove an electron from the particles, creating positive ions.

Question 6

(b)

Mass spectrometer: Acceleration

Answer 6

The positive ions are accelerated to a high speed by an electric field so they all have the same kinetic energy.

Question 7

(b)

Mass spectrometer:
Deflection

Answer 7

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.

Question 8

(b)

Mass spectrometer:
Detection

Answer 8

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.

Question 9

(b)

RAM

Answer 9

Ar = sum of isotope mass x isotope abundance / 100

Question 10

(c)

Simple mass spectra: Chlorine

Answer 10

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

Question 11

(d)

Empirical and molecular formulae determined from given data

Answer 11

1. Find the amount in moles of each element (divide by the molar mass)
2. Find the ratio of the number of atoms present (divide by the smallest value in step 1)
3. Convert these numbers into whole numbers (atoms combine together in whole number ratios)

Question 12

(e)

Relationship between the Avogadro constant, the mole and molar mass

Answer 12

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

Question 13

(f)

Relationship between grams and moles

Answer 13

moles= molarmass(g/mol)/
mass(g)

Question 14

(g)

Concentration in terms of grams formula

Answer 14

Concentration (g/dm3 ) = Mass (g) / Volume (dm3 )

Question 15

(g)

Concentration in terms of moles per unit volume

Answer 15

Concentration = amount of moles of solute/volume of solution

Question 16

(h)

The molar gas volume

Answer 16

Number of moles (n) = volume given/ molar gas volume (24.5 dm3)

Question 17

(i)

Ideal gas equation

Answer 17

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)

Question 18

(i) Conversions

Kpa = Pa
1atm = Pa
m = dm = cm
°C = K

Answer 18

1 kPa = 1 × 1000 Pa
1 atm = 1.01 × 100000 Pa (data booklet)
m = dm = cm (x1000)
°C (+273) = K

Question 19

(k)

Atom economy formula

Answer 19

(total Mr of required product/ total Mr of recatants) × 100

Question 20

(k)

Percentage yield

Answer 20

% yield = (mass of product obtained/ maximum therotical yield) × 100

Question 21

(l)

Percentage error

Answer 21

percentage error = 0.1 / x × 100

Question 22

Water of crystallisation

Answer 22

1. Calculate the moles of anhydrous
2. Find moles of water
3. Divide both number of moles by the smallest number to find ratio
4. Formula