Question

In chemical research we often send newly synthesized compounds to commercial laboratories for analysis. These laboratories determine the weight percent of \(\mathrm{C}\) and \(\mathrm{H}\) by burning the compound and collecting the evolved \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} .\) They determine the molar mass by measuring the osmotic pressure of a solution of the compound. Calculate the empirical and molecular formulas of a compound, \(\mathrm{C}_{x} \mathrm{H}_{y}\) Cr, given the following information: (a) The compound contains \(73.94 \%\) C and \(8.27 \%\) H; the remainder is chromium. (b) At \(25^{\circ} \mathrm{C},\) the osmotic pressure of a solution containing 5.00 mg of the unknown dissolved in exactly 100 mL of chloroform solution is \(3.17 \mathrm{mm}\) Hg.

Short answer

The empirical and molecular formulas are both \(\text{C}_{18}\text{H}_{24}\text{Cr}\).

Step-by-step solution

Calculate the Mass Percent of Chromium

We know that the sum of the mass percentages of the elements in a compound is 100%. Given that the compound contains 73.94% C and 8.27% H, the percentage of Chromium (Cr) can be found by subtracting these from 100%.\[\text{Mass percent of Cr} = 100\% - 73.94\% - 8.27\% = 17.79\%\]

Calculate the Empirical Formula

Assume a 100 g sample of the compound. This means you have 73.94 g of C, 8.27 g of H, and 17.79 g of Cr.- Convert grams to moles: \[\text{Moles of C} = \frac{73.94}{12.01} = 6.16 \text{ moles} \] \[\text{Moles of H} = \frac{8.27}{1.008} = 8.20 \text{ moles} \] \[\text{Moles of Cr} = \frac{17.79}{51.996} = 0.342 \text{ moles} \]- Determine the simplest mole ratio by dividing each mole value by the smallest number of moles calculated (0.342). \[\text{Ratio C: } \frac{6.16}{0.342} \approx 18\] \[\text{Ratio H: } \frac{8.20}{0.342} \approx 24\] \[\text{Ratio Cr: } \frac{0.342}{0.342} = 1\]The empirical formula is \(\text{C}_{18}\text{H}_{24}\text{Cr}\).

Calculate the Molar Mass Using Osmotic Pressure

Use the formula for osmotic pressure (\( \Pi = \frac{n}{V}RT \)) to find the molar mass, where \( \Pi \) is osmotic pressure, \( R = 0.08206 \text{ L atm/mol K} \), and \( T = 298 \text{ K} \) (25°C).- Convert 3.17 mm Hg to atm: \( \frac{3.17}{760} = 0.004171 \text{ atm} \).- Convert 5.00 mg to g: 0.005 g.- Volume \( V = 0.1 \text{ L} \).Using osmotic pressure formula:\[\Pi = \frac{n}{V}RT \Rightarrow 0.004171 = \frac{0.005}{M \times 0.1} \times 0.08206 \times 298 \]Solving for \( M \) (molar mass):\[M = \frac{0.005 \times 0.08206 \times 298}{0.004171 \times 0.1} \approx \text{ 418.00 g/mol} \]

Calculate the Molecular Formula

Calculate the molar mass of the empirical formula \(\text{C}_{18}\text{H}_{24}\text{Cr}\):- Total molar mass = \(18 \times 12.01 + 24 \times 1.008 + 51.996 = 420.432 \text{ g/mol}\)Since the calculated molar mass of about 418 g/mol is quite close to 420.432 g/mol, we conclude that the empirical formula \(\text{C}_{18}\text{H}_{24}\text{Cr}\) is also the molecular formula.

Key concepts

Osmotic Pressure

Osmotic pressure is a vital concept when calculating molar mass. It refers to the pressure required to prevent the flow of a solvent into a solution through a semipermeable membrane. Essentially, it's a measure of a solution's ability to draw in water.
Using osmotic pressure, one can determine various properties of a solution, such as molar mass.

• In our exercise, the unknown compound's molar mass is calculated by measuring the solution's osmotic pressure.
• The osmotic pressure formula is given as \( \Pi = \frac{n}{V}RT \), where \( \Pi \) is osmotic pressure, \( n \) is the number of moles, \( V \) is volume in liters, \( R \) is the gas constant, and \( T \) is temperature in Kelvin.

By applying this formula, the number of moles can be calculated for a given sample, shedding light on the compound's molar mass.

Mole Ratio

Understanding mole ratio is crucial when determining the empirical formula of a compound. A mole ratio expresses the proportions of elements in a compound.
These ratios enable chemists to determine the simplest whole-number ratio of atoms present.

• In the provided solution, the amounts of \( \text{C} \), \( \text{H} \), and \( \text{Cr} \) are converted to moles.
• The mole ratio is achieved by dividing each element's mole count by the smallest number of moles present.
• This ensures the smallest whole numbers form the ratio, vital for correct chemical representation.

Mastery of mole ratios is essential for deriving the empirical formula, reflecting the simplest atomic composition of a molecule.

Molar Mass Calculation

Molar mass calculation bridges the empirical and molecular formulas. It involves determining a compound's total mass per mole.
Using the empirical formula mass and the experimental value, we can ascertain if the empirical formula matches the molecular formula.

• The process begins by estimating the empirical formula mass using the atomic weights of each constituent element.
• Comparisons are drawn between this calculated mass and the molar mass obtained through osmotic pressure measurements.
• If close or equal, the empirical formula serves as the molecular formula too.
• In cases where these values differ, multiplying the empirical formula by a whole number resolves the discrepancy.

This exercise utilized an estimated empirical molar mass to verify the correctness of the empirical formula derived.

Elemental Analysis

Elemental analysis is the foundation for determining a compound's composition.
It involves quantifying the percentage of each element present, often using combustion analysis.

• In this exercise related to \text{C_x H_y Cr}, the weight percent of \( \text{C} \), \( \text{H} \), and \( \text{Cr} \) were given from elemental analysis.
• The combustion method is commonly employed to release and measure \( \text{CO}_2 \) and \( \text{H}_2 \text{O} \), providing necessary data about \( \text{C} \) and \( \text{H} \).
• By subtracting these percentages from 100%, the percentage of other elements like chromium are inferred.

Elemental analysis ensures accuracy in deriving empirical formulas, aiding in understanding chemical compositions more clearly.