205k views
2 votes
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.)

User Achimnol
by
7.0k points

1 Answer

1 vote

The question displayed below shows the missing information which therefore completes the question.

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.)

Answer:

the empirical formula =
\mathbf {C_3H_6O_(12)N}

the molecular formula =
\mathbf {C_3H_6O_(12)N}

Step-by-step explanation:

From the given information:


\bigg ( 0.2587 \ g \ of CO_2 \bigg) * (1 \ mol \ of CO_2)/(44 \ of \ CO_2) * (1 \ mol \ of \ C)/(1 \ mol \ of CO_2)


= 0.00588 \ mol \ of \ C * (12.01 \ g \ of \ C)/(1 \ mol \ of \ C )

= 0.0706g of C


\bigg ( 0.0861\ g \ of H_2O \bigg) * (1 \ mol \ of H_2O)/(18.02 \ g \ of \ H_2O) * (2 \ mol \ of \ H)/(1 \ mol \ of H_2O)


=0.0096 \ mol * (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 = (1 \ atm * 0.0389 \ of \ H_2)/(0.0821 \ L.atm /mol.K * 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 = (0.02)/(16)

= 0.001375 mol of O


O \ in \ product = (0.00588 \ mol \ of \ C ) * (2 \ mol \ of \ O )/(1 \ mol \ of \ C )+ \bigg ( 0.0096 \ mol \ of \ H ) * (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 = (0.00588)/(0.00173) \simeq 3


H = 0.0096 = (0.0096)/(0.00173) \simeq 6


O = 0.0199= (0.0199)/(0.00173) \simeq 12


N = 0.00173= (0.00173)/(0.00173) \simeq 1

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

To estimate the molecular formula; we have:


MM = (dRT)/(P)


MM = (2.80 \ g/ L * 0.0821 \ L.atm /mol.K * 400 \ K )/(0.337 \ atm)

MM = 272.86 g/mol

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


= (272.86 \ g/mol)/(248 \ g/mol)


=1

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

User Stypox
by
7.1k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.