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  • Hard Chemistry Gas Stoich question? Can anyone help me out with this question. I am struggling with it. A small sample of drug with a mass of 0.25 g undergoes a series of reactions that changed all of nitrogen in the compound into N2. This gas had a volume of 18.9 ml when collected over water at 25 degrees celsius and a pressure of 101.3 kPa a.) Calculate the percentage of nitrogen in the sample b.) When 6.8 mg of the compound was burned in pure oxygen, 17.6 mg CO2, and 4.6 mg of H20 were obtained. What are the percentages of C and H in the compound. Assuming that any undetermined element is oxygen, write an empirical formula of the compound. Mostly need help with b and c, c.) The molecular mass of the compound was found to be 324. What is its molecular formula?

  • Answer:

    b) 17.6 mg CO2 is 17.6/44.0 = 0.400 mmol of CO2, also 0.400 mmol of C: 0.400 mmol of C has a mass of 0.400*12.0 = 4.8 mg 4.6 mg of H2O = 4.6/18 = 0.256 mm H2O or 0.511 mmol of H; mass of H = 0.511*1.0 = 0.51 mg mass fraction of C = 4.8/6.8 = 0.71 and of H = 0.51/6.8 = 0.075 C = 71%; H = 7.5% Mass of O = 6.8 - 4.8 -0.51 = 1.49 mg; mmoles of O = 1.49/16 = 0.093 H = 0.511 mmol ≈ 0.5 C = 0.400 mmol O = 0.093 mmol ≈ 0.1 Emperical formula H5C4O 324 = (5 + 4*12 + 16)*n = 69*n n = 4.7 ?? ≈ 5 H20C16O4 ??

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b) 17.6 mg CO2 is 17.6/44.0 = 0.400 mmol of CO2, also 0.400 mmol of C: 0.400 mmol of C has a mass of 0.400*12.0 = 4.8 mg 4.6 mg of H2O = 4.6/18 = 0.256 mm H2O or 0.511 mmol of H; mass of H = 0.511*1.0 = 0.51 mg mass fraction of C = 4.8/6.8 = 0.71 and of H = 0.51/6.8 = 0.075 C = 71%; H = 7.5% Mass of O = 6.8 - 4.8 -0.51 = 1.49 mg; mmoles of O = 1.49/16 = 0.093 H = 0.511 mmol ≈ 0.5 C = 0.400 mmol O = 0.093 mmol ≈ 0.1 Emperical formula H5C4O 324 = (5 + 4*12 + 16)*n = 69*n n = 4.7 ?? ≈ 5 H20C16O4 ??

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