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 =

the molecular formula =

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

= 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 =

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.