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A sample of titanium was ionised by electron impact in a time of flight (TOF) mass spectrometer - AQA - A-Level Chemistry - Question 4 - 2017 - Paper 1

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A sample of titanium was ionised by electron impact in a time of flight (TOF) mass spectrometer. Information from the mass spectrum about the isotopes of titanium in... show full transcript

Worked Solution & Example Answer:A sample of titanium was ionised by electron impact in a time of flight (TOF) mass spectrometer - AQA - A-Level Chemistry - Question 4 - 2017 - Paper 1

Step 1

Calculate the relative atomic mass of titanium in this sample.

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Answer

To calculate the relative atomic mass, use the formula:

Relative Atomic Mass=∑(mass×abundance)∑abundance\text{Relative Atomic Mass} = \frac{\sum (\text{mass} \times \text{abundance})}{\sum \text{abundance}}

Substituting the values from the table:

Relative Atomic Mass=(46×9.1)+(47×7.8)+(48×74.6)+(49×8.5)9.1+7.8+74.6+8.5\text{Relative Atomic Mass} = \frac{(46 \times 9.1) + (47 \times 7.8) + (48 \times 74.6) + (49 \times 8.5)}{9.1 + 7.8 + 74.6 + 8.5}

Calculating the numerator:

  • 46×9.1=418.646 \times 9.1 = 418.6
  • 47×7.8=366.647 \times 7.8 = 366.6
  • 48×74.6=3580.848 \times 74.6 = 3580.8
  • 49×8.5=416.549 \times 8.5 = 416.5

Adding these gives: 418.6+366.6+3580.8+416.5=4361.5418.6 + 366.6 + 3580.8 + 416.5 = 4361.5

For the denominator: 9.1+7.8+74.6+8.5=1009.1 + 7.8 + 74.6 + 8.5 = 100

Thus, Relative Atomic Mass=4361.5100=43.615\text{Relative Atomic Mass} = \frac{4361.5}{100} = 43.615

Rounded to one decimal place, the relative atomic mass is 47.4.

Step 2

Write an equation, including state symbols, to show how an atom of titanium is ionised by electron impact.

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The ionisation of titanium can be expressed as:

Ti(g)+e−→Ti+(g)+2e−\text{Ti(g)} + e^- \rightarrow \text{Ti}^+(g) + 2e^-

The m/z value for the ion that will reach the detector first is the one for the ion Ti which has a m/z of 48.

Step 3

Calculate the mass, in kg, of one atom of 48Ti.

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To find the mass of one atom of 48Ti, use the formula:

mass=ML\text{mass} = \frac{M}{L}

Where M is the molar mass of 48Ti (approximately 48 g/mol), and L is Avogadro's number:

L=6.022×1023 mol−1L = 6.022 \times 10^{23} \text{ mol}^{-1}

Thus, mass=48 g/mol6.022×1023extmol−1=7.96×10−26 kg\text{mass} = \frac{48 \text{ g/mol}}{6.022 \times 10^{23} ext{ mol}^{-1}} = 7.96 \times 10^{-26} \text{ kg}.

Step 4

Calculate the time of flight of the '48Ti' ion.

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Answer

Using the time of flight equation:

t=dm2Et = d \sqrt{\frac{m}{2E}}

First, let's find the mass of the ion: Using m for 48Ti which is approximately 48×10−348 \times 10^{-3} kg,

Given:

  • d=1.5d = 1.5 m (assumed from previous parts)
  • Kinetic energy, E=1.013×10−3 JE = 1.013 \times 10^{-3}\text{ J}

Substituting: t=1.548×10−32×1.013×10−3t = 1.5 \sqrt{\frac{48 \times 10^{-3}}{2 \times 1.013 \times 10^{-3}}}

Calculating: t≈1.523.65≈1.5×4.86≈7.29 st \approx 1.5 \sqrt{23.65} \approx 1.5 \times 4.86 \approx 7.29 \text{ s}

Thus, the time of flight of the "48Ti" ion is approximately 7.3 s (to 2 significant figures).

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