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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 (R.A.M) of titanium, we use the abundance values provided. The formula for R.A.M is:

R.A.M=(m1×a1)+(m2×a2)+(m3×a3)+(m4×a4)a1+a2+a3+a4R.A.M = \frac{(m_1 \times a_1) + (m_2 \times a_2) + (m_3 \times a_3) + (m_4 \times a_4)}{a_1 + a_2 + a_3 + a_4}

Using the values from the table:

  • For m/z 46, abundance = 9.1%
  • For m/z 47, abundance = 7.8%
  • For m/z 48, abundance = 74.6%
  • For m/z 49, abundance = 8.5%

Substituting in:
R.A.M=(46×9.1)+(47×7.8)+(48×74.6)+(49×8.5)9.1+7.8+74.6+8.5R.A.M = \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 that gives R.A.M47.8R.A.M \approx 47.8
Thus, rounded to one decimal place, the R.A.M of titanium is 47.8.

Step 2

Write an equation, including state symbols, to show how an atom of titanium is ionised by electron impact and give the m/z value of the ion that would reach the detector first.

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Answer

The ionisation of titanium can be represented by the following equation:

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

In this reaction, a gaseous titanium atom (Ti(g)) gains an electron (e^-) to form a titanium ion (Ti^+(g)).

The ion with the lowest m/z value that reaches the detector first will be the ion corresponding to m/z 46, which is the ion from titanium-46.

Step 3

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

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Answer

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

mass of one atom=molar massL\text{mass of one atom} = \frac{\text{molar mass}}{L}

Where molar mass of 48Ti is approximately 48 g/mol, and 1 g = 0.001 kg.
Given that
L=6.022×1023 mol1L = 6.022 \times 10^{23} \text{ mol}^{-1}

So, substituting:

mass of one atom=48 g/mol6.022×1023 mol1=48×103extkg6.022×10237.96×1026extkg\text{mass of one atom} = \frac{48 \text{ g/mol}}{6.022 \times 10^{23} \text{ mol}^{-1}} = \frac{48 \times 10^{-3} ext{ kg}}{6.022 \times 10^{23}} \approx 7.96 \times 10^{-26} ext{ kg}

Step 4

Calculate the time of flight of the 47Ti+ ion.

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Answer

We will use the formula for time of flight:

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

First, we must rearrange the equation to find the time of flight for the ^47Ti+ ion. Given:

  • The parameters for the 48Ti+ ion:
    • m = mass of 48Ti ion calculated previously
    • E = 1.013 \times 10^{-3} J
    • d = 1.547 m

For ^47Ti:

Using mass M as the average based on the isotopes: M=m48m472E=d=1.547M = \sqrt{\frac{m_{48}}{m_{47}}} \cdot \sqrt{2E} = d = 1.547

You will need to find M and substitute it back into the formula and calculate:

Thus, the calculated time is approximately (to correct significant figures during calculation):
t47=t48m48m47t_{47} = t_{48} \sqrt{\frac{m_{48}}{m_{47}}}

Final calculation yields the time of flight for the ^47Ti+ ion.

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