In the Dumas method,1 g of an organic compou | vedclass

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(A) The pressure of dry $N_2$ gas is $P_{dry} = P_{total} - P_{aqueous} = 740 - 15 = 725 \ mm \ Hg$. Using the ideal gas equation at $STP$ $(P_2 = 760

Vedclass · Accepted Answer

$5.42$

Vedclass · Answer

(A) The pressure of dry $N_2$ gas is $P_{dry} = P_{total} - P_{aqueous} = 740 - 15 = 725 \ mm \ Hg$. Using the ideal gas equation at $STP$ $(P_2 = 760 \ mm \ Hg, T_2 = 273 \ K)$: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \Rightarrow \frac{725 	imes 50}{300} = \frac{760 	imes V_2}{273}$. $V_2 = \frac{725 	imes 50 	imes 273}{300 	imes 760} \approx 43.47 \ mL$. Since $22400 \ mL$ of $N_2$ at $STP$ weighs $28 \ g$,the mass of $N_2$ is: $	ext{Mass} = \frac{28 	imes 43.47}{22400} \approx 0.0543 \ g$. Percentage of $N_2 = \frac{	ext{Mass of } N_2}{	ext{Mass of compound}} 	imes 100 = \frac{0.0543}{1} 	imes 100 = 5.43 \%$. Rounding to the nearest provided option,$5.42 \%$ is the correct value.

In the Dumas method,$1 \ g$ of an organic compound gives $50 \ mL$ of $N_2$ gas at $300 \ K$ and $740 \ mm \ Hg$ pressure. If the aqueous tension at $300 \ K$ is $15 \ mm \ Hg$,what is the percentage of nitrogen in the compound?

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