It states, “mass is neither gained nor lost during a chemical reaction” or that “in a chemical reaction, the total mass of reactants consumed equals the total mass of products formed.”<\/em><\/strong><\/p>\n\n\n\nIt is also known as law of indestructibility of matter. <\/p>\n\n\n\n
If reactants A and B combine to form products C and D, then the sum of the masses of A and B equals the sum of the masses of C and D.<\/p>\n\n\n\n
Exceptions: Nuclear Reaction, Radioactive Disintegration<\/em><\/p>\n\n\n\nLimitation: Not entirely applicable for nuclear reactions in which the mass of the products formed is slightly less than the mass of the reactants due to the conversion of some mass into energy according to the equation E=mc2<\/sup>.<\/p>\n\n\n\n<\/span>Law of definite\/constant proportions<\/strong><\/span><\/h3>\n\n\n\nProposed By<\/strong>: French chemist Joseph Louis Proust in 1799<\/p>\n\n\n\nVerified By<\/strong>: Stats and Richards<\/p>\n\n\n\nIt states, “The same chemical compound always contains the same elements combined together in a definite proportion by mass, regardless of the compound’s origin or mode of formation.”<\/em><\/strong><\/p>\n\n\n\nCarbon and oxygen atoms are combined in the ratio 3:8 by mass in carbon dioxide gas obtained either by heating limestone or by combustion of carbon or hydrocarbon.<\/p>\n\n\n\n
Exception: Isotopic reaction, non-stoichiometric compounds (Berthollide compounds) e.g. Wustite: Fe0.94<\/sub>O or (VOx), x can be any number between 0.6 and 1.3.<\/em><\/p>\n\n\n\nLimitations: This law does not apply if the element involved in compound formation has multiple isotopes. In contrast to the first case, CO2 <\/sub>formed by the combination of C-14 isotope and oxygen contains carbon and oxygen atoms in the ratio of 7:16 by mass. Also, this law does not apply to non-stoichiometric compounds such as wustite, where the iron-oxygen ratio ranges from 0.83:1 to 0.95:1.<\/p>\n\n\n\n<\/span>Law of multiple proportions<\/strong><\/span><\/h3>\n\n\n\nProposed By<\/strong>: John Dalton in 1803<\/p>\n\n\n\nVerified By<\/strong>: Berzelius<\/p>\n\n\n\nIt states, “When two or more different elements combine to form two or more chemical compounds, the mass of one of the elements that combines with a fixed mass of the other bears a simple whole number ratio.”<\/em><\/strong><\/p>\n\n\n\nNitrogen, for example, combines with oxygen to form a series of nitrogen oxides, with the ratio of masses of oxygen that combine with 28 parts by mass of nitrogen being 1:2:3:4:5, demonstrating the law of multiple proportion.<\/p>\n\n\n\n
<\/span>Law of reciprocal proportions<\/strong><\/span><\/h3>\n\n\n\nPropose By<\/strong>: Richter in 1792<\/p>\n\n\n\nVerified By<\/strong>: Stas<\/p>\n\n\n\nIt states, \u201cThe ratio of the masses of two elements which combine separately with a fixed mass of the third element is either same or some simple multiple of the ratio of the masses in which these elements combine with each other\u201d<\/strong><\/em><\/p>\n\n\n\nConsider the elements hydrogen, oxygen, and sulfur as an example. Separately, oxygen and hydrogen react with sulfur to form SO2 <\/sub>and H2<\/sub>S. They also combine to form H2<\/sub>O, as shown in the figure.<\/p>\n\n\n\nLaw of reciprocal proportions<\/strong> (Stoichiometry)<\/figcaption><\/figure>\n\n\n\nIllustration:<\/em><\/strong> <\/p>\n\n\n\n32 parts by mass of sulfur combine with 32 parts by mass of oxygen to form sulfur dioxide (SO2<\/sub>).<\/p>\n\n\n\n2 parts hydrogen by mass combine with 32 parts sulphur by mass to form hydrogen sulphide (H2<\/sub>S).<\/p>\n\n\n\nAs a result, the ratio of hydrogen and oxygen masses that combine with fixed mass, i.e. 32 parts by mass of sulphur, is<\/p>\n\n\n\n
2:32 i.e. 1:16\u2026\u2026\u2026\u2026\u2026.. (i)<\/p>\n\n\n\n
In water (H2<\/sub>O), two parts hydrogen combine with 16 parts oxygen. The mass ratio of hydrogen to oxygen is<\/p>\n\n\n\n2:16 i.e. 1:8\u2026\u2026\u2026\u2026\u2026.. (ii)<\/p>\n\n\n\n
These two ratios are related as 1\/16:1\/8, which is a simple multiple ratio of 1:2.<\/p>\n\n\n\n
<\/span>Law of gaseous volume<\/strong><\/span><\/h3>\n\n\n\nProposed By<\/strong>: Gay Lussac in 1808<\/p>\n\n\n\nIt states, “when gases react together under similar conditions of temperature and pressure, they do so in volumes that bear a simple ratio to one another and the volumes of the products as well if they are also gases.”<\/em><\/strong><\/p>\n\n\n\nFor example, if steam is formed by combining hydrogen and oxygen gas, there is a simple volume ratio.<\/p>\n\n\n\n
2H2<\/sub> (g) + O2<\/sub> \u2192 2H2<\/sub>O (g)<\/strong><\/p>\n\n\n\nThe volume ratio of reactants and products is 2:1:2.<\/p>\n\n\n\n
<\/span>Stoichiometric coefficient<\/strong><\/span><\/h2>\n\n\n\nThe stoichiometric coefficient, also known as the stoichiometric number, is the number of molecules involved in the reaction. Any balanced reaction has an equal number of components on both sides of the equation, as you can see if you take a closer look.<\/p>\n\n\n\n
Stoichiometric coefficients can be either fractions or whole numbers. The coefficients essentially assist us in determining the mole ratio of reactants to products.<\/p>\n\n\n\n
We can deduce from:<\/p>\n\n\n\n
2Na (s) + 2HCl (aq) \u2192<\/strong> 2NaCl (aq) + H2<\/sub> (g) <\/strong><\/p>\n\n\n\n2 moles of HCl will react with 2 moles of Na(s) to form 2 moles of NaCl(aq) and 1 mole of H2<\/sub>(g). If we know how many moles of Na reacted, we can use the ratio of 2 moles of NaCl to 2 moles of Na to calculate how many moles of NaCl were produced, or we can convert to NaCl using the ratio of 1 mole of H2<\/sub> to 2 moles of Na. This is referred to as the coefficient factor. The balanced equation allows you to convert information about one reactant or products change into quantitative data about another reactant or product. Understanding this is critical in order to solve stoichiometric problems.<\/p>\n\n\n\n