Chemistry, 03.11.2020 17:00 miaaa0

An organic compound contains , , , and . Combustion of 0.1023 g of the compound in excess oxygen yielded 0.2587 g and 0.0861 g . A sample of 0.4831 g of the compound was analyzed for nitrogen by the Dumas method. The compound is first reacted by passage over hot : The product gas is then passed through a concentrated solution of to remove the . After passage through the solution, the gas contains and is saturated with water vapor. At STP, 38.9 mL of dry was obtained. In a third experiment, the density of the compound as a gas was found to be 2.86 g/L at 127°C and 256 torr. What are the empirical and molecular formulas of the compound? (Enter the elements in the order: C, H, N, O.)

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An organic compound contains C, H, N and O. Combustion of 0.1023 g of the compound in excess oxygen yielded 0.2587 g of CO2 and 0.0861 g of H2O. A sample of 0.4831 g of the compound was analyzed for nitrogen by the Dumas method. The compound is first reacted by passage over hot: The product gas is then passed through a concentrated solution of to remove the. After passage through the solution, the gas contains and is saturated with water vapor. At STP, 38.9 mL of dry N2 was obtained. In a third experiment, the density of the compound as a gas was found to be 2.86 g/L at 127°C and 256 torr. What are the empirical and molecular formulas of the compound? (Enter the elements in the order: C, H, N, O.)

the empirical formula = $\mathbf {C_3H_6O_{12}N}$

the molecular formula = $\mathbf {C_3H_6O_{12}N}$

Explanation:

From the given information:

$\bigg ( 0.2587 \ g \ of CO_2 \bigg) \times \dfrac{1 \ mol \ of CO_2}{44 \ of \ CO_2} \times \dfrac{1 \ mol \ of \ C}{1 \ mol \ of CO_2}$

$= 0.00588 \ mol \ of \ C \times \dfrac{12.01 \ g \ of \ C}{1 \ mol \ of \ C }$

= 0.0706g of C

$\bigg ( 0.0861\ g \ of H_2O \bigg) \times \dfrac{1 \ mol \ of H_2O}{18.02 \ g \ of \ H_2O} \times \dfrac{2 \ mol \ of \ H}{1 \ mol \ of H_2O}$

$=0.0096 \ mol \times \dfrac{1.008 \ g \ of \ H}{1 mol \ H}$

0.0097g of H

Given that N2 at STP = 1 atm, 273 K and V = 0.0389 L

PV = nRT

n = PV/RT

$n = \dfrac{1 \ atm \times 0.0389 \ of \ H_2}{0.0821 \ L.atm /mol.K \times 273 \ K }$

n = 0.00173 mol of N2

The oxygen in the sample = The total grams in sample - gram in H - gram in C

The oxygen in the sample = 0.1023 g - 0.0097 g - 0.706 g

The oxygen in the sample = 0.022 g of O

The number of moles of $O_2 = \dfrac{0.02}{16}$

= 0.001375 mol of O

$O \ in \ product = (0.00588 \ mol \ of \ C ) \times \dfrac{2 \ mol \ of \ O }{1 \ mol \ of \ C }+ \bigg ( 0.0096 \ mol \ of \ H ) \times \dfrac{1 \ mol \ of \ O }{1 \ mol \ of \ H}$

O in product = 0.02136 mol of O

∴

we are meant to divide the moles of each compound by the smallest number of moles; we have:

$C = \dfrac{0.00588}{0.00173} \simeq 3$

$H = 0.0096 = \dfrac{0.0096}{0.00173} \simeq 6$

$O = 0.0199= \dfrac{0.0199}{0.00173} \simeq 12$

$N = 0.00173= \dfrac{0.00173}{0.00173} \simeq 1$

Thus; the empirical formula = $\mathbf {C_3H_6O_{12}N}$

To estimate the molecular formula; we have:

$MM = \dfrac{dRT}{P}$

$MM = \dfrac{2.80 \ g/ L \times 0.0821 \ L.atm /mol.K \times 400 \ K }{0.337 \ atm}$

MM = 272.86 g/mol

Also; the molar mass of $\mathbf {C_3H_6O_{12}N}$ = 248 g/mol

∴

$= \dfrac{272.86 \ g/mol}{248 \ g/mol}$

Thus; we can conclude that empirical formula as well as the molecular formula are the same.

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An organic compound contains , , , and . Combustion of 0.1023 g of the compound in excess oxygen yie...

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