Differenciate between law of conservation of mass and law of constant proportion
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In any given chemical compound, the elements always combine in the same proportion with each other. This is the law of constant composition.
The law of constant composition says that, in any particular chemical compound, all samples of that compound will be made up of the same elements in the same proportion or ratio. For example, any water molecule is always made up of two hydrogen atoms and one oxygen atom in a 2:12:1 ratio. If we look at the relative masses of oxygen and hydrogen in a water molecule, we see that 94%94% of the mass of a water molecule is accounted for by oxygen and the remaining 6%6% is the mass of hydrogen. This mass proportion will be the same for any water molecule.
This does not mean that hydrogen and oxygen always combine in a 2:12:1 ratio to form H2OH2O. Multiple proportions are possible. For example, hydrogen and oxygen may combine in different proportions to form H2O2H2O2 rather than H2OH2O. In H2O2H2O2, the H:OH:O ratio is 1:11:1and the mass ratio of hydrogen to oxygen is 1:161:16. This will be the same for any molecule of hydrogen peroxide.
Law of constant composition
Aim
To investigate the ratio in which compounds combine.
Apparatus
0,10,1 mol⋅dm−3mol·dm−3 silver nitrate (AgNO3AgNO3)
0,10,1 mol⋅dm−3mol·dm−3 sodium chloride (NaClNaCl)
0,10,1 mol⋅dm−3mol·dm−3 lead nitrate (PbNO3PbNO3)
0,10,1 mol⋅dm−3mol·dm−3 sodium iodide (NaINaI)
0,10,1 mol⋅dm−3mol·dm−3 iron (III) chloride (FeCl3FeCl3)
0,10,1 mol⋅dm−3mol·dm−3 sodium hydroxide (NaOHNaOH)
9 large test tubes
3 propettes

Method
Reaction 1: Prepare three test tubes with 55 mLmL, 1010 mLmL and 1515 mLmL of silver nitrate respectively. Using a clean propette add 55 mLmLof sodium chloride to each one and observe what happens.
Reaction 2: Prepare three test tubes with 55 mLmL, 1010 mLmL and 1515 mLmL of lead nitrate respectively. Using a clean propette add 55 mLmLof sodium iodide to each one and observe what happens. Write a balanced equation for this reaction.
Reaction 3: Prepare three test tubes with 55 mLmL, 1010 mLmL and 1515 mLmL of sodium hydroxide respectively. Add 55 mLmL of iron(III) chloride to each one and observe what happens.
Discussion and conclusion
Regardless of the amount of reactants added, the same products, with the same compositions, are formed (i.e. the precipitate observed in the reactions). However, if the reactants are not added in the correct ratios, there will be unreacted reactants that will remain in the final solution, together with the products formed.
Volume relationships in gases (ESADX)
In a chemical reaction between gases, the relative volumes of the gases in the reaction are present in a ratio of small whole numbers if all the gases are at the same temperature and pressure. This relationship is also known as Gay-Lussac's Law.
For example, in the reaction between hydrogen and oxygen to produce water, two volumes of H2H2 react with 1 volume of O2O2 to produce 2 volumes of H2OH2O.
2H2(g)+O2(g)→2H2O (l)2H2(g)+O2(g)→2H2O (l)
In the reaction to produce ammonia, one volume of nitrogen gas reacts with three volumes of hydrogen gas to produce two volumes of ammonia gas.
N2(g)+3H2(g)→2NH3(g)
The law of constant composition says that, in any particular chemical compound, all samples of that compound will be made up of the same elements in the same proportion or ratio. For example, any water molecule is always made up of two hydrogen atoms and one oxygen atom in a 2:12:1 ratio. If we look at the relative masses of oxygen and hydrogen in a water molecule, we see that 94%94% of the mass of a water molecule is accounted for by oxygen and the remaining 6%6% is the mass of hydrogen. This mass proportion will be the same for any water molecule.
This does not mean that hydrogen and oxygen always combine in a 2:12:1 ratio to form H2OH2O. Multiple proportions are possible. For example, hydrogen and oxygen may combine in different proportions to form H2O2H2O2 rather than H2OH2O. In H2O2H2O2, the H:OH:O ratio is 1:11:1and the mass ratio of hydrogen to oxygen is 1:161:16. This will be the same for any molecule of hydrogen peroxide.
Law of constant composition
Aim
To investigate the ratio in which compounds combine.
Apparatus
0,10,1 mol⋅dm−3mol·dm−3 silver nitrate (AgNO3AgNO3)
0,10,1 mol⋅dm−3mol·dm−3 sodium chloride (NaClNaCl)
0,10,1 mol⋅dm−3mol·dm−3 lead nitrate (PbNO3PbNO3)
0,10,1 mol⋅dm−3mol·dm−3 sodium iodide (NaINaI)
0,10,1 mol⋅dm−3mol·dm−3 iron (III) chloride (FeCl3FeCl3)
0,10,1 mol⋅dm−3mol·dm−3 sodium hydroxide (NaOHNaOH)
9 large test tubes
3 propettes

Method
Reaction 1: Prepare three test tubes with 55 mLmL, 1010 mLmL and 1515 mLmL of silver nitrate respectively. Using a clean propette add 55 mLmLof sodium chloride to each one and observe what happens.
Reaction 2: Prepare three test tubes with 55 mLmL, 1010 mLmL and 1515 mLmL of lead nitrate respectively. Using a clean propette add 55 mLmLof sodium iodide to each one and observe what happens. Write a balanced equation for this reaction.
Reaction 3: Prepare three test tubes with 55 mLmL, 1010 mLmL and 1515 mLmL of sodium hydroxide respectively. Add 55 mLmL of iron(III) chloride to each one and observe what happens.
Discussion and conclusion
Regardless of the amount of reactants added, the same products, with the same compositions, are formed (i.e. the precipitate observed in the reactions). However, if the reactants are not added in the correct ratios, there will be unreacted reactants that will remain in the final solution, together with the products formed.
Volume relationships in gases (ESADX)
In a chemical reaction between gases, the relative volumes of the gases in the reaction are present in a ratio of small whole numbers if all the gases are at the same temperature and pressure. This relationship is also known as Gay-Lussac's Law.
For example, in the reaction between hydrogen and oxygen to produce water, two volumes of H2H2 react with 1 volume of O2O2 to produce 2 volumes of H2OH2O.
2H2(g)+O2(g)→2H2O (l)2H2(g)+O2(g)→2H2O (l)
In the reaction to produce ammonia, one volume of nitrogen gas reacts with three volumes of hydrogen gas to produce two volumes of ammonia gas.
N2(g)+3H2(g)→2NH3(g)
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