We are given information about the combustion of an organic compound containing carbon, hydrogen, and oxygen (no nitrogen). The mass of the original substance is 0.290g. The combustion products are 0.270g of water ($H_2O$) and 0.66g of carbon dioxide ($CO_2$). We also have information about the V. Meyer method for determining the molar mass: 0.15g of the substance yields 62.4 $cm^3$ of air at 17$^\circ$C and 737 mmHg. The density of air is given as 1.29 x 10$^{-3}$ g/cm$^3$. The goal is to determine the density, approximate molar mass, and empirical formula of the substance.

1. Problem Description

We are given information about the combustion of an organic compound containing carbon, hydrogen, and oxygen (no nitrogen). The mass of the original substance is 0.290g. The combustion products are 0.270g of water ($H_2O$) and 0.66g of carbon dioxide ($CO_2$). We also have information about the V. Meyer method for determining the molar mass: 0.15g of the substance yields 62.4 $cm^3$ of air at 17$^\circ$C and 737 mmHg. The density of air is given as 1.29 x 10$^{-3}$ g/cm$^3$. The goal is to determine the density, approximate molar mass, and empirical formula of the substance.

2. Solution Steps

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

Mass of carbon in $CO_2$:

$m_C = \frac{12.01 \ g/mol}{44.01 \ g/mol} \times 0.66 \ g = 0.180 \ g$

Mass of hydrogen in $H_2O$:

$m_H = \frac{2.016 \ g/mol}{18.015 \ g/mol} \times 0.270 \ g = 0.0302 \ g$

Mass of oxygen in the original sample:

$m_O = 0.290 \ g - 0.180 \ g - 0.0302 \ g = 0.0798 \ g$

Now, we convert the masses to moles:

$n_C = \frac{0.180 \ g}{12.01 \ g/mol} = 0.0150 \ mol$

$n_H = \frac{0.0302 \ g}{1.008 \ g/mol} = 0.0300 \ mol$

$n_O = \frac{0.0798 \ g}{16.00 \ g/mol} = 0.00499 \ mol$

Divide by the smallest number of moles to find the empirical formula ratios:

$C: \frac{0.0150}{0.00499} \approx 3$

$H: \frac{0.0300}{0.00499} \approx 6$

$O: \frac{0.00499}{0.00499} = 1$

Therefore, the empirical formula is $C_3H_6O$.

Now, we calculate the approximate molar mass using the V. Meyer method. We are given the volume of air, temperature, and pressure. We will use the ideal gas law to find the number of moles of air.

$PV = nRT$

$P = 737 \ mmHg = \frac{737}{760} \ atm = 0.970 \ atm$

$V = 62.4 \ cm^3 = 0.0624 \ L$

$T = 17^\circ C = 290 \ K$

$R = 0.0821 \frac{L \cdot atm}{mol \cdot K}$

$n = \frac{PV}{RT} = \frac{0.970 \ atm \times 0.0624 \ L}{0.0821 \frac{L \cdot atm}{mol \cdot K} \times 290 \ K} = 0.00254 \ mol$

Since 0.15 g of the substance displaced 0.00254 mol of air, we assume that these moles represent the moles of the vaporized substance. Therefore, the approximate molar mass is:

$M = \frac{0.15 \ g}{0.00254 \ mol} = 59.1 \ g/mol$

The molar mass of $C_3H_6O$ is $3(12.01) + 6(1.008) + 16.00 = 36.03 + 6.048 + 16.00 = 58.078 \ g/mol$. Since the approximate molar mass calculated from the V. Meyer method is 59.1 g/mol, the empirical formula is likely the molecular formula.

The density of the vapor relative to air is $\frac{\text{Molar mass of substance}}{\text{Molar mass of air}}$. Molar mass of air is approximated by $\frac{density_{air}*R*T}{P}$ = $\frac{1.29*10^{-3} g/cm^3*0.0821 L\cdot atm/mol\cdot K*290K}{0.97atm}$ = 0.0338 g/cm^

3.

However, we can't obtain the density of the substance from the given data without additional information.

3. Final Answer

Empirical Formula: $C_3H_6O$

Approximate Molar Mass: 59.1 g/mol

We cannot determine the density of the substance.