Thursday, 17 September 2015

The Law of Conservation of Mass


 
What is the Law of Conservation of Mass?
 
  • When elements and compounds react to form new products, mass cannot be lost or gained.

  • "The Law of Conservation of Mass" definition states that mass cannot be created or destroyed, but changed into different forms.

  • So, in a chemical change, the total mass of reactants must equal the total mass of products.
  • By using this law, together with atomic and formula masses, you can calculate the quantities of reactants and products involved in a reaction and the simplest formula of a compound





  • ... before tackling the first calculations based on the Law of Conservation of mass, its worth describing a simple experiment to demonstrate the validity of the law. The experiment is illustrated in the diagram above. You prepare solutions of copper sulfate (blue) and sodium hydroxide (colourless, light grey in diagram!). The most impressive way to demonstrate this is to use a sealed system on an accurate electronic one pan balance. You can use 50 cm3 of 1 molar copper sulfate solution and pour into conical flask. The concentrated sodium hydroxide solution is suspended by a string in a suitable container - small test tube or weighing/sample bottle. The whole lot is weighed (fictitiously 67.25g) with the rubber bung on sealing the 'system'. Then, releasing the bung and string, the sodium hydroxide container is lowered into the copper sulfate solution and shaken gently to thoroughly mix the reactants. The reaction is immediate and a dark blue precipitate of copper hydroxide is formed and the solution eventually turns colourless because only colourless sodium sulfate is left in solution. The recorded mass will still be 67.25g showing that no mass was created or destroyed in the chemical reaction, though to observe the law in action, you must do the experiment in a sealed system where nothing can get in or get out i.e. no atoms have been gained or lost.

    The equation for this reaction is ...

    copper sulfate + sodium hydroxide ==> copper hydroxide + sodium sulfate

    CuSO4 +  2NaOH  ===>  Cu(OH)2  +  Na2SO4

    Teacher note

    50 cm3 of 1 molar copper sulfate = 1.0 x 50 / 1000 = 0.05 mol CuSO4, Mr(NaOH) = 40, you need 2x 0.05 = 0.10 mol NaOH,

    which equals 0.10 x 40 = 4.0g NaOH pellets dissolved in the minimum volume of water, 4.1g should complete the precipitation.

     

    NOTE that in calculations ...

    (1) the symbol equation must be correctly balanced to get the right answer!

    (2) You convert all the formula in the equations into their formula masses AND take into account any balancing numbers to get the true theoretical reactiing masses.

    (2) There are good reasons why, when doing a real chemical preparation-reaction to make a substance you will not get 100% of what you theoretically calculate.

    • Law of conservation of mass calculation Example 3.1
      • Magnesium + Oxygen ==> Magnesium oxide
      • 2Mg + O2 ==> 2MgO (atomic masses required: Mg=24 and O=16)
      • think of the ==> as an = sign, so the mass changes in the reaction are:
      • (2 x 24) + (2 x 16) = 2 x (24 + 16)
      • 48 + 32 = 2 x 40 and so 80 mass units of reactants = produces 80 mass units of products.
        • You can work with any mass units such as g, kg or tonne (1 tonne = 1000 kg), as long as you use the same units for all the masses involved.
    • Law of conservation of mass calculation Example 3.2
      • iron + sulphur ==> iron sulphide (see the diagram at the top of the page!)
      • Fe + S ==> FeS (atomic masses: Fe = 56, S = 32)
      • If 59g of iron is heated with 32g of sulphur to form iron sulphide, how much iron is left unreacted? (assuming all the sulphur reacted)
      • From the atomic masses, 56g of Fe combines with 32g of S to give 88g FeS.
      • This means 59 - 56 = 3g Fe unreacted.
    • Law of conservation of mass calculation Example 3.3
      • When limestone (calcium carbonate) is strongly heated, it undergoes thermal decomposition to form lime (calcium oxide) and carbon dioxide gas.
      • CaCO3 ==> CaO + CO2 (relative atomic masses: Ca = 40, C = 12 and O = 16)
      • Calculate the mass of calcium oxide and the mass of carbon dioxide formed by decomposing 50 tonnes of calcium carbonate.
      • (40 + 12 + 3x16) ==> (40 + 16) + (12 + 2x16)
      • 100 ==> 56 + 44
      • scaling down by a factor of two
      • gives
      • 50 ==> 28 + 22
      • so decomposing 50 tonnes of limestone produces 28 tonnes of lime and 22 tonnes of carbon dioxide gas.
    • Example 3.4:
    • For more complicated examples and more practice of calculations based on reacting masses in accordance with the Law of Conservation of Mass ...
     
    credit to : doc brown chemistry as level

    DEFINING & CALCULATING RELATIVE ATOMIC MASS part 1 (#docbrown)

      
    Bonjour!
     
    Assalamualaikum wr wb
     
     
    Dear My As Level students,
     
    I wish you have a great term break as well as me ...Lol
     
    Kindly find review from doc brown on this post... just to remind you that we will have review test after term break over...see yaa
     
    Waalaikumsalam wr wb
     
     
     
        DEFINING& CALCULATING RELATIVE ATOMIC MASS
     
     
    Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations
     
     Explaining and calculating relative atomic mass RAM or Ar and relative isotopic mass
     
    Quantitative Chemistry calculations online Help for problem solving in relative atomic mass calculations. Definitions of relative atomic mass and relative isotopic mass (A level students only) Practice revision questions on working out relative atomic mass from isotopic composition (% isotopes, A level students will learn about very accurate mass spectrometer data). What is relative atomic mass? How do you calculate the relative atomic mass of an element. What is the standard mass unit? Relative atomic mass is explained below, with reference to the carbon-12 atomic mass scale and the relevance of isotopes and 'u' the unified atomic mass unit is explained. Detailed examples of the method of how to calculate relative atomic mass from the isotopic composition are fully explained with reference to the definition of the relative atomic mass of a compound. For A level students, how to define and use relative isotopic masses to calculate relative atomic mass. These notes on defining, explaining and calculating relative atomic mass and defining relative isotopic mass are designed to meet the highest standards of knowledge and understanding required for students/pupils doing GCSE chemistry, IGCSE chemistry, O Level chemistry, KS4 science courses and A Level chemistry courses.
     
    1. Explaining and how to calculate the relative atomic mass RAM or Ar of an elementstudy examples carefully

     
    How to calculate relative atomic mass
     
    Introduction
     
     
    Every atom has its own unique relative atomic mass (RAM) based on a standard comparison or relative scale e.g. it has been based on hydrogen H = 1 amu and oxygen O = 16 amu in the past (amu = relative atomic mass unit).
    The relative atomic mass scale is now based on an isotope of carbon, namely, carbon-12, nuclide symbol , which is given the value of 12.0000 amu.
    The unit 'amu' is now being replaced by a lower case u, where u is the symbol for the unified atomic mass unit.
    1. Therefore one atom of carbon, isotopic mass 12, equals 12 u, or,
    2. 1 u = 1/12th the mass of one atom of the carbon-12 isotope.
    Note that for the standard nuclide notation, http://www.docbrown.info/page04/4_73calcs/ns55.gif, the top left number is the mass number (12) and the bottom left number is the atomic/proton number (6).
    In other words the relative atomic mass of an element is now based on the arbitrary value of the carbon-12 isotope being assigned a mass of 12.0000 by international agreement!
    Examples are shown in the Periodic Table diagram above.
    Note
    1. (i) Because of the presence of neutrons in the nucleus, the relative atomic mass is usually at least double the atomic/proton number because there are usually more neutrons than protons in the nucleus (mass proton = 1, neutron = 1). Just scan the periodic table above and examine the pairs of numbers.
      1. You should also notice that generally speaking the numerical difference between the atomic/proton number and the relative atomic mass tends to increase with increasing atomic number. This has consequences for nuclear stability.
    2. (ii) For many calculation purposes, relative atomic masses are usually quoted and used at this academic level to zero or one decimal place eg.
      1. e.g. hydrogen H = 1.0 or ~1, calcium Ca= 40.0 or ~40, chlorine Cl = 35.5, copper Cu = 63.6 or ~64, silver Ag 107.9 or ~108 etc.
    3. At A level, values of relative atomic masses may be quoted to one or two decimal places.
      1. Many atomic masses are known to an accuracy of four decimal places, but for some elements, isotopic composition varies depending on the mineralogical source, so four decimal places isn't necessarily more accurate!
    In using the symbol Ar for RAM, you should bear in mind that the letter A on its own usually means the mass number of a particular isotope and amu is the acronym shorthand for atomic mass units.
    However there are complications due to isotopes and so very accurate atomic masses are never whole integer numbers.
    Isotopes are atoms of the same element with different masses due to different numbers of neutrons. The very accurate relative atomic mass scale is based on a specific isotope of carbon, carbon-12, 12C = 12.0000 units exactly, for most purposes C = 12 is used for simplicity.
    For example http://www.docbrown.info/page04/4_73calcs/Image416.gifhydrogen-1, http://www.docbrown.info/page04/4_73calcs/Image417.gifhydrogen-2, and http://www.docbrown.info/page04/4_73calcs/Image418.gifhydrogen-3, are the nuclide notation for the three isotopes of hydrogen, though the vast majority of hydrogen atoms have a mass of 1. When their accurate isotopic masses, and their % abundance are taken into account the average accurate relative mass for hydrogen = 1.008, but for most purposes H = 1 is good enough!
    The strict definition of relative atomic mass (Ar) is that it equals the average mass of all the isotopic atoms present in the element compared to 1/12th the mass of a carbon-12 atom (relative isotopic mass of 12.0000).
    So, in calculating relative atomic mass you must take into account the different isotopic masses of the same elements, but also their % abundance in the element.
    Therefore you need to know the percentage (%) of each isotope of an element in order to accurately calculate the element's relative atomic mass.
    For approximate calculations of relative atomic mass you can just use the mass numbers of the isotopes, which are obviously all integers ('whole numbers'!) e.g. in the two calculations below.
    To the nearest whole number, isotopic mass = mass number for a specific isotope.
     
     

    Examples of relative atomic mass calculations for GCSE/IGCSE/AS level students
    How do I calculate relative atomic mass?
    • Example 1.1 Calculating the relative atomic mass of bromine http://www.docbrown.info/page04/4_71atom/Br_79.gifand http://www.docbrown.info/page04/4_71atom/Br_81.gif
      • bromine consists of two isotopes, 50% 79Br and 50% 81Br, calculate the Ar of bromine from the mass numbers (top left numbers).
      • Ar = [ (50 x 79) + (50 x 81) ] /100 = 80
      • So the relative atomic mass of bromine is 80 or RAM or Ar(Br) = 80
      • Note the full working shown. Yes, ok, you can do it in your head BUT many students ignore the %'s and just average all the isotopic masses (mass numbers) given, in this case bromine-79 and bromine-81.
      • The element bromine is the only case I know where averaging the isotopic masses actually works! so beware!
      Example 1.2 Calculating the relative atomic mass of chlorine
      • chlorine consists of two isotopes, 75% chlorine-35 and 25% chlorine-37, so using these two mass numbers ...
      • ... think of the data based on 100 atoms, so 75 have a mass of 35 and 25 atoms have a mass of 37.
      • The average mass = [ (75 x 35) + (25 x 37) ] / 100 = 35.5
      • So the relative atomic mass of chlorine is 35.5 or RAM or Ar(Cl) = 35.5
      • Note: 35Cl and 37Cl are the most common isotopes of chlorine, but, there are tiny percentages of other chlorine isotopes which are usually ignored at GCSE/IGCSE and Advanced GCE AS/A2 A level.
      Example 1.3: 
    Examples for Advanced Level Chemistry students only
    How to calculate relative atomic mass with accurate relative isotopic masses
    Using data from modern very accurate mass spectrometers
     (a) Accurate calculation of relative atomic mass (need to know and define what relative isotopic mass is)
    Relative isotopic mass is defined as the accurate mass of a single isotope of an element compared to 1/12th the mass of a carbon-12 atom e.g. the accurate relative isotopic mass of the cobalt-5 http://www.docbrown.info/page04/4_73calcs/Image435.gifis 58.9332
    This definition of relative isotopic mass is a completely different from the definition of relative atomic mass, except both are based on the same international standard of atomic mass i.e. 1 unit (1 u) = 1/12th the mass of a carbon-12 isotope (12C).
    If we were to redo the calculation of the relative atomic mass of chlorine (example 1.1 above), which is quite adequate for GCSE purposes (and maybe A level too), but more accurately at A level, we might do ....
    chlorine is 75.77% 35Cl of isotopic mass 34.9689 and 24.23% 37Cl of isotopic mass 36.9658
    so Ar(Cl) = [(75.77 x 34.9689) + (24.23 x 36.9658)] / 100
    = 35.4527 (but 35.5 is usually ok in calculations pre-university!)
    See also (c) doc bMass Spectrometer and isotope analysis on the GCSE-AS(basic) Atomic Structure Notes, with further RAM calculations.
    (b) Calculations of % composition of isotopes
    It is possible to do the reverse of a relative atomic mass calculation if you know the Ar and which isotopes are present.
    It involves a little bit of arithmetical algebra.
    The Ar of boron is 10.81 and consists of only two isotopes, boron-10 and boron-11
    The relative atomic mass of boron was obtained accurately in the past from chemical analysis of reacting masses but now mass spectrometers can sort out all of the isotopes present and their relative abundance.
    If you let X = % of boron 10, then 100-X is equal to % of boron-11
    Therefore Ar(B) = (X x 10) + [(100-X) x 11)] / 100 = 10.81
    so, 10X -11X +1100 =100 x 10.81
    -X + 1100 = 1081, 1100 - 1081 = X (change sides change sign!)
    therefore X = 19
    so naturally occurring boron consists of 19% 10B and 81% 11B
    (the data books actually quote 18.7 and 81.3, but we didn't use the very accurate relative isotopic masses mentione above!)

     
    credit to Docbrown