Mr Cole Chemistry

Mass Spec Calculations Questions

TOF Calculations

Worksheet

Practice Questions

1. Calculate the length of a mass spectrometer that was used to analyse a 40Ar+ ion. The ions were accelerated to a kinetic energy of 1.60×10-15 J and took 2.44×10-6 seconds to reach the detector.
Working Out
Step 1: Calculate Mass of Ion (kg)
Mass = 40 / (1000 × 6.022×1023)
Mass = 6.642 × 10-26 kg
Step 2: Calculate Velocity (v)
KE = ½mv2  →  v = √(2KE / m)
v = √((2 × 1.60×10-15) / 6.642×10-26)
v = 219,468 ms-1
Step 3: Calculate Length (d)
d = v × t
d = 219,468 × 2.44×10-6
d = 0.535 m
2. What was the kinetic energy of a 70Ga+ ion that took 1.40×10-6 seconds to travel through a mass spectrometer that was 1.60m long.
Working Out
Step 1: Calculate Mass of Ion (kg)
Mass = 70 / (1000 × 6.022×1023)
Mass = 1.162 × 10-25 kg
Step 2: Calculate Velocity (v)
v = d / t
v = 1.60 / 1.40×10-6
v = 1,142,857 ms-1
Step 3: Calculate Kinetic Energy (KE)
KE = ½mv2
KE = 0.5 × 1.162×10-25 × (1,142,857)2
KE = 7.59 × 10-14 J
3. Determine the identity of an ion that took 2.45×10-6 seconds to travel down a mass spectrometer that was 96.4cm long and imparted 2.438×10-15 joules of energy into the ions.
Working Out
Step 1: Convert Units
d = 96.4 cm = 0.964 m
Step 2: Calculate Velocity (v)
v = d / t
v = 0.964 / 2.45×10-6
v = 393,469 ms-1
Step 3: Calculate Mass (kg)
m = 2KE / v2
m = (2 × 2.438×10-15) / (393,469)2
m = 3.15 × 10-26 kg
Step 4: Calculate Relative Mass
Mr = m × 1000 × 6.022×1023
Mr = 18.97
Identity: Fluorine (19F+)
4. Calculate the length of time that a 96Mo+ ion would take to travel down mass spectrometer that is 80cm long, given a kinetic energy of 4.5×10-15 J.
Working Out
Step 1: Mass and Distance
Mass = 96 / (1000 × 6.022×1023) = 1.594 × 10-25 kg
d = 0.80 m
Step 2: Calculate Velocity (v)
v = √(2KE / m)
v = √((2 × 4.5×10-15) / 1.594×10-25)
v = 237,635 ms-1
Step 3: Calculate Time (t)
t = d / v
t = 0.80 / 237,635
t = 3.37 × 10-6 s
5. Calculate the length of time that it would take a 79Br+ ion to reach the detector given that a 81Br+ ion took 4.5×10-5 seconds.
Working Out

Tip: Since KE and distance are constant, time is proportional to the square root of the mass.

Ratio Method
t1 / t2 = √(m1 / m2)
t(79Br) = t(81Br) × √(79 / 81)
t = 4.5×10-5 × √(0.9753)
t = 4.5×10-5 × 0.9876
t = 4.44 × 10-5 s
6. Determine the identity of an ion that took 1.6×10-6 seconds to travel down an 80cm long spectrometer, in doing so gaining 2.92×10-15 J of energy.
Working Out
Step 1: Variables
d = 0.80 m
t = 1.6×10-6 s
KE = 2.92×10-15 J
Step 2: Calculate Velocity
v = 0.80 / 1.6×10-6 = 500,000 ms-1
Step 3: Calculate Mass (kg)
m = 2KE / v2
m = (2 × 2.92×10-15) / (500,000)2
m = 2.336 × 10-26 kg
Step 4: Identify
Mr = 2.336×10-26 × 1000 × 6.022×1023
Mr = 14.06
Identity: Nitrogen (14N+)
7. Calculate the length of a mass spectrometer in which a 121Sb+ ion was given 3.2×10-15 Joules and took 8.4×10-5 seconds to reach the detector.
Working Out
Step 1: Mass of Antimony-121
m = 121 / (1000 × 6.022×1023)
m = 2.009 × 10-25 kg
Step 2: Calculate Velocity
v = √(2KE / m)
v = √((2 × 3.2×10-15) / 2.009×10-25)
v = 178,480 ms-1
Step 3: Calculate Length
d = v × t
d = 178,480 × 8.4×10-5
d = 14.99 m (approx 15m)
8. A sample of chlorine was analysed. The 35Cl+ ions took 2.46×10-6 seconds to reach the detector in a 0.8m spectrometer. How long would a 37Cl2+ ion take to reach the detector.
Working Out

Note: A 2+ ion has twice the charge of a + ion, so it receives twice the Kinetic Energy in the electric field.

Variables
t1 = 2.46×10-6 s
Mass Ratio = 37 / 35
KE Ratio = 2 / 1 (due to 2+ charge)
Calculation
t ∝ √(m / KE)
t2 = t1 × √(Mass Ratio) × √(1 / KE Ratio)
t2 = 2.46×10-6 × √(37/35) × √(1/2)
t2 = 2.46×10-6 × 1.028 × 0.707
t = 1.79 × 10-6 s
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