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Chemistry 11 2010-2011

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There are some compounds that we know have elements in fixed mass proportions. Water H 2 O 2 g : 16 g Carbon dioxide CO 2 12 g : 32 g

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The elements in a given chemical compound are always present in the same proportions by mass.

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The mass of an element in a compound, expressed as a percent of the total mass of the compound. Note: The Law of Definite proportions does not imply that elements in compounds are always present in the same relative amount.

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% = Part__ Whole What percent of 92 is 42? In other words, the Part “42” of the whole “92” is what percent? 42 x 100 = 45.7 % 92

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% Mass of Element = Mass of Element x 100% Mass of Compound Example 1: CO 2 C = 12.00 g = 12.00 g O = 2 x 16.00 g = + 32.00 g_ 44.00 g = Molar Mass % Mass of C = 12.00 g_ x 100 = 27.3% 44.00 g % Mass of O = 32.00 g_ x 100 = 72.7% 44.00 g

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% Mass of Element = Mass of Element x 100% Mass of Compound Example 2: C 6 H 12 O 6 C = 6 x 12.00 g = 72.00 g H = 12 x 1.01 g = 12.12 g O = 6 x 16.00 g = + 96.00 g_ 180.12 g = Molar Mass

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% Mass of Element = Mass of Element x 100% Mass of Compound Ex. C 6 H 12 O 6 % Mass of C = 72.00 g_ x 100 = 40.0% 180.12 g % Mass of H = 12.12 g_ x 100 = 6.7% 180.12 g % Mass of O = 96.00 g_ x 100 = 53.3% 180.12 g

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Example 3: A sample of a compound has a mass of 48.72 g. The sample is found to contain 32.69 g of zinc and 16.03 g of sulfur. What is the percentage composition of the compound?

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Page 82 Question’s 1, 2, 3, & 4.

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1: In a homogeneous sample, the size of the sample does NOT matter. 2: Assume one mole of compound. Therefore, you can calculate % composition using the molar mass and chemical formula.

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Find the % composition of ZnO.

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Page 85 Question’s 5, 6, 7, & 8.

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In chemistry, empirical means “by experiment”. Thus, an empirical formula is a formula devised from experimental data (rather than theory).

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When an unknown compound is synthesized in the laboratory, an elemental analysis is performed to determine its percent composition. This gives the percent (by mass) of each element of the compound. From this data, the empirical formula of the compound can be determined.

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A compound is 82.6% carbon and 17.4% hydrogen by mass. What is the empirical formula? Step 1: Assume a 100 gram sample 82.6% of 100 g = 82.6 g 17.4% of 100 g = 17.4 g

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Step 2: Convert grams to moles 82.6 g C x 1 mol = 6.883 mol C 12.0g 17.4 g H x 1 mol = 17.4 mol H 1.01g C 6.883 H 17.4 Although this is a mole ratio of C to H, we want a WHOLE NUMBER Mole Ratio.

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Step 3: Get the WHOLE NUMBER mole ratio by dividing each value by the smallest 6.883 mol = 1 6.883 mol 17.4 mol = 2.528 = 2.5 6.883 mol This is still not a whole number ratio, so we need to determine the least common multiple (LCM)

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Step 4: Find the least common multiple 6.883 mol = 1 x 2 = 2 6.883 mol 17.4 mol = 2.528 = 2.5 x 2 = 5 6.883 mol Since this is the lowest possible whole number ratio between carbon and hydrogen, the empirical formula of the compound is C 2 H 5.

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Calculate the empirical formula of a compound that is 85.6% carbon and 14.4% hydrogen.

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