LAW of CONSERVATION of MASS

Doc Brown's Chemistry - GCSE 9-1, IGCSE, O Level (and very basic GCE A level) O Level  Online Chemical Calculations

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.

These are 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. These revision notes and practice questions on the law of conservation of mass chemical calculations and worked examples should prove useful for the new AQA, Edexcel and OCR Gateway and 21st Century GCSE (9–1) chemistry science courses.

Spotted any careless error? EMAIL query ? comment or request a type of GCSE calculation not covered?

QUIZ on Law of Conservation of Mass and simple reacting mass calculations

Reminder! What is chemistry?

Chemistry is basically taking 'stuff' (the reactants) and changing it and separating out 'different stuff' (the products).

In a chemical change, the atoms of the reactants are rearranged to give the products.

The atoms remain the same elements BUT are arranged or bonded in a different way in the products.

In the 'picture equation' above, just look at how the copper, carbon, oxygen, hydrogen and sulfur atoms are changed in their arrangement from reactants on the left, to products to the right of the arrow - indicating the direction of chemical change.

So, what about the relative mass of all the products compared to the total mass of the original reactants?

Read on .... !!!

• 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 whatever the physical state of the reactants and products.
• The law of conservation of mass can also be stated as "no atoms can be lost or made in a chemical reaction", which is why the total mass of products must equal the total mass of reactants you started with.
• 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
  • One consequence of the law of conservation of mass is that In a balanced chemical symbol equation, the total of relative formula masses of the reactants is equal to the total relative formula masses of the products.
  • You can see this in the examples worked out for you  ...
  • ... so, this page just explains how to do simple reacting mass calculations based on the reaction equation and applying the Law of Conservation of Mass,
  • but first, by at least one clearly observed experiment, that the Law of Conservation of Mass holds good, even in the humble school or college laboratory! see diagram below and read on. ...
  • See also Section 5. which shows how to use this law to get to a compound's formula too

... 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 and represents an enclosed system, where nothing can escape !

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

CuSO_4(aq) +  2NaOH_(aq)  ===>  Cu(OH)_2(s)  +  Na₂SO_4(aq)

Teacher note

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

You can do exactly the same sort of experiment using the same apparatus and procedure as above using lead nitrate solution and potassium iodide solution. It doesn't matter which solution is in the little test tube or the conical flask.

Both are colourless solutions BUT on mixing you get a bright yellow precipitate of lead iodide, an obvious chemical change has taken place.

And, again, you find the total mass at the start (unmixed solutions) equals the total mass at the end (including the residual solution and precipitate).

The chemistry of this demonstration is ..

lead nitrate  +  potassium iodide  ===>  lead iodide  +  potassium nitrate

Pb(NO₃)₂(aq)  + 2KI(aq)  ===>  PbI₂(s)  +  2KNO₃(aq)

Note: (i) Here I've included the state symbols, (aq) meaning an aqueous solution (solvent water), and (s) to denote the solid precipitate formed.

(ii) The two 2s are needed to balance the equation, a consequence of the law of conservation of mass and the rules on balancing chemical equations.

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Experiments in which a mass change is observed

- where there doesn't seem to be conservation of mass, we need an explanation!

Reactions that do not go to completion might give false results and if the reaction involves a reacting gas OR a gaseous product it is difficult to make accurate measurements to confirm the validity of the law of conservation of mass.

Three typical situations you will encounter:

(i) If a solid reacts with a gas to give a solid product you appear to gain mass e.g. heating a metal in air to form an oxide, there is no direct way to measure the mass of oxygen used from the air outside of the crucible to show the law of conservation of mass is obeyed. You can only measure directly the mass of one of the reactants, but not the other.

(ii) If a solid decomposes on strong heating to give a solid and a gas e.g. decomposition of a metal carbonate to a metal oxide and carbon dioxide, there is no direct way to measure the mass of carbon dioxide lost to the air outside of the crucible/test tube to show the law of conservation of mass is obeyed. This means you can only measure directly the mass of one of the products, but not the other.

(iii) If solid reacts with an acid to form a gas e.g. a metal + acid ==> hydrogen or a carbonate + acid ==> carbon dioxide, there is no direct way to measure the mass of the gaseous products lost to the air outside of the conical flask to show the law of conservation of mass is obeyed. All you can measure is the mass of the products that were not gases.

BUT, as you will see, theoretical calculations based on the law of conservation of mass can get round the situation!

Some reactions may appear to involve a change in mass as measured with limited school laboratory apparatus. However, these can usually be explained because a reactant or product is a gas and its mass not been taken into account.

For example: when a metal reacts with oxygen the mass of the oxide produced is greater than the mass of the metal, so when you heat magnesium ribbon in a crucible there is gain in mass because the 'weightless' oxygen in air has combined with the magnesium to form magnesium oxide. It is impossible to directly measure the mass of the oxygen used from the air.

2Mg(s)  + O₂(g)  ====>  2MgO(s)   mass gained

BUT, using the law of conservation of mass, you can do a reacting mass calculation from the mass of magnesium oxide formed to find out the mass of oxygen used in making the magnesium oxide.

Using the atomic mass: Mg = 24 and O = 16, you can write the equation in terms of mass ...

(2 x 24)  + (2 x 16)  ====> 2 x (24 + 16), both sides of the equation arrow should add up to 80, check it out!

In thermal decompositions of metal carbonates, the carbon dioxide is produced and escapes into the atmosphere leaving the metal oxide as the only solid product. Again, it is impossible to directly measure the carbon dioxide evolved from the decomposing carbonate. However, you can clearly observe the chemical change because of the colour changes.

green copper carbonate  ====>  black copper oxide  +  colourless carbon dioxide.

CuCO₃(s)  ====>  CuO(s)  +  CO₂(g)   lost mass

BUT, again, using the law of conservation of mass, you can do a reacting mass calculation from the residual mass of the oxide to find out the mass of carbon dioxide produced by the thermal decomposition of the carbonate e.g. copper carbonate decomposing heating to give solid copper oxide and gaseous carbon dioxide.

Using atomic masses: Cu = 63.5,  C = 12  and O = 16 to show mass is conserved writing the equation in masses

{63.5 + 12 + (3 x 16)}  ====> (63.5 + 16)  +  {12 + (2 x 16)}, both sides of the equation add up to 123.5

When an acid reacts with a metal, hydrogen gas is produced, or reacting an acid with a carbonate produces carbon dioxide gas. The gas escapes into the air. You can follow the mass loss by carrying out the reaction in a flask placed on an electronic balance. You can even use the mass loss to see how fast the reaction is going e.g.

hydrogen from a metal-acid reaction

carbon dioxide from a carbonate-acid reaction

illustration of a non-enclosed system

In the case above, carbon dioxide is produced when limestone reacts with acid forming carbon dioxide gas which escapes through the cotton wool plug. The other products of the reaction remain in the flask. If you could weigh the CO₂ formed, all would add up to comply with the law of conservation of mass!

If a gas escapes from a non-enclosed reaction vessel you cannot observe the law of conservation of mass, BUT, since we have a verified scientific law (of mass conservation) we can calculate mass changes that we cannot observe directly.

No matter what happens, the law of conservation of mass must be obeyed!

You need to be able to explain any observed changes in mass in non-enclosed systems during a chemical reaction given the balanced symbol equation for the reaction and explain these changes in terms of the particle model. If you carry out an experiment that produces a gas in a non-enclosed system (i.e. the gas can escape), you will observe a mass loss, BUT, if you could somehow weigh the gas, you would find that the total mass of reactants and products had remained constant. So, even in reactions producing a gas, the law of conservation still holds good. The same arguments applies to when a gas is a reactant producing a solid product.

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 reacting masses i.e. as ratios to enable calculations to be done with any given 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 of % yield

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Examples of 'atom counting' to illustrate the law of conservation of mass

(these are from my page on how to balance chemical equations)

Any correctly balanced equation, especially in diagrammatic form, illustrates the law of conservation of mass, ie the number of atoms for each element MUST be the same on both sides of the equation. No atoms lost or gained and neither do atoms change their atomic mass in a reaction.

ATOMS at the START = ATOMS at the END (atoms conserved, just arranged differently!) and

TOTAL MASS REACTANTS = TOTAL MASS of PRODUCTS (the law of conservation of mass)

You also need to be able to read chemical formula and balance chemical equations, at least appreciate why an equation is balanced - which after all is a symbolic or diagrammatic representation of the Law of Conservation of Mass.

The 'Law of Conservation of Mass' means you can do theoretical calculations on the relative amounts of reactants and products involved in a chemical reaction, as long as you have the correct formulae and a correctly balanced chemical equation.

Atomic masses are quoted and listed on a periodic table at the bottom of the page. Incidentally, the examples are worked out in terms of atomic masses, but in real calculations you may use g, kg or tonnes, as long as you use the same mass units for each mass value involved.

AND you must be able to working out formula masses

You will follow the arguments for the more complex examples if you can already balance awkward equations!

Atom counting and balancing the reactant mass units and product mass units from the equation ratios

I am assuming you can work out the relative formula mass of an element or compound.

(1)  iron  +  sulfur  ==>  iron sulfide

Fe  +  S  ====>  FeS

Atomic masses: Fe = 56, S = 32

One atom of each element on each side of the equation

Law of conservation of mass balance:  56  +  32  ====>  88,  (calculation check 56 + 32 = 88)

Important note on units:

I haven't specified the units of mass in these 7 examples BUT ...

... it doesn't matter whether you are working in mass units of g, kg or tonnes!

as long as you 'work' the final ratio numbers from the equation in the SAME mass units.

(2)  sodium hydroxide  +  hydrochloric acid  ===> sodium chloride +  water

 

NaOH  +  HCl  ====>  NaCl  + H₂O

Atomic masses: Na = 23, O = 16, H = 1, Cl = 35.5

one atom of Na, one atom of oxygen, two atoms of hydrogen and one atom of chlorine on both sides of the equation

Law of conservation of mass balance: 40  +  36.5  ====>  58.5  +  18,  (both sides equal 76.5 mass units)

Note the subscript 2 after the H in water means two atoms of that element.

(3)  magnesium  +  hydrochloric acid  ====>  magnesium chloride  +  hydrogen

Mg + 2HCl ====> MgCl₂ + H₂

Atomic masses: Mg = 24, H = 1, Cl = 35.5

one atom of Mg, 2 atoms of H and 2 atoms of Cl on both sides of the equation

Law of conservation of mass balance: {24 + (2 x (1 + 35.5)}  ===> {24 + (2 x 35.5)} + (2 x 1),   (both sides equal 97)

Note the subscript 2 after the Cl in magnesium chloride or the 2 after the H in the hydrogen molecule, means two atoms of that element.

The 2 before the HCl doubles the number of hydrochloric acid molecules.

(4)  methane  +  oxygen  ====>  carbon dioxide  +  water

CH₄  +  2O₂  ====>  CO₂  +  2H₂O

Atomic masses: C = 12, H = 1, O = 16

One atom of carbon, 4 atoms of hydrogen and four atoms of oxygen on both sides of the equation

Law of conservation of mass balance: 16 + (2 x 32)  ===> 44 + (2 x 18), both sides of equation equal 80 mass units

Note the 2 before the O₂ and H₂O doubles the number of these molecules to balance the equation.

The subscript 4 in methane means 4 atoms of hydrogen in the methane molecule.

(5)  copper carbonate  +  sulfuric acid  ===>  copper sulfate  +  water  + carbon dioxide

CuCO₃ + H₂SO₄ ====> CuSO₄ + H₂O + CO₂

Atomic masses: Cu = 63.5, C = 12, O = 16, H = 1, S = 32

Law of conservation of mass balance: one atom of copper, one atom of carbon, one atom of sulfur and seven atoms of oxygen on both sides of the equation, no atoms lost or gained, no mass lost or gained!

Law of conservation of mass balance: I'm assuming you can work out formula masses by now.

(123.5) + (98)  ====> (159.5) + (18) + (44),  both sides of equation equal 221.5 mass units in total

Reminders on reading a formula:

The formula H₂SO₄ means 2 atoms of H, 1 atom of S and four atoms of O

The formula CuCO₃ means 1 atom of Cu, 1 atom of C and 3 atoms of O.

(6)  magnesium hydroxide + nitric acid  ====> magnesium nitrate + water

Mg(OH)₂  +  2HNO₃  ====>  Mg(NO₃)₂  +  2H₂O

Atomic masses: Mg = 24, O = 16, H = 1. N = 14

Law of conservation of mass balance: 1 atom of Mg, 8 atoms of O, 4 atoms of H and 2 atoms of N on both sides of the equation, no atoms lost or gained, no mass lost or gained!

Work this one out for yourself using the formula masses for practice.

Note that here the subscript 2 after the (NO₃) in the magnesium nitrate means everything in the brackets is doubled.

(7)  aluminium oxide  + sulfuric acid  ====>  aluminium sulfate  +  water

Al₂O₃  +  3H₂SO₄  ====>  Al₂(SO₄)₃  +  3H₂O

Atomic masses: Al = 27, O = 16, H = 1, S = 32

Law of conservation of mass balance: 2 atoms of Al, 15 atoms of O, 6 atoms of H and 3 atoms of S on both sides of the equation, no atoms lost or gained, no mass lost or gained!

Work this one out for yourself using the formula masses for practice.

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Examples of using the 'Law of Conservation of Mass' in reacting mass calculations.

By converting the atoms or formulae into atomic masses or formula/molecular masses you can then use the law of conservation of mass to reacting mass calculations. (More reacting mass calculations in section 6.)

• 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 = 80, so 80 mass units of reactants = producing 80 mass units of products.
    • You can then state things like  ..
      • 48 g of Mg gives in excess oxygen, 80 g of MgO
      • or 24 tonnes of Mg produces 40 tonnes of MgO
    • 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.
    • You are quite simply 'working' the equation in terms of mass units, but taking into account the balancing numbers in the symbol equation.
    • -
• Law of conservation of mass calculation Example 3.2
  • sodium  +  chlorine  ===> sodium chloride
  • 2Na  +  Cl₂  ===>  2NaCl   (atomic masses required: Na=23 and Cl=35.5)
  • If 23 g of sodium is burned in excess chlorine gas, theoretically, what mass of sodium chloride is produced?
  • The equation in terms of mass units is ...
  • (2 x 23) + (2 x 35.5)  ==> 2(23 +35.5) = (2 x 58.5)
  • 46 + 71 = 117
  • therefore, dividing by two: 23 + 35.5  ==> 58.5
  • So 58.5 g of sodium chloride is formed.
  • When you get more experienced, you can just pick out the formula mass ratio required
  • e.g 1 Na  ==>  1 NaCl giving a working ratio of 23 : 58.5
  • -
• Law of conservation of mass calculation Example 3.3

  • CuCO₃ + H₂SO₄ ====> CuSO₄ + H₂O + CO₂

  • How much copper sulfate can you make from 494 kg of copper carbonate
  • The mass ratios are ...
  • (123.5) + (98)  ====> (159.5) + (18) + (44),  both sides of equation equal 221.5 mass units in total
  • The crucial ratio is 1 CuCO₃  ==> 1 CuSO₄
  • In mass units:  123.5  ===>  159.5
  • 494/123.5 = 4, so we need to scale up by 4,
  • so 494 g CuCO₃ will produce 4 x 159.5 = 638 kg of CuSO₄
  • -
• Law of conservation of mass calculation Example 3.4
  • iron + sulphur ==> iron sulphide (see the diagram at the top of the page!)
  • Fe + S ==> FeS (atomic masses: Fe = 56, S = 32)
  • If 59 g of iron is heated with 32 g of sulphur to form iron sulphide, how much iron is left unreacted? (assuming all the sulphur reacted)
  • From the atomic masses, 56 g of Fe combines with 32 g of S to give 88g FeS.
  • This means 59 - 56 = 3 g Fe unreacted.
  • This is an example of a 'limiting reactant' calculation - where the reactant NOT in excess limits the maximum amount of product.
  • More on how much of a reactant is needed? limiting reactant
  • -
• Law of conservation of mass calculation Example 3.5
  • 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 + (3 x 16)}  ===> (40 + 16) + {12 + (2 x 16)}
  • 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.
  • -
• For more complicated examples and more practice of calculations based on reacting masses in accordance with the Law of Conservation of Mass ...
• See Reacting mass ratio calculations of reactants & products from equations (NOT using moles)

Self-assessment Quiz on the 'Law of Conservation of Mass' and simple reacting mass calculations

QUIZ on Law of Conservation of Mass and simple reacting mass calculations

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cross-multiplying - apparently, according to maths departments, the naughty way' to solve ratios!

As pupil in the late 1950s and (very) early1960s I was taught to solve ratios by cross-multiplying, wrote learning!

Suppose you have the ratio situation of A : B and C : D as in the reacting mass ratio questions on this page.

You can also express these ratios as

A		C
–	=	–
B		D

Therefore, logically, by cross-multiplying you get A x D = B x C

and rearranging, as you do in simple algebra you get the following relationship by dividing through by A, B, C or D appropriately

A = B x C / D,    B = A x D / C,   C = A x D / B  and  D = B x C / A

and if you don't believe me, just put some numbers in e.g 2 : 5 for A : B and 6 : 15  for C : D

2/5 = 6/15 and 2 x 15 = 5 x 6

I find its by far the quickest general route to solving two ratios that match, but its frowned on!

It does actually amount to the same as the methods described above, personally, I just find it quicker!

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Above is typical periodic table used in GCSE science-chemistry specifications in doing reacting mass and conservation of mass chemical calculations, and I've 'usually' used these values in my exemplar calculations to cover most syllabuses

Doc Brown's School Science Website

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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

   • Reacting masses, concentration of solution and volumetric titration calculations (NOT using moles)

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

    • 14.1 % purity of a product 14.2a % reaction yield 14.2b atom economy 14.3 dilution of solutions

    • 14.4 water of crystallisation calculation  14.5 how much of a reactant is needed? limiting reactant

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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QUIZ on Law of Conservation of Mass and simple reacting mass calculations

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