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Laws of Proportion

Laws of Proportion. Law of Definite Proportion Law of Multiple Proportions. Law o f Definite Proportions: Why Do the Formulas Work?. For formulas to work, chemistry of elements has to follow a pattern. John Dalton came up with the pattern in the early 19 th Century.

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Laws of Proportion

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  1. Laws of Proportion Law of Definite Proportion Law of Multiple Proportions

  2. Law of Definite Proportions: Why Do the Formulas Work? • For formulas to work, chemistry of elements has to follow a pattern. • John Dalton came up with the pattern in the early 19th Century. • The Law of Definite Proportions states that the mass ratios of elements in a particular compound are fixed. • If the ratios are not the same in two samples then they are different compounds.

  3. Importance of Dalton’s Observation • No standard symbols for elements or an organized method of naming compounds or even writing formulae. • Seeing the pattern evident in comparing the measurements of elemental compositions from other researchers, Dalton’s Law of Definite Proportions formalized the relationships observed. • There was no knowledge of how compounds actually formed during that part of the century.

  4. Determining Elemental Proportions • Must analyze the sample of a compound to determine its elemental composition and the mass of each element in the compound. • Example: A sample of a reddish colored material is found to have a mass of 14.650 g. • The sample has 10.246 g of iron and the remainder of the mass of the sample was analyzed as oxygen. • In this case you need to divide the mass of iron by the mass of the sample: 10.246g/14.650=0.6994. • Multiply the result by 100% to get the percent composition: 69.94% Fe and 30.06% O.

  5. Practice ProblemsPercent Composition • 1. Sample Mass: 9.468g; P= 2.1550g; K= 2.7202g; H=0.14025g; O= 4.4526g • 2. Sample mass: 11.7456g; Sn= 4.6058g; N= 2.1738g; O=4.9660g • 3. Sample Mass: 7.3894g; Br= 2.9310g; Cl= 1.3005g; H=0.22184g; O= 1.1738g; C= 1.7623g • 4. Sample Mass: 146.9674g; Cr= 51.996; O=63.9976; P= 30.97376 • 5.

  6. Law of Multiple Proportions: • When two different compounds, as confirmed by their different analyses, are made of the same elements, then, when normalized for one of the elements, another element in the two compounds will be related by a small whole number ratios. • This is obvious when we look a formulae of two different compounds with the same elements: i.e. FeSO4 and Fe2(SO4)3. • In the first, normalized to Iron, the sulfur ratio is 2:3 and the oxygen is also 2:3.

  7. Examples of what will be on the test; I • An analysis of a pair of samples yielded the following results listed below. Are the samples of the same compound? If not what is the ratio of the oxygen in the two samples? • sample A: Cl 3.990801 gO 4.502469 g • sample B: Cl 9.648141 gO 4.354059 g • We will walk through this problem on the board.

  8. Examples of what will be on the test; II • An analysis of a pair of samples yielded the following results listed below. Are the samples of the same compound? If not what is the ratio of the sulfur in the two samples? • sample A: Mo 1.968603 gS 1.973836 g • sample B: Mo4.296019 gS 3.589531 g • An analysis of a pair of samples yielded the following results listed below. Are the samples of the same compound? If not what is the ratio of the iodine in the two samples? • sample A: Tl 5.3542 gI 9.973494 g • sample B: Tl 3.739101 gI 6.964982 g

  9. Reviewing steps for solving • 1. Find the atomic masses of the two elements in the problem. • 2. Normalize the masses of each sample to the mass of the element in the sample you are NOT finding the ratio for: e.g. chlorine in the first example. • 3. Divide the results of each calculation: sample A unknown by sample B unknown. • 4. Determine the small whole number ratio by inspection; multiplying the result of 3 by a factor that gives you a small whole number ratio. • 5. If the result of 3 is one, then the samples are the same.

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