Chapter 12: Problem 58

The elemental analysis of an organic solid extracted from gum arabic (a gummy substance used in adhesives, inks, and pharmaceuticals) showed that it contained 40.0 percent \(\mathrm{C}, 6.7\) percent \(\mathrm{H},\) and 53.3 percent O. A solution of \(0.650 \mathrm{~g}\) of the solid in \(27.8 \mathrm{~g}\) of the solvent diphenyl gave a freezing-point depression of \(1.56^{\circ} \mathrm{C}\). Calculate the molar mass and molecular formula of the solid. \(\left(K_{\mathrm{f}}\right.\) for diphenyl is \(8.00^{\circ} \mathrm{C} / \mathrm{m} .\)

Short Answer

Expert verified

The molar mass of the solid is 120 g/mol and the molecular formula is \(C4H8O4\)

Step by step solution

Derive the Empirical Formula

First, let's assume that the total mass of the compound is 100 g. This way, the given percentages translate directly to masses: 40.0 g carbon, 6.7 g hydrogen, and 53.3 g oxygen. The number of moles for each element can be calculated as follows; For carbon, \(40.0 \, g \, C \times \frac{1 \, mol}{12.01 \, g} = 3.33 \, mol \, C\), for Hydrogen, \(6.7 \, g \, H \times \frac{1 \, mol}{1.01 \, g} = 6.63 \, mol \, H\), and for Oxygen, \(53.3 \, g \, O \times \frac{1 \, mol}{16.00 \, g} = 3.33 \, mol \, O\). Dividing each value by the smallest number of moles, we get the empirical formula \(C1H2O1\), or \(CH2O\).

Calculate Molality of Solution

Next, we can use the freezing point depression formula ΔTf = iKfm to calculate the molality of the solution. In this problem, 'i' is the number of ions that the solute separates into in solution; for molecular compounds, this value would be 1. ΔTf is the change in freezing point which is 1.56°C, Kf is the cryoscopic constant of diphenyl, which is 8.00 °C/m. Using these values we can solve for m (molality): \(1.56\, °C = 1 \times 8.00 \, °C/m \times m\). So, the molality, \(m = \frac{1.56}{8.00} = 0.195 \, mol/kg\).

Find Molecular Weight of Solute

To find the molar mass, we remember that molality is moles of solute per kilogram of solvent. So, rearranging for moles and then using moles = mass/molar mass to find the molar mass, the calculations will be: Moles of solute = molality x mass of solvent in kg = \(0.195 \, mol/kg \times 27.8 g \times \frac{1 \, kg}{1000 \, g} = 0.00542 \, mol\). Now the molar mass of solute = \(\frac{mass \, of \, solute}{moles \, of \, solute} = \frac{0.650 \, g}{0.00542 \, mol} = 120 \, g/mol\).

Calculating Molecular Formula

We now use the ratio of molar mass to empirical formula mass (E.F.M) to find the molecular formula. The E.F.M. for \(CH2O\) is \(12.0 + 2*1 + 16.0 = 30.0 g/mol\). The ratio (molar mass/E.F.M.) is \(120 g/mol/30 g/mol = 4\). Thus, multiply each subscript in the empirical formula by 4, yielding the molecular formula \(C4H8O4\).

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Short Answer

Step by step solution

Derive the Empirical Formula

Calculate Molality of Solution

Find Molecular Weight of Solute

Calculating Molecular Formula

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