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LAW of CONSERVATION of MASS

Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations

study examples carefully3. Law of Conservation of Mass Experiments and Reacting Mass Calculationsstudy examples carefully

Quantitative Chemistry calculations online A demonstration experiment of the Law of Conservation of Mass is described and explained. Help for problem solving in doing law of conservation of mass calculations, using experiment data, making predictions. Practice revision questions on the law of conservation of mass in chemical reactions using the balanced equation. The Law of Conservation of Mass is defined and explained using examples of reacting mass calculations using the law are fully explained with worked out examples using the balanced symbol equation. The method involves reacting masses deduced from the balanced symbol equation. Online practice exam chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB courses.

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study examples carefully3. Law of Conservation of mass calculationsstudy examples carefully

  • 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. See discussion in section 14.2

  • 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 ...
  • top sub-indexSelf-assessment Quizzes

    [com] type in answer click me for QUIZ!for F and H  or  multiple choice click me for QUIZ!for F and H

    OTHER CALCULATION PAGES

    1. What is relative atomic mass?, relative isotopic mass and calculating relative atomic mass

    2. Calculating relative formula/molecular mass of a compound or element molecule

    3. Law of Conservation of Mass and simple reacting mass calculations (this page)

    4. Composition by percentage mass of elements in a compound

    5. Empirical formula and formula mass of a compound from reacting masses (easy start, not using moles)

    6. Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination

    7. Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations (relating reacting masses and formula mass)

    8. Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)

    9. Moles and the molar volume of a gas, Avogadro's Law

    10. Reacting gas volume ratios, Avogadro's Law and Gay-Lussac's Law (ratio of gaseous reactants-products)

    11. Molarity, volumes and solution concentrations (and diagrams of apparatus)

    12. How to do acid-alkali titration calculations, diagrams of apparatus, details of procedures

    13. Electrolysis products calculations (negative cathode and positive anode products)

    14. Other calculations e.g. % purity, % percentage & theoretical yield, dilution of solutions (and diagrams of apparatus), water of crystallisation, quantity of reactants required, atom economy

    15. Energy transfers in physical/chemical changes, exothermic/endothermic reactions

    16. Gas calculations involving PVT relationships, Boyle's and Charles Laws

    17. Radioactivity & half-life calculations including dating materials

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