In Dumas' method 292text mg$292\text{ mg}$ of an organic compound released 50text mL$50\text{ mL}$ of nitrogen gas (textN_2)$(\text{N}_2)$ at 300text K$300\text{ K}$ temperature and 715text mm Hg$715\text{ mm Hg}$ pressure. [cite: 435, 451] The percentage composition of 'N' in the organic compound is dots$\dots$ % (Nearest integer) [cite: 452, 455] (Aqueous tension at 300text K = 15text mm Hg$300\text{ K} = 15\text{ mm Hg}$)

### Related Formula P_textdry textN_2 = P_texttotal - textAqueous Tension $$P_{\text{dry } \text{N}_2} = P_{\text{total}} - \text{Aqueous Tension} $$ PV = nRT implies n = fracPVRT $$PV = nRT \implies n = \frac{PV}{RT} $$ \%textN = fractextMass of NitrogentextMass of organic compound times 100 $$\%\text{N} = \frac{\text{Mass of Nitrogen}}{\text{Mass of organic compound}} \times 100 $$ ### Core Logic First, isolate the pressure contribution of the dry nitrogen gas: P_textdry textN2 = 715 - 15 = 700text mm Hg = frac700760text atm $$P_{\text{dry } \text{N}2} = 715 - 15 = 700\text{ mm Hg} = \frac{700}{760}\text{ atm} $$ Using the ideal gas parameters: - V = 50text mL = 0.050text L$V = 50\text{ mL} = 0.050\text{ L}$ [cite: 1073, 1076] - T = 300text K$T = 300\text{ K}$ - R = 0.0821text L atm mol^-1textK^-1$R = 0.0821\text{ L atm mol}^{-1}\text{K}^{-1}$ ### Step 1: Compute Moles and Mass Calculate total moles of textN_2$\text{N}_2$ molecules collected: ntextN2 = fracleft(frac700760 ight) times 0.0500.0821 times 300 approx 1.868 times 10^-3text mol $$n{\text{N}2} = \frac{\left(\frac{700}{760} ight) \times 0.050}{0.0821 \times 300} \approx 1.868 \times 10^{-3}\text{ mol} $$ Compute corresponding mass of atomic Nitrogen elements (2 times 14 = 28text g/mol$2 \times 14 = 28\text{ g/mol}$): [cite: 1016, 1017] textMass of N = ntextN2 times 28 = 1.868 times 10^-3 times 28 approx 0.0523text g = 52.3text mg$$\text{Mass of N} = n{\text{N}2} \times 28 = 1.868 \times 10^{-3} \times 28 \approx 0.0523\text{ g} = 52.3\text{ mg}$$ ### Step 2: Calculate Percentage Composition Applying the fraction formulation against total sample mass: [cite: 1021, 1022] \%textN = frac52.3text mg292text mg times 100 approx 17.91\% approx 18\% $$\%\text{N} = \frac{52.3\text{ mg}}{292\text{ mg}} \times 100 \approx 17.91\% \approx 18\% $$ ### Pattern Recognition Dumas analysis safety check: Always strip away the vapor pressure of water (aqueous tension) from the measured barometric value before computing the chemical molar counts. For standard conditions shortcuts, remember that 1text mole = 22400text mL$1\text{ mole} = 22400\text{ mL}$ at STP can act as an alternate path if values are normalized. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques