Rachel Maddiee Posted December 26, 2019 Share Posted December 26, 2019 I need help finishing this problem. I’m not sure if the steps are correct of what I have done so far. Percent by mass C = 40.60% Percent by mass H = 5.18% Percent by mass O = 54.22% Empirical formula = ? (unknown) Inverse molar mass C = 1 mol/12.01 g Inverse molar mass H = 1 mol/1.008 g Inverse molar mass O = 1 mol/16.00 g Convert the masses to moles. C = 40.60 g/12.01 g/mol = 3.380 mol H = 5.18 g/1.008 g/mol = 5.138 mol O = 54.22 g/16.00 g/mol = 3.388 mol Divide by the lowest, seeking the smallest whole-number ratio Link to comment Share on other sites More sharing options...
Sensei Posted December 26, 2019 Share Posted December 26, 2019 (edited) 3 hours ago, Rachel Maddiee said: I need help finishing this problem. I’m not sure if the steps are correct of what I have done so far. Percent by mass C = 40.60% Percent by mass H = 5.18% Percent by mass O = 54.22% Empirical formula = ? (unknown) Compound composed of Carbon, Hydrogen and Oxygen will have generic formula CxHyOz where x,y,z are your unknowns. 3 hours ago, Rachel Maddiee said: Convert the masses to moles. C = 40.60 g/12.01 g/mol = 3.380 mol H = 5.18 g/1.008 g/mol = 5.138 mol O = 54.22 g/16.00 g/mol = 3.388 mol ..from the above you know that z=x ... what with y? Edited December 26, 2019 by Sensei Link to comment Share on other sites More sharing options...
Rachel Maddiee Posted December 26, 2019 Author Share Posted December 26, 2019 Can you show me using this method? Link to comment Share on other sites More sharing options...
Sensei Posted December 26, 2019 Share Posted December 26, 2019 1 hour ago, Rachel Maddiee said: Can you show me using this method? Looks to me, that everything is given on the plate... Link to comment Share on other sites More sharing options...
Rachel Maddiee Posted December 26, 2019 Author Share Posted December 26, 2019 Percent by mass C = 40.60% Percent by mass H = 5.18% Percent by mass O = 54.22% Empirical formula = ? (unknown) Inverse molar mass C = 1 mol/12.01 g Inverse molar mass H = 1 mol/1.008 g Inverse molar mass O = 1 mol/16.00 g Convert the masses to moles. C = 40.60 g/12.01 g/mol = 3.380 mol H = 5.18 g/1.008 g/mol = 5.138 mol O = 54.22 g/16.00 g/mol = 3.388 mol Divide by the lowest, seeking the smallest whole-number ratio C = 3.380/3.380 = 1 = 1 mol C H = 5.138/3.380 = 1.52 = 1.5 mol H O = 3.388/3.380 = 1.00 = 1 mol O The simplest mole ratio is (1 mol C):(1.5 mol H):(1 mol O). 2 x 1 mol C = 2 mol C 2 x 1.5 mol H = 3 mol H 2 x 1 mol O = 2 mol O The simplest whole-number ratio of atoms is (2 atoms C):(3 atoms H):(2 atoms O). The empirical formula of succinic acid is C2H3O2. Link to comment Share on other sites More sharing options...
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