Question

A quantity of \(7.480 \mathrm{~g}\) of an organic compound is dissolved in water to make \(300.0 \mathrm{~mL}\) of solution. The solution has an osmotic pressure of \(1.43 \mathrm{~atm}\) at \(27^{\circ} \mathrm{C}\). The analysis of this compound shows that it contains 41.8 percent \(\mathrm{C}, 4.7\) percent \(\mathrm{H}, 37.3\) percent \(\mathrm{O},\) and 16.3 percent \(\mathrm{N}\). Calculate the molecular formula of the compound.

Short answer

The molecular formula is \( C_{15}H_{20}O_{10}N_5 \).

Step-by-step solution

Convert Osmotic Pressure Units

First, ensure that all units are appropriate for the van't Hoff equation. Osmotic pressure \( \Pi \) is already given in atm. The temperature needs to be converted to Kelvin using the formula: \[ T(K) = T(^{\circ}C) + 273.15 = 27 + 273.15 = 300.15 \; K. \]

Use Van't Hoff Equation to Find Molarity

The van't Hoff equation states that \( \Pi = i \times M \times R \times T \), where \( \Pi \) is the osmotic pressure, \( M \) is the molarity, \( R \) is the ideal gas constant (0.0821 L·atm/mol·K), and \( T \) is the temperature in Kelvin. Assuming the compound does not dissociate, \( i = 1 \). Solve for molarity \( M \): \[ 1.43 = 1 \times M \times 0.0821 \times 300.15. \]Rearranging gives: \[ M = \frac{1.43}{0.0821 \times 300.15} = 0.0587 \; \text{mol/L}. \]

Calculate Moles of Solute

Use the molarity to find the moles of solute in 300.0 mL of solution. Since 300.0 mL is 0.300 L:\[ \text{moles of solute} = M \times \text{Volume in L} = 0.0587 \times 0.300 = 0.01761 \; \text{mol}. \]

Determine Molar Mass

The molar mass \( M_w \) is calculated using the initial mass of the compound and the moles found:\[ M_w = \frac{\text{mass}}{\text{moles}} = \frac{7.480}{0.01761} = 424.73 \; \text{g/mol}. \]

Calculate Empirical Formula Mass

Given the percentages, assume you have 100 g of the compound and convert these to moles of each element:- Carbon: \(41.8 \; \text{g} \times \frac{1 \; \text{mol}}{12.01 \; \text{g/mol}} = 3.479 \; \text{mol C}\)- Hydrogen: \(4.7 \; \text{g} \times \frac{1 \; \text{mol}}{1.008 \; \text{g/mol}} = 4.662 \; \text{mol H}\)- Oxygen: \(37.3 \; \text{g} \times \frac{1 \; \text{mol}}{16.00 \; \text{g/mol}} = 2.331 \; \text{mol O}\)- Nitrogen: \(16.3 \; \text{g} \times \frac{1 \; \text{mol}}{14.01 \; \text{g/mol}} = 1.164 \; \text{mol N}\).

Find Empirical Formula

Determine the ratios by dividing each mole amount by the smallest mole amount:- Carbon: \( \frac{3.479}{1.164} = 2.99 \approx 3 \)- Hydrogen: \( \frac{4.662}{1.164} = 4.01 \approx 4 \)- Oxygen: \( \frac{2.331}{1.164} = 2.00 \)- Nitrogen: \( \frac{1.164}{1.164} = 1.00 \).Empirical formula is thus \( C_3H_4O_2N \) (rounded appropriately to simplest whole number ratio).

Find Molecular Formula

Calculate the empirical formula mass and compare it with the molar mass:\[ \text{Empirical Formula Mass} = 3(12.01) + 4(1.008) + 2(16.00) + 1(14.01) = 87.09 \; \text{g/mol}. \]Determine the multiplier to obtain the molecular formula:\[ \frac{424.73}{87.09} \approx 4.88 \approx 5. \]Thus, the molecular formula is \( \text{5} \times C_3H_4O_2N = C_{15}H_{20}O_{10}N_5 \).

Key concepts

Osmotic Pressure

Osmotic pressure is a crucial chemical concept that describes the pressure difference needed to stop the flow of a solvent through a semipermeable membrane. This pressure arises when there is a concentration gradient between two solutions. In simpler terms, osmotic pressure is the force exerted by a solution to draw in water across a membrane to balance out solute concentrations on each side.
In the given exercise, the osmotic pressure of the solution is used to calculate the molarity of an organic compound dissolved in water. It's important to understand that this pressure depends on factors such as the number of solute particles, temperature, and volume of the solution. Hence, knowing osmotic pressure helps in determining molecular information about dissolved substances in a solution.

Van't Hoff Equation

The Van't Hoff equation is a fundamental formula in chemistry, especially when dealing with solutions and their osmotic properties. It describes the relationship between osmotic pressure and the concentration of solutions. The equation is represented as \( \Pi = i \times M \times R \times T \), where

• \(\Pi\) is the osmotic pressure,
• \(M\) is the molarity of the solution,
• \(R\) is the ideal gas constant, typically 0.0821 L·atm/mol·K,
• \(T\) is the absolute temperature in Kelvin,
• and \(i\) is the van 't Hoff factor, representing ionization or dissociation.

This equation allows the calculation of the molarity of a solution if the osmotic pressure, temperature, and ideal gas constant are known. This calculation was used in the exercise to determine the molarity of the organic compound, reflecting the practical application of the Van't Hoff equation in determining molecular properties.

Empirical Formula

The empirical formula provides the simplest whole-number ratio of atoms of each element in a compound. It does not represent the exact number of atoms like the molecular formula, but rather the simplest integer ratio of all elements present.
To find the empirical formula, one typically starts with the percent composition of each element in the compound. In the exercise, these percentages were converted into moles to understand the mole ratio of the elements present. The smallest mole value among the elements is used to divide all the mole amounts, thereby yielding a basic ratio of elements.
In this practical example, converting percent composition into moles led to determining the empirical formula which offers insights into the basic chemical structure of the compound.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, typically measured in grams per mole (g/mol). It is a critical concept in chemistry for converting between mass of a substance and the moles of substances, which are used in chemical equations and reactions.
The process of calculating the molar mass often involves taking the mass of a known amount of substance and dividing it by the number of moles. In the exercise, this was done by utilizing the weight of the organic compound and the moles calculated from the osmotic pressure analysis. Understanding how to convert mass to moles and vice-versa is essential for working with chemical reactions and formulas. It also helps compare the empirical formula mass to the actual molar mass to find the molecular formula, showcasing the interdependence of various calculations in determining molecular makeup.