Question

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.)

Answer 1

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.