A level chemistry time of flight mass spectrometry questions
I am stuck with these questions, need help:
6. Bromine has 2 principle isotopes, 79Br and 81Br and under standard conditions the element exists as a diatomic molecule. Find the isotope(s) present in the ion with kinetic energy of 4.440 x 10-16 J and a time of flight of
3.524 x 10-5s in a mass spectrometer of drift length 2.050 m.
7. An ionised sample contained the ions 27Al+ and 54Fe+. The 27Al+ ions had a time of flight of 2.345 x 10-5 s. Find the time of flight for the 54Fe + ions.
8. An ionised sample contained the ions 23Na+ which had a time of flight of 1.134 x 10 -5 s and an unkown element X whose ions had a time of flight of 1.689 x 10 -5 s. Find the Ar of the unknown element.
6. Bromine has 2 principle isotopes, 79Br and 81Br and under standard conditions the element exists as a diatomic molecule. Find the isotope(s) present in the ion with kinetic energy of 4.440 x 10-16 J and a time of flight of
3.524 x 10-5s in a mass spectrometer of drift length 2.050 m.
7. An ionised sample contained the ions 27Al+ and 54Fe+. The 27Al+ ions had a time of flight of 2.345 x 10-5 s. Find the time of flight for the 54Fe + ions.
8. An ionised sample contained the ions 23Na+ which had a time of flight of 1.134 x 10 -5 s and an unkown element X whose ions had a time of flight of 1.689 x 10 -5 s. Find the Ar of the unknown element.
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Are you sure you've posted in the right place? Here's a link to our subject forum which should help get you more responses if you post there. For question 7, because both samples are going through the same flight tube, we can say: KE = KE
Now we can also say: 1/2 mv² = 1/2 mv²
If we rearrange the equation (I've named them 1 and 2, so we know which one is first sample and second sample), we get:
m1/(t1)² = m2/(t2)²
If we substitute our values it will give us:
27/(2.345×10^-10) = 54/(t2)²
We work out that and it gives us:
4.91×10¹⁰ = 54/(t2)²
We can now rearrange this to make (t2)² the subject because we are trying to find the t (time) of the 2nd sample. So this will look like:
(t2)² = 54/4.91×10¹⁰
So this gives us:
(t2)² = 1.1×10^-9
So to find only t2 (without the square) we square root the other side, like this:
t2 = √1.1×10^-9
And if we out it in the calculator (always put brackets so we don't make any mistakes:
t2 = 3.32×10^-5
Therefore, the time of flight of ⁵⁴Fe is 3.32×10^-5 seconds.
Hope it helped idk how to do the 6th or the 8th but I'm going to try now
Now we can also say: 1/2 mv² = 1/2 mv²
If we rearrange the equation (I've named them 1 and 2, so we know which one is first sample and second sample), we get:
m1/(t1)² = m2/(t2)²
If we substitute our values it will give us:
27/(2.345×10^-10) = 54/(t2)²
We work out that and it gives us:
4.91×10¹⁰ = 54/(t2)²
We can now rearrange this to make (t2)² the subject because we are trying to find the t (time) of the 2nd sample. So this will look like:
(t2)² = 54/4.91×10¹⁰
So this gives us:
(t2)² = 1.1×10^-9
So to find only t2 (without the square) we square root the other side, like this:
t2 = √1.1×10^-9
And if we out it in the calculator (always put brackets so we don't make any mistakes:
t2 = 3.32×10^-5
Therefore, the time of flight of ⁵⁴Fe is 3.32×10^-5 seconds.
Hope it helped
idk how to do the 6th or the 8th but I'm going to try nowOriginal post by Miongo
Hi did you end up getting 1.658x10^-5 for question 7? The answer to question 6 is 79Br and 79 Br. I could send you the working out. No clue about 8
For question 7, because both samples are going through the same flight tube, we can say: KE = KE
Now we can also say: 1/2 mv² = 1/2 mv²
If we rearrange the equation (I've named them 1 and 2, so we know which one is first sample and second sample), we get:
m1/(t1)² = m2/(t2)²
If we substitute our values it will give us:
27/(2.345×10^-10) = 54/(t2)²
We work out that and it gives us:
4.91×10¹⁰ = 54/(t2)²
We can now rearrange this to make (t2)² the subject because we are trying to find the t (time) of the 2nd sample. So this will look like:
(t2)² = 54/4.91×10¹⁰
So this gives us:
(t2)² = 1.1×10^-9
So to find only t2 (without the square) we square root the other side, like this:
t2 = √1.1×10^-9
And if we out it in the calculator (always put brackets so we don't make any mistakes:
t2 = 3.32×10^-5
Therefore, the time of flight of ⁵⁴Fe is 3.32×10^-5 seconds.
Hope it helped
idk how to do the 6th or the 8th but I'm going to try nowOriginal post by learn233
I am stuck with these questions, need help:
6. Bromine has 2 principle isotopes, 79Br and 81Br and under standard conditions the element exists as a diatomic molecule. Find the isotope(s) present in the ion with kinetic energy of 4.440 x 10-16 J and a time of flight of
3.524 x 10-5s in a mass spectrometer of drift length 2.050 m.
7. An ionised sample contained the ions 27Al+ and 54Fe+. The 27Al+ ions had a time of flight of 2.345 x 10-5 s. Find the time of flight for the 54Fe + ions.
8. An ionised sample contained the ions 23Na+ which had a time of flight of 1.134 x 10 -5 s and an unkown element X whose ions had a time of flight of 1.689 x 10 -5 s. Find the Ar of the unknown element.
6. Bromine has 2 principle isotopes, 79Br and 81Br and under standard conditions the element exists as a diatomic molecule. Find the isotope(s) present in the ion with kinetic energy of 4.440 x 10-16 J and a time of flight of
3.524 x 10-5s in a mass spectrometer of drift length 2.050 m.
7. An ionised sample contained the ions 27Al+ and 54Fe+. The 27Al+ ions had a time of flight of 2.345 x 10-5 s. Find the time of flight for the 54Fe + ions.
8. An ionised sample contained the ions 23Na+ which had a time of flight of 1.134 x 10 -5 s and an unkown element X whose ions had a time of flight of 1.689 x 10 -5 s. Find the Ar of the unknown element.
For Question number 6, here is what you have to do:
Ek = 4.440x10^-16 J
distance = 2.050 m
TOF = 3.524 x 10^-5 S
1) find the velocity - as you would need it in the equation for Ek to find the mass:
Velocity = 2.050 / 3.524 x 10^-5 = 58172.53121 m/s
2) Find the mass (kg) by changing the equation:
Ek = 0.5mv^2
4.440x10^-16 = 0.5 x m x (58172.53121)^2
4.440x10^-16 / 0.5 x (58172.53121)^2
=2.624079831x10^-25 kg
3) To find the Mr the equation is m(g) x Avogadro's number.
(2.624079831x10^-25)x1000 = 2.624079831x10^-22 g
4) Find the Mr:
(2.624079831x10^-22) x (6.022x10^23)
= 158.022 = 158 (3sf)
79 Br + 79 Br isotopes add up to make the Mr 158 so it his basically 79Br2
Original post by learn233
I am stuck with these questions, need help:
6. Bromine has 2 principle isotopes, 79Br and 81Br and under standard conditions the element exists as a diatomic molecule. Find the isotope(s) present in the ion with kinetic energy of 4.440 x 10-16 J and a time of flight of
3.524 x 10-5s in a mass spectrometer of drift length 2.050 m.
7. An ionised sample contained the ions 27Al+ and 54Fe+. The 27Al+ ions had a time of flight of 2.345 x 10-5 s. Find the time of flight for the 54Fe + ions.
8. An ionised sample contained the ions 23Na+ which had a time of flight of 1.134 x 10 -5 s and an unkown element X whose ions had a time of flight of 1.689 x 10 -5 s. Find the Ar of the unknown element.
6. Bromine has 2 principle isotopes, 79Br and 81Br and under standard conditions the element exists as a diatomic molecule. Find the isotope(s) present in the ion with kinetic energy of 4.440 x 10-16 J and a time of flight of
3.524 x 10-5s in a mass spectrometer of drift length 2.050 m.
7. An ionised sample contained the ions 27Al+ and 54Fe+. The 27Al+ ions had a time of flight of 2.345 x 10-5 s. Find the time of flight for the 54Fe + ions.
8. An ionised sample contained the ions 23Na+ which had a time of flight of 1.134 x 10 -5 s and an unkown element X whose ions had a time of flight of 1.689 x 10 -5 s. Find the Ar of the unknown element.
For Question 8 you basically do the same as question 7!
Na+ ions = TOF of 1.32x10^-5 s
Na+ ions = Mr of 23
X (unknown) ions = TOF of 1.689x10^-5s
X (unknown) ions = Mr of ?
1) Ke = Ke because the isotopes are accelerated to the same speed.
1/2mv^2 = 1/2mv^2
You have to substitute your values into the equation. The 1/2 in both equations cancel out. Because we don't have velocity we will divide the mass (which we will consider to be the Mr) with TOF^2
23/(1.132 x 10^-5)^2 = m of x / (1.689 x 10^-5)^2
1.78855 x 10^11 = m/(1.689 x 10^-5)^2
You know how to multiply by (1.689 x 10^-5)^2 to get 'm' alone.
m = (1.78855 x 10^11) x (1.689 x 10^-5)^2
m = 51.022... = 51 (2sf)
51 is the Ar of the unknown element!
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