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# AQA A1 RAM Mass Spec

### AQA A1 isotopes, RAM, mass spectrometer

Question | Answer |
---|---|

What are isotopes? | ATOMS of the same element with the same number of p+ but different number of n0 |

Explain the difference between 12C and 14C | same number of p+ (6) but 14C has two MORE n0 than 12C. You must COMPARE. ("different" is too vague when you are given a specific example). |

What is recorded on a mass spectrum? What are the labels on the mass spectrum | Mass/charge ratio on the x-axis and Abundance on the y-axis |

24Mg, 25Mg, 26Mg are three isotopes. There is 10% of 25Mg. The RAM is 24.3. Work out the abundances of the other two isotopes. | Set up the equation first: (mass x abundance + …)/total abundance. Replace with values. Here there is one unknown. (24*x +25*10 + 26*(100-10-x))/100=24.3. Rearrange. x=80 |

RAM of Sulfur is 32.16. Abundance for 32S is 91.0%, 33S=1.8%; the third isotope is 7.2%. Work out the mass number of the third isotope | Set up the equation first: (mass x abundance + …)/total abundance. Replace with values. Here there is one unknown. (91.0*32+1.8*33+x*7.2)/100=32.16. Rearrange. x=34 |

Chlorine has two isotopes 35Cl and 37Cl with respective abundances of 75 and 25. Work out the RAM. | Set up the equation first: (mass x abundance + …)/total abundance. Replace with values. (35*75+37*25)/100=35.5 |

Describe how ions are detected and how the abundance of an isotope is determined. | Ions pick up an electrons, a current is generated and the current is proportional to the number of ions. |

Define relative atomic mass | Weighed average mass of an atom of an element, compared to 1/12th of a 12C carbon atom |

CHALLENGE: in the mass spectrum of Magnesium, there are 3 peaks at 24, 25 and 26 with respective abundances of 80, 10, 10%. But there are also very small peaks at m/z=12, 12.5 and 13. Explain the existence of these peaks. | The atoms will have lost TWO electrons in the ionisation chambers so the charge is +2 instead of +1 and the m/z ratio is half of a +1 ion. |

Why are the chemical properties of two isotopes the same? | Same number of electrons in the outer shell. |

There is 3 times more 85Rb than 87Rb. Work out the RAM of Rubidium. | We are given a ratio of abundance not a percentage! RAM=(mass x abundance + …)/total abundance. (85*3+87*1)/(3+1)= |

Write the equation, including state symbols, for the first ionisation of Carbon. | C(g) --> C+(g) + e-..... Did you remember the state symbols? |

CHALLENGE: Why do we not use more than the minimum energy to ionise an element? | Because there is a risk that the atoms would lose TWO electrons and the m/z ratios would be halved. |

Describe how atoms are ionised | By firing HIGH ENERGY electrons at the atoms; the atoms then lose an electron |

Give two reasons why atoms must be ionised? | 1) to be attracted to the negative plate in the acceleration chamber; 2) to pick up an electron at the detector (and generate a current) |

Explain why lighter ions arrive first at the detector | When they enter the drifting chamber, all ions have the same kinetic energy. As Ke=1/2mv2, the lighter the ions, the faster its velocity, so lighter ions arrive at the detector first |

Rearrange the equations Ke=1/2mv2 and v=d/t to make d the subject | d=t * V(2Ke/m). Practice. |

Rearrange the equations Ke=1/2mv2 and v=d/t to make t the subject | t=d * V(m/2Ke). Practice. |

In the Ke=1/2mv2 questions, what must you remember? | Convert the grams into kilograms! |

What is the mass of ONE ion for 24Mg+? (L=6.022*10^23 mole-1) | Mass of one particle=mass of one mole of particles / avogadro= 24 / 6.022*10^23 |

How do you work out the sketch of the mass spectrum for a molecule? | 1) Work out all the mass (m/z) combination by adding the mass numbers of the isotopes; 2) work out the abundance by multiplying the abundances of each isotopes |

Describe the spectrum for a CH3Cl molecule; assume that there is only one isotope of C: 12C and one for H: 1H but two isotopes for Cl:35Cl and 37Cl with respective abundances of 75% and 25%. | 1) Mass combinations: only two possible 15+35=50 or 15+37=52; 2) Abundances: for Mass=50: 100%*100%*75% and for Mass=52: 100%*100%*25% so ratio of 3:1 |

Describe the spectrum for a Br2 molecule; two isotopes for Br: 79-Br and 81-Br with equal abundances (50%). | 1) Mass combinations: four possible 79+79=158; 79+81=160; 81+79=160 81+81=162 2) Abundances: for Mass=158: 50%*50% and for Mass=162:50%*50%; and for mass=160: 50%*50% PLUS 50%*50% (remember two combinations) so ratio of 1:2:1 |

Describe the spectrum for a Cl2 molecule; two isotopes for Cl: 35-Cl and 37-Cl with respective abundances 75% to 25%. | 1) Mass combinations: four possible 35+35=70; 35+37=72; 37+35=72; 37+37=74 2) Abundances: for Mass=70: 75%*75% and for Mass=74:25%*25%; and for mass=72: 75%*25% PLUS 25%*75% (remember two combinations) so ratio of 56.25:37.5:6.25 or 9:6:1 |