GCSE & A level chemistry calculations: Defining-using the 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
3. Law of Conservation of Mass Experiments and Simple Reacting Mass 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.
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3. Law of Conservation of 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 .... !!!
... 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 !
The equation for this reaction is ...
copper sulfate + sodium hydroxide ==> copper hydroxide + sodium sulfate
CuSO4(aq) + 2NaOH(aq) ===> Cu(OH)2(s) + Na2SO4(aq)
50 cm3 of 1 molar copper sulfate = 1.0 x 50 / 1000 = 0.05 mol CuSO4, Mr(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.
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:
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.
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.
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.
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.
(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
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
(2) sodium hydroxide + hydrochloric acid ===> sodium chloride + water
(3) magnesium + hydrochloric acid ====> magnesium chloride + hydrogen
(4) methane + oxygen ====> carbon dioxide + water
(5) copper carbonate + sulfuric acid ===> copper sulfate + water + carbon dioxide
(6) magnesium hydroxide + nitric acid ====> magnesium nitrate + water
(7) aluminium oxide + sulfuric acid ====> aluminium sulfate + water
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.)
Self-assessment Quiz on the 'Law of Conservation of Mass' and simple reacting mass calculations
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
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!
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
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