Relative Mass

Relative Mass

1. Relative Masses are Masses of Atoms Compared to Carbon - 12

   • The actual mass of an atom is very, very tiny.
   • Don't worry about exactly how tiny for now, but it's far too small to weigh.
   • So, the mass of one atom is compared to the mass of a ddifferent atom.
   • This is its relative mass
   • Relative atomic mass is an average, so it's not usually a whole number. Relative isotopic mass is always a whole number.
     • The relative atomic mass, Ar, is the averge mass of an atom of an element on a scale where an atom of carbon-12 is 12
     • Relative isotopic mass is the mass of an atom of an isotope of an element on a scale where an atom of carbon-12 is 12
   • The relative molecular mass (or relative formula mass), Mr, is the average mass of a molecule or formula unit on a scale where an atom of carbon-12 is 12
     • Relative formula mass is used for compounds that are ionic (or giant covalent, such as SiO2). To find the relative formula mass, just add up the relative atomic masses (Ar) of allthe ions in the formula unilt. (Ar of ion = Ar of atom. The electrons make no difference to the mass.) E.g. Mr(CaF2) 40 + (2 x 19) = 78.
     • To find the relative molecular mass, just ass up the relative atomic mass walues of all the atoms in the molecule, e.g. Mr(C2H6O) = (2 x 12) + (6 x 1) + 16 = 46

2. Relative Masses can be Measured Using a Mass Spectrometer

   • You can use a mass spectormeter to find out loads of stuff. It can tell you the relative atomic mass, relative moleccular mass, relative isotopic abundance and Moleculat structure.
   • There are 4 things that happen when a sample is squited into a so-called time-of-flight mass spwctrometer.
     • 2) Ionisation - the gas particles are bombarded with high-energy electrons to ionise them. Electrons are knocked off the particles, leaving positive ions.
     • 4) Detection - the time taken for the positive ions to reach the detector is measured. This depends on an ion's mass and charge - light, highly charged ions will reach the detectio first, while heavier ions with a smaller charge will take longer. For each sample analysed, a mass spectum is produced.
     • 1) Vaporisation - the sample is turned into gas (waporised) using an electrical heater
     • 3) Acceleration - the positive ions are accelerated by an electric field.
   • A Mass Spectrum
     • The y-axis gives the abundance of ions, often as a percentage.
       • For an element, the height of each peak gives the relative isotopic abundance.
       • If the sample is an element, each line will represent a different isotope of the element.
     • The x-axis units are given as a 'mass/charge' ratio.
       • Since the charge on the ions is mostly 1+, you can often assume the x-axis is simply the relative mass.

3. Ar and Relative Isotopic Abundance can be Worked Out form a Mass Spectrum

   • You need to know how to calculate the relative atomic mass (Ar) of an element from the mass spectum.
   • Here's how to calculate Ar for magnesium, using the mass spectrum below -
     • Step 1: For each peak, read the % relative isotopic abundance from the y-axis and the relative isotopic mass from the x-axis. Multiply them together to get the total mass for each isotope. 79 x 24 = 1896; 10 x 25 = 250; 11 x 26 = 286
     • Step 2: Add up these totals. 1896 + 250 + 286 = 2432
     • Step 3: Divid by 100 (since percentages were used). Ar (Mg) = 2432/100 = 24.32 = 24.3
   • If the relative abundance is not given as a percentage, the total abundance may not add up to 100. In this case, don't panic.
   • Just do steps 1 and 2 as above, but then divide by the total relative abundance instead of 100 - like this:
     • Ar(Ne) = (114 x 20) + (0.2 x 21) + (11.2 x 22) / 114 + 0.2 + 11.2 = 20.18
   • You can also calculate the % relative isotopic abundance from this type of data. E.g., for neon - 20: % relative isotopic abundance = relative abundance/ total relative abundance x 100% = 90.91%

4. Mass Spectometry can be used to Find Out Mr

   • You can also get a mass spwctrum for a molecular sample, such as ethanol (CH3CH2OH)
     • 1) A molecular ion, M+ (g), is formed when the bonbarding electrons remove 1 electrom from the molecule. This gives the peak in the wpectrum with the highest mass (furthest to the right, ignoring isotopes). The mass of M+ gives Mr for the molecule, e.g. CH3CH2OH+ has Mr = 46
     • 2) But it's not that simple - bombarding with electrons makes some molecules break up into fragments. These all show up on the mass spectrum, making a fragmentation pattern.
       • For ethanol, the fragments you get include: CH3+ (Mr = 15), CH3CH2+ (Mr = 29) and CH2OH+ (MR = 31)
     • Fragmentation patterns are actually pretty cool because you can use them to identify molecules and even their structure.

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