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Determining Molar Mass of Substances: A Comparison of Ethanol and Methanol

This analysis explores the molar mass of substances derived from measured weights and gas laws. Using the Ideal Gas Law (PV=nRT), calculations reveal two distinct molar masses: 46.05 g/mol as an average of two samples and their respective changes in masses. The substances analyzed include ethanol (C2H5OH) and methanol (CH3OH), with individual calculations showing their molar masses as approximately 32.05 g/mol for methanol and around 46.08 g/mol for ethanol. This study highlights the significance of precise measurements in determining molecular weights.

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Determining Molar Mass of Substances: A Comparison of Ethanol and Methanol

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  1. 83.73g - 83.32g = 0.41g 83.82g - 83.39g = 0.43g 1 L 267 mL x = 0.267 L 1000 mL 99.0 + 273 = 372.0 K 99.0 + 273 = 372.0 K 773.5mm Hg x 101.3 kPa = 103.1 kPa 760 mm Hg 812.0mm Hg x 101.3 kPa = 108.2 kPa 760 mm Hg

  2. 1 L 267 mL x = 0.267 L 1000 mL 99.0 + 273 = 372.0 K 99.0 + 273 = 372.0 K 773.5mm Hg x 101.3 kPa = 103.1 kPa 760 mm Hg 812.0mm Hg x 101.3 kPa = 108.2 kPa 760 mm Hg PV = nRT PV RT (103.1)(0.267) (8.31 x 372.0) n = = = 0.00890 moles (108.2)(0.267) (8.31 x 372.0) n = = 0.00935 moles

  3. 83.73g - 83.32g = 0.41g 83.82g - 83.39g = 0.43g grams mole Molar mass = 1 L 267 mL x = 0.267 L 1000 mL 99.0 + 273 = 372.0 K 0.41 g 0.00890 moles = 46.1 g/mol 99.0 + 273 = 372.0 K 0.43 g 0.00935 moles 773.5mm Hg x 101.3 kPa = 46.0 g/mol = 103.1 kPa 760 mm Hg 812.0mm Hg x 101.3 kPa = 108.2 kPa 760 mm Hg PV RT (103.1)(0.267) (8.31 x 372.0) n = = = 0.00890 moles (108.2)(0.267) (8.31 x 372.0) n = PV = nRT = 0.00935 moles

  4. H H O-H H H O-H H H H H C C C grams mole 0.41 g 0.00890 moles Molar mass = = 46.1 g/mol 0.43 g 0.00935 moles = 46.0 g/mol (46.1 + 46.0)/2 = 46.05 g/mol C = 12.01 Ethanol H = 1.01 O = 16.00 16 1.01 1.01 1.01 12.01 12.01 12.01 1.01 1.01 1.01 1.01 16 1.01 1.01 1.01 methanol ethanol =32.05 g/mol = 46.08 g/mol Which one has a molar mass of 46 g/mol?

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