The Grass Doctor

empirical and chemical formulas from elemental analysis, help!?

my professor went through this briefly, and i still do not some what understand. here is the question. Urea is used as a comercial fertilizer because of the nitrogen content. An analysis of 25.0 mg of Urea showed that it contains 5.0mg C, 11.68mg N, 6.65mg O, and the remainder, hydrogen. What is the empirical formula of Urea? if someone, anyone could help me with this problem, i do not even know where to begin to solve this. i would like the answer in detail please so i understand it better. thanks

Public Comments

1. 0.0050 g C / 12.011 g/mol = 4.16x10^-4 mol C atoms 0.01168 g N / 14.0067 g/mol = 8.34x10^-4 mol N atoms 0.00665 g O / 15.9994 g/mol = 4.06x10^-4 mol O atoms 0.025 g sample - 0.0050 g - 0.01168 g - 0.00665 g = 0.00167 g H 0.00167 g H / 1.0079 g/mol = 16.57x10^-4 mol H atoms Find the ratio of the moles of atoms by dividing by the smallest- C: 4.16 / 4.06 = 1.0 N: 8.34 / 4.06 = 2.0 O: 4.06 / 4.06 = 1.0 H: 16.57 / 4.06 = 4.1 CH4N2O (H2NCONH2)