We are given that 0.290 g of an organic substance (containing C, H, and O) is combusted. The combustion produces 0.270 g of water ($H_2O$) and 0.66 g of carbon dioxide ($CO_2$). We want to find the density, approximate molar mass, and empirical formula of the substance. We are given that 0.15 g of the substance occupies 62.4 $cm^3$ of air at 17°C and 737 mm Hg. The density of air is 1.29 * $10^{-3}$ g/cm³.

1. Problem Description

We are given that 0.290 g of an organic substance (containing C, H, and O) is combusted. The combustion produces 0.270 g of water ($H_2O$) and 0.66 g of carbon dioxide ($CO_2$). We want to find the density, approximate molar mass, and empirical formula of the substance. We are given that 0.15 g of the substance occupies 62.4 $cm^3$ of air at 17°C and 737 mm Hg. The density of air is 1.29 * $10^{-3}$ g/cm³.

2. Solution Steps

First, determine the mass of carbon and hydrogen in the original sample.

Mass of carbon = mass of $CO_2$ * (molar mass of C / molar mass of $CO_2$)

Mass of carbon = 0.66 g * (12.01 g/mol / 44.01 g/mol) = 0.1802 g

Mass of hydrogen = mass of $H_2O$ * (2 * molar mass of H / molar mass of $H_2O$)

Mass of hydrogen = 0.270 g * (2 * 1.008 g/mol / 18.015 g/mol) = 0.0302 g

Now, calculate the mass of oxygen in the original sample.

Mass of oxygen = total mass - mass of carbon - mass of hydrogen

Mass of oxygen = 0.290 g - 0.1802 g - 0.0302 g = 0.0796 g

Next, calculate the number of moles of C, H, and O.

Moles of carbon = 0.1802 g / 12.01 g/mol = 0.0150 mol

Moles of hydrogen = 0.0302 g / 1.008 g/mol = 0.0300 mol

Moles of oxygen = 0.0796 g / 16.00 g/mol = 0.0050 mol

Now, find the ratio of moles to determine the empirical formula.

Divide each mole value by the smallest mole value (0.0050).

C: 0.0150 / 0.0050 = 3

H: 0.0300 / 0.0050 = 6

O: 0.0050 / 0.0050 = 1

The empirical formula is $C_3H_6O$.

Next, determine the molar mass using the V. Meyer method.

We have 0.15 g of the substance occupying 62.4 $cm^3$ at 17°C and 737 mm Hg.

Convert the pressure to atm: 737 mm Hg / 760 mm Hg/atm = 0.97 atm

Convert the volume to L: 62.4 $cm^3$ = 0.0624 L

Convert the temperature to Kelvin: 17°C + 273.15 = 290.15 K

Use the ideal gas law: PV = nRT

$n = \frac{PV}{RT} = \frac{(0.97 atm)(0.0624 L)}{(0.0821 L \cdot atm / (mol \cdot K))(290.15 K)} = 0.00252 mol$

Molar mass = mass / moles = 0.15 g / 0.00252 mol = 59.52 g/mol

Now, we determine the density. Density = mass/volume. We have that 0.15g of the substance displaces 62.4 $cm^3$ of air. Therefore, we need the density in this state. Density = 0.15 g / 62.4 $cm^3$ = 0.0024 g/$cm^3$

3. Final Answer

The empirical formula is $C_3H_6O$.

The approximate molar mass is 59.52 g/mol.

The density is 0.0024 g/$cm^3$.