Question

The iron in an organometallic compound is determined by treating a \(0.4873-\mathrm{g}\) sample with \(\mathrm{HNO}_{3}\) and heating to volatilize the organic material. After ignition, the residue of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) weighs \(0.2091 \mathrm{~g}\). (a) What is the \(\% \mathrm{w} / \mathrm{w}\) Fe in this compound? (b) The carbon and hydrogen in a second sample of the compound are determined by a combustion analysis. When a \(0.5123-\mathrm{g}\) sample is carried through the analysis, \(1.2119 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.2482 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) re collected. What are the \(\% \mathrm{w} / \mathrm{w} \mathrm{C}\) and \(\% \mathrm{w} / \mathrm{w} \mathrm{H}\) in this compound and what is the compound's empirical formula?

Short answer

(a) %Fe = 30.01%; (b) %C = 64.56%, %H = 5.42%; empirical formula C10H10Fe.

Step-by-step solution

Calculate the Moles of Fe in Fe2O3

First, calculate the moles of Fe in the Fe2O3 residue. The molar mass of Fe2O3 is calculated as follows:\[ M(\mathrm{Fe}_2\mathrm{O}_3) = 2 \times 55.85 + 3 \times 16.00 = 159.70 \, \text{g/mol} \]Now, determine the moles of Fe2O3 in the residue:\[ n(\mathrm{Fe}_2\mathrm{O}_3) = \frac{0.2091 \, \text{g}}{159.70 \, \text{g/mol}} = 0.001309 \, \text{mol} \]In Fe2O3, there are two moles of Fe for each mole of Fe2O3. So, the moles of Fe are:\[ n(\mathrm{Fe}) = 2 \times 0.001309 = 0.002618 \, \text{mol} \]

Calculate Mass and Percentage of Fe

Convert moles of Fe to grams using the molar mass of Fe (55.85 g/mol):\[ m(\mathrm{Fe}) = 0.002618 \, \text{mol} \times 55.85 \, \text{g/mol} = 0.1462 \, \text{g} \]Now calculate the percentage by mass of Fe in the original sample:\[ \%\,\mathrm{w/w}\,\mathrm{Fe} = \left(\frac{0.1462 \, \text{g}}{0.4873 \, \text{g}}\right) \times 100 = 30.01\% \]

Calculate Moles of C from CO2

Start by calculating the moles of CO2 produced:\[ n(\mathrm{CO}_2) = \frac{1.2119 \, \text{g}}{44.01 \, \text{g/mol}} = 0.02754 \, \text{mol} \]Each mole of CO2 corresponds to one mole of C. Thus, moles of C = 0.02754 mol.Convert moles of C to grams (molar mass of C = 12.01 g/mol):\[ m(\mathrm{C}) = 0.02754 \, \text{mol} \times 12.01 \, \text{g/mol} = 0.3308 \, \text{g} \]

Calculate Moles of H from H2O

Calculate the moles of H2O produced:\[ n(\mathrm{H}_2\mathrm{O}) = \frac{0.2482 \, \text{g}}{18.02 \, \text{g/mol}} = 0.01377 \, \text{mol} \]Each mole of H2O contains two moles of H. Thus, moles of H = 2 x 0.01377 mol = 0.02754 mol.Convert moles of H to grams (molar mass of H = 1.008 g/mol):\[ m(\mathrm{H}) = 0.02754 \, \text{mol} \times 1.008 \, \text{g/mol} = 0.02776 \, \text{g} \]

Calculate Percentage of C and H

Calculate the percentage by mass of C in the sample:\[ \%\,\mathrm{w/w}\,\mathrm{C} = \left(\frac{0.3308 \, \text{g}}{0.5123 \, \text{g}}\right) \times 100 = 64.56\% \]Calculate the percentage by mass of H in the sample:\[ \%\,\mathrm{w/w}\,\mathrm{H} = \left(\frac{0.02776 \, \text{g}}{0.5123 \, \text{g}}\right) \times 100 = 5.42\% \]

Determine Empirical Formula

Find moles of each element per 100g of the compound and divide by smallest moles.For C:\[ \frac{64.56 \, \text{g}}{12.01 \, \text{g/mol}} = 5.38 \, \text{mol} \]For H:\[ \frac{5.42 \, \text{g}}{1.008 \, \text{g/mol}} = 5.38 \, \text{mol} \]For Fe:\[ \frac{30.01 \, \text{g}}{55.85 \, \text{g/mol}} = 0.537 \text{mol} \]Divide each by smallest (0.537): This gives \( \text{C}_{10} \text{H}_{10} \text{Fe} \). Therefore, the empirical formula is C10H10Fe.

Key concepts

Empirical Formula Determination

In chemistry, an empirical formula represents the simplest whole-number ratio of atoms present in a compound. It is derived from the percent composition or mass of each element within the compound. To determine an empirical formula, follow these steps:
First, calculate the number of moles of each element using their respective masses and molar masses. Then, find the smallest number of moles calculated among the elements.
Divide the mole value for each element by this smallest number. The resulting numbers are the subscripts for each element in the empirical formula. If necessary, multiply these numbers to get whole numbers.
In the provided exercise, you calculate atoms of carbon, hydrogen, and iron to find the empirical formula. This method highlights how an empirical formula provides a basic understanding of a compound's structure.

Gravimetric Analysis

Gravimetric analysis is a classic method used to quantify the amount of an analyte based on its mass. It involves transforming the analyte into a compound of known composition that can be easily quantified.
This technique offers great precision and accuracy as it is based on mass measurements. In the exercise, gravimetric analysis is employed to find the percentage of iron present in the sample. The sample undergoes several chemical processes to yield iron oxide (\(\mathrm{Fe}_2\mathrm{O}_3\)).
The mass of this residue is then measured and used to calculate the mass and percentage of iron in the initial sample. Gravimetric analysis is potent because it doesn't require sophisticated instruments, just precise weight measurements.

Combustion Analysis

Combustion analysis is a method used to determine the elemental composition of a compound, especially of organic compounds containing carbon and hydrogen. In this method, a sample is burned in excess oxygen, allowing carbon and hydrogen to convert into carbon dioxide and water, respectively.
By measuring the amounts of these products, one can calculate the amounts of carbon and hydrogen in the original compound. In the given exercise, the collected masses of CO2 and H2O are used to deduce the number of moles of carbon and hydrogen.
This approach is beneficial when identifying the formula of an organic compound. The results from combustion analysis, when combined with other analyses, can also help deduce empirical formulas.

Percent Composition Calculations

Percent composition involves expressing each element's mass percentage in a compound. It is an essential concept for describing a compound's composition in terms of each constituent element's contribution.
To calculate percent composition, take the mass of the element in the compound, divide it by the total mass of the compound, and multiply by 100. This provides a clear picture of how much of each element is present in compound form.
In the exercise, this calculation is used to determine the percentages of iron, carbon, and hydrogen in the different samples. These percentages are crucial for further analyses, including determining empirical formulas and understanding compound properties.