Question

A manganese compound has the formula \(\mathrm{Mn}(\mathrm{CO})_{x}\left(\mathrm{CH}_{3}\right)_{y}\) To find the empirical formula of the compound, you burn 0.225 g of the solid in oxygen and isolate \(0.283 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(0.0290 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula for the compound? That is, what are the values of \(x\) and \(y ?\)

Short answer

The empirical formula is \( \text{Mn(CO)}_4\text{(CH}_3)_2 \).

Step-by-step solution

Find Moles of CO2

Given: Mass of \( \text{CO}_2 = 0.283 \, \text{g} \).The molar mass of \( \text{CO}_2 \) is approximately \( 44.01 \, \text{g/mol} \).Find the moles of \( \text{CO}_2 \):\[\text{moles of CO}_2 = \frac{0.283 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.00643 \, \text{moles}\]

Determine Moles of Carbon

Each mole of \( \text{CO}_2 \) contains 1 mole of carbon. Thus, the moles of carbon are:\[\text{moles of carbon} = 0.00643 \, \text{moles}\]

Find Moles of H2O

Given: Mass of \( \text{H}_2\text{O} = 0.0290 \, \text{g} \).The molar mass of \( \text{H}_2\text{O} \) is approximately \( 18.02 \, \text{g/mol} \).Find the moles of \( \text{H}_2\text{O} \):\[\text{moles of H}_2\text{O} = \frac{0.0290 \, \text{g}}{18.02 \, \text{g/mol}} \approx 0.00161 \, \text{moles}\]

Determine Moles of Hydrogen

Each mole of \( \text{H}_2\text{O} \) contains 2 moles of hydrogen. Thus, the moles of hydrogen are:\[\text{moles of hydrogen} = 2 \times 0.00161 \, \text{moles} = 0.00322 \, \text{moles}\]

Determine Molar Ratio of CH3

The empirical formula contains groups of \( \text{CH}_3 \), consisting of 1 carbon and 3 hydrogens. Find the ratio of carbon to hydrogen for \( \text{CH}_3 \):Ratio of carbon to hydrogen = \( \frac{0.00643}{0.00322} \approx 2 \).Since this ratio matches \( \text{CH}_3 \) (1 carbon : 3 hydrogens), \( y = 2 \).

Calculate Mass of CO

Given: Mass of solid = \( 0.225 \, \text{g} \).Mass from carbon and hydrogen products:- Mass of carbon from \( \text{CO}_2 \) = \( 0.00643 \, \text{moles} \times 12.01 \, \text{g/mol} = 0.0773 \, \text{g} \).- Mass of hydrogen from \( \text{H}_2\text{O} \) = \( 0.00322 \, \text{moles} \times 1.01 \, \text{g/mol} = 0.00325 \, \text{g} \).Total mass from carbon and hydrogen = \( 0.0773 \, \text{g} + 0.00325 \, \text{g} = 0.08055 \, \text{g} \).Mass from \( \text{CO} \):\[0.225 \, \text{g} - 0.08055 \, \text{g} = 0.14445 \, \text{g}\]

Find Moles of CO

The molar mass of \( \text{CO} \) is approximately \( 28.01 \, \text{g/mol} \).Find the moles of \( \text{CO} \):\[\text{moles of CO} = \frac{0.14445 \, \text{g}}{28.01 \, \text{g/mol}} \approx 0.00516 \, \text{moles}\]

Determine Values of x and y

With \( 0.00516 \, \text{moles} \text{ of CO} \) and \( 0.00643 \, \text{moles} \text{ of CH}_3 \):The ratio of \( \text{CO} \) to \( \text{CH}_3 \) is \( \frac{0.00516}{0.00643} \approx 0.8 \). Simplifying to a whole number, \( x = 4 \) for \( \text{CO} \).Thus, the empirical formula is \( \text{Mn(CO)}_4\text{(CH}_3)_2 \).

Key concepts

Understanding Combustion Analysis

Combustion analysis is an essential technique used in chemistry to determine the empirical formula of a compound. This method involves burning a sample in the presence of oxygen to produce combustion products like carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)). By measuring these products, we can deduce the amounts of carbon and hydrogen originally present in the compound. In our problem, burning the manganese compound yields \(0.283\,\text{g}\) of \(\text{CO}_2\) and \(0.0290\,\text{g}\) of \(\text{H}_2\text{O}\).

• First, by analyzing CO₂, we can determine the amount of carbon in the compound.
• Next, by examining the water produced, we can find out the amount of hydrogen present.

These steps help us in calculating the empirical formula by enabling the calculation of moles. This is a crucial first step, as it allows the determination of the percentage composition of the compound.

The Role of Molar Ratio Calculation

Calculating molar ratios is a significant aspect of empirical formula determination, particularly in combustion analysis. After finding the moles of each element, we must determine the simplest whole-number ratio of atoms in the compound. In the exercise, we found:

• \(0.00643\) moles of carbon from \(\text{CO}_2\)
• \(0.00322\) moles of hydrogen from \(\text{H}_2\text{O}\)

These values were then used to find the molar ratio of carbon to hydrogen, which in this case was approximately 2:1, matching the atomic ratio in \(\text{CH}_3\) groups. Calculating molar ratios allows us to hypothesize about how atoms are bonded together, like how carbon and hydrogen form methyl groups (\(\text{CH}_3\)). This calculation step helps ensure our deduced formula aligns with chemical reality.

Exploring Molecular Composition

The molecular composition of a compound reveals the type and number of atoms present. Determining it accurately is key in chemistry for identifying compounds. In our scenario, after combustion and various calculations, we determined which parts of the compound formed during combustion. The empirical formula, \(\text{Mn(CO)}_4\text{(CH}_3)_2\), tells us that the compound consists of manganese, carbonyl, and methyl groups.

• The carbonyl groups (\(\text{CO}\)) are determined by calculating the leftover mass and deducing its composition after accounting for carbon and hydrogen mass.
• The methyl groups (\(\text{CH}_3\)) are confirmed by finding that the carbon to hydrogen ratio corresponds perfectly to what is found in each methyl group.

Overall, knowing the exact molecular composition is vital for understanding the properties and behavior of chemical substances.

Chemical Reaction Stoichiometry Simplified

Chemical reaction stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It ensures that the law of conservation of mass is obeyed, meaning that the mass and types of atoms on the reactants side must equal those on the products side. In our problem, we use stoichiometry to balance the equation for combustion and to transform masses into moles. This transformation allows us to calculate empirical formulas.
In the solution:

• We dealt with transforming the mass of combustion products to moles (like \(\text{CO}_2\) and \(\text{H}_2\text{O}\)).
• Then, we used these moles to ensure that the calculated ratios and masses correctly accounted for the input and output of the reaction.

Mastering stoichiometry is crucial as it allows chemists to predict yields, scale reactions, and design processes effectively.