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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Ответ:
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.)
the empirical formula =![\mathbf {C_3H_6O_{12}N}](/tpl/images/0863/4310/da6b2.png)
the molecular formula =![\mathbf {C_3H_6O_{12}N}](/tpl/images/0863/4310/da6b2.png)
Explanation:
From the given information:
= 0.0706g of C
0.0097g of H
Given that N2 at STP = 1 atm, 273 K and V = 0.0389 L
PV = nRT
n = PV/RT
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}](/tpl/images/0863/4310/c4d5f.png)
= 0.001375 mol of O
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:
Thus; the empirical formula =![\mathbf {C_3H_6O_{12}N}](/tpl/images/0863/4310/da6b2.png)
To estimate the molecular formula; we have:
MM = 272.86 g/mol
Also; the molar mass of
= 248 g/mol
∴
Thus; we can conclude that empirical formula as well as the molecular formula are the same.
Ответ:
Because it would lower the amount in the sample causing it to be heated up faster in the process, duh.