Problem 106 - General Chemistry: Principles and Modern Applications

Question

The insecticide dieldrin contains carbon, hydrogen, oxygen, and chlorine. When burned in an excess of oxygen, a 1.510 g sample yields \(2.094 \mathrm{g} \mathrm{CO}_{2}\) and \(0.286 \mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) The compound has a molecular mass of 381 u and has half as many chlorine atoms as carbon atoms. What is the molecular formula of dieldrin?

Step-by-Step Solution

Answer

The molecular formula of dieldrin is \(C_{10}H_{20}O_{4}Cl\).

Step 1: Calculation of moles of Carbon and Hydrogen

The yield of \(CO_2\) and \(H_2O\) can be used to calculate the moles of Carbon (C) and Hydrogen (H). The molar mass of \(CO_2\) is approximately 44.01g/mol and \(H_2O\) is approximately 18.02g/mol. So, the moles of \(CO_2\) (hence C) can be calculated as, \[moles = \frac{2.094g}{44.01g/mol} = 0.0476 mol\]. Each \(CO_2\) molecule contains one C atom, so the sample contains 0.0476 mol of C. Similarly, the moles of \(H_2O\) (hence H) can be calculated as, \[moles = \frac{0.286g}{18.02g/mol} = 0.0159 mol\]. Each \(H_2O\) molecule contains two H atoms, hence the sample contains \(0.0159 mol \times 2 = 0.0318 mol\) of H.

Step 2: Calculation of mass of Carbon and Hydrogen

Next, we calculate the mass of Carbon and Hydrogen in the dieldrin sample. The mass of Carbon (C) can be calculated from its molar mass (12.01g/mol) as, \(0.0476 mol \times 12.01g/mol = 0.571 g\). Similarly, the mass of Hydrogen (H) can be calculated from its molar mass (1.01g/mol) as, \(0.0318 mol \times 1.01g/mol = 0.032 g\).

Step 3: Calculation of mass of Oxygen and Chlorine

The rest of the mass of the sample is accounted for by Oxygen (O) and Chlorine (Cl). Their total mass can be calculated as, \(1.510 g - (0.571 g + 0.032 g) = 0.907 g\). The problem specifies that the compound has half as many chlorine atoms as carbon atoms. The molar mass of chlorine is 35.45 g/mol. From Step 1, we know there are 0.0476 mol of carbon in the compound, so there must be \(0.0476 mol/2 = 0.0238 mol\) of chlorine in the compound. From the molar mass of chlorine, this implies a mass of,\(0.0238 mol x 35.45 g/mol = 0.843 g\). The rest of the mass must therefore be due to oxygen, so the mass of oxygen is \((0.907 - 0.843) g = 0.064 g\).

Step 4: Calculation of molecular formula

The molecular formula can be found by dividing each quantity of atoms in the empirical formula by the smallest quantity of \(2.984 \times 10^{-2} mol\), so the ratio of atoms becomes: {C:H:O:Cl}={5:10:2:0.5}. Multiplying all atomic ratios by 2 then gives {C:H:O:Cl}={10:20:4:1}. The molecular formula of dieldrin is then determined to be \(C_{10}H_{20}O_{4}Cl\).

Key Concepts

Combustion Analysis

When chemists want to determine the composition of a compound, they often use a technique called combustion analysis. This method is quite useful in breaking down organic compounds composed mainly of carbon, hydrogen, and oxygen. During combustion analysis, the sample is burned in the presence of excess oxygen. This results in the production of gases, predominantly carbon dioxide \( (CO_2) \) and water \( (H_2O) \), which are then measured.

• By measuring the mass of \( CO_2 \) and \( H_2O \), one can deduce the amounts of carbon and hydrogen in the original substance.
• For each mole of \( CO_2 \) produced, one mole of carbon was present in the compound.
• Similarly, for each mole of \( H_2O \), two moles of hydrogen were originally present.

In our dieldrin example, combustion analysis helped determine the amount of carbon and hydrogen by converting the given masses of \( CO_2 \) and \( H_2O \) into moles. This is a vital first step in establishing the compound's composition.

Empirical Formula

The empirical formula of a compound gives the simplest whole-number ratio of atoms of each element in the compound. It is derived after determining the relative number of atoms in each element based on mass from combustion analysis.
To find the empirical formula of dieldrin, we need to understand the ratios of its atoms:

• First, calculate the number of moles of each element based on their mass, using their molar masses.
• Next, divide the number of moles of each element by the smallest number of moles computed across all elements to simplify the ratio.
• The resulting values give us the empirical formula.

For dieldrin, the calculation showed a base ratio, which through adjustment and multiplication, was derived into a more complex ratio reflecting its actual structure.

Molecular Mass

Molecular mass, also referred to as molecular weight, is the sum of the atomic masses of all atoms in a molecule. It's essential in deducing the molecular formula from the empirical formula.
The compound's molecular mass is often determined experimentally, as in our example where dieldrin has a molecular mass of 381 u.

• Once the empirical formula is determined, its mass can be calculated by summing the atomic masses of each element in the empirical formula.
• The actual molecular formula is found by comparing the empirical formula mass with the known molecular mass.
• If necessary, multiply the subscripts in the empirical formula to match the known molecular mass.

In our case with dieldrin, the empirical formula mass led us to adjust until the molecular formula matched the given molecular mass.

Stoichiometry

Stoichiometry is a branch of chemistry that involves the calculation of reactant and product quantities in chemical reactions. It's instrumental in transitions involving chemical formulas, such as moving from empirical formulas to molecular ones.
In problems like the dieldrin exercise, stoichiometry helps relate masses from combustion products to amounts in the original compound:

• Based on combustion product masses, calculate the initial amounts of each element.
• Apply stoichiometry principles to ascertain any leftover elements, like oxygen or chlorine.
• Through mass balance, discern the composition of the sample.

Using stoichiometry, we found a balanced way to include all elements and bring together the empirical and molecular formulas systematically.