1.- Combustion analysis of a 25.70mg sample of an organiccompound containing onl

Question

1.- Combustion analysis of a 25.70mg sample of an organiccompound containing only C, H, and N yielded 71.49mg of carbondioxide and 14.64mg of water. What is the mass % of nitrogen in theorganic compound? a)17.7%   b)75.9%  c)6.4%   d)8.9% e)12.8% 2.- The % composition by mass of a compound is 76.0% C,12.8% H, and 11.2% O. The3 molar mass of this compound is 569g/mol. What are the empirical and molecular formulas of thecompound?   Empirical:           Molecular: a)C9H18O             C9H18O   b)C9H18O             C36H32O4 c)C9H18O              C18H36O2 d)C8H16O2            C9H18O   e)C8H16O2            C18H36O2 1.- Combustion analysis of a 25.70mg sample of an organiccompound containing only C, H, and N yielded 71.49mg of carbondioxide and 14.64mg of water. What is the mass % of nitrogen in theorganic compound? a)17.7%   b)75.9%  c)6.4%   d)8.9% e)12.8% 2.- The % composition by mass of a compound is 76.0% C,12.8% H, and 11.2% O. The3 molar mass of this compound is 569g/mol. What are the empirical and molecular formulas of thecompound?   Empirical:           Molecular: a)C9H18O             C9H18O   b)C9H18O             C36H32O4 c)C9H18O              C18H36O2 d)C8H16O2            C9H18O   e)C8H16O2            C18H36O2 d)C8H16O2            C9H18O   e)C8H16O2            C18H36O2

Explanation / Answer

2) 76 g *(1mol C/12g) = 6.33 mol C 12.8g *(1mol H/1g) = 12.8 mol H 11.2g * (1mol O/16g) = 0.7 mol O Divide all numbers by 0.7 6.33/0.7 = 9 mol C 12.8/0.7 = 18.2 mol H 0.7/0.7 = 1 mol O empirical : C9H18O formula weight: 9*12 + 1*18 + 1*16 = 142 569/142 = 4 Thus actual formula : (C9H18O)4 =C36H32O4

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