Question

A major oil company has used a gasoline additive called MMT to boost the octane rating of its gasoline. What is the empirical formula of MMT if it is \(49.5 \% \mathrm{C}\) \(3.2 \% \mathrm{H}, 22.0 \% \mathrm{O},\) and \(25.2 \% \mathrm{Mn} ?\)

Short answer

The empirical formula of MMT is \( \text{C}_9\text{H}_7\text{O}_3\text{Mn} \).

Step-by-step solution

Convert Percentages to Grams

Assume 100 g of MMT for simplicity, which makes the percentages equal to the mass in grams. Thus, the mass of carbon (C) is 49.5 g, hydrogen (H) is 3.2 g, oxygen (O) is 22.0 g, and manganese (Mn) is 25.2 g.

Calculate Moles of Each Element

Calculate the number of moles for each element by dividing the mass by the atomic mass: - Moles of C = \( \frac{49.5 \text{ g}}{12.01 \text{ g/mol}} = 4.12 \text{ mol} \)- Moles of H = \( \frac{3.2 \text{ g}}{1.01 \text{ g/mol}} = 3.17 \text{ mol} \)- Moles of O = \( \frac{22.0 \text{ g}}{16.00 \text{ g/mol}} = 1.38 \text{ mol} \)- Moles of Mn = \( \frac{25.2 \text{ g}}{54.94 \text{ g/mol}} = 0.459 \text{ mol} \).

Determine Smallest Whole Number Ratio

Divide each of the moles by the smallest value obtained:- C: \( \frac{4.12}{0.459} = 8.98 \approx 9 \)- H: \( \frac{3.17}{0.459} = 6.91 \approx 7 \)- O: \( \frac{1.38}{0.459} = 3.01 \approx 3 \)- Mn: \( \frac{0.459}{0.459} = 1 \).

Write the Empirical Formula

Using the ratios found, the empirical formula of MMT is \( \text{C}_9\text{H}_7\text{O}_3\text{Mn} \).

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that involves the calculation of reactants and products in chemical reactions. It essentially lets us know how much of one substance we need to fully react with a certain amount of another. In empirical formula calculations, stoichiometry helps in converting mass percentages into moles, enabling us to find the simplest whole-number ratios of atoms in a compound.
Stoichiometry works by using the atomic masses of the elements involved. Once we have the mass of each element from the initial problem, formatted as percentages, stoichiometry guides us to convert these percentages into grams. This is typically done by assuming a 100 g sample for simplicity. Then, the mass is converted to moles using the atomic mass of each element.

• This process helps in determining the proportion of each element in the compound.
• The mole ratios are crucial because they define the structure of the empirical formula.

Understanding stoichiometry aids in grasping how different elements bond and interact with each other to form compounds.

Mole Concept

The mole concept is fundamental in chemistry as it provides a bridge between the atomic scale and the laboratory scale. One mole is defined as exactly 6.02214076 × 10²³ entities (atoms, molecules, etc.), known as Avogadro's number. This number allows chemists to count atoms by weighing them.
In our exercise, the mole concept is used to convert the mass of each element to the number of moles:

• Moles of Carbon, Hydrogen, Oxygen, and Manganese were calculated by dividing their mass by their respective atomic masses.
• This step is vital because reactions occur on a molecular level and moles allow chemists to tally these interactions in a practical way that aligns with observable laboratory studies.

By using the mole concept, one can understand and predict the outcome of a chemical reaction based on the quantities of the reactants.

Chemical Composition

Chemical composition expresses the constituent elements of a compound and their relative ratios. It forms the basis of empirical formulas, which only reflect the simplest ratio of elements, not actual numbers in molecules.
For example, in the empirical formula calculation for MMT, we derived the chemical composition:

• 49.5% Carbon
• 3.2% Hydrogen
• 22.0% Oxygen
• 25.2% Manganese

Understanding the chemical composition helps in identifying and characterizing substances. It reveals the elements involved and their relative abundance, essential for exploring the compound's properties and behavior. By determining the relative masses of the constituents, we can compare different substances on a common basis, making the study and classification of chemical substances more systematic and uniform.

Elemental Analysis

Elemental analysis is a process used to determine the elemental composition of a chemical compound. This process is crucial in deriving empirical formulas as it provides the qualitative and quantitative data needed.
During the elemental analysis, the compound is decomposed, and the elemental percentages are measured. The results are typically presented in mass percentages, which are then converted to moles in the empirical formula process.

• It involves techniques such as combustion analysis or spectroscopy to identify and quantify elements present in a sample.
• The precision of elemental analysis influences the accuracy of the empirical formula obtained.
• Knowing the precise elemental makeup of a compound allows chemists to predict its reactivity, stability, appearance, and other physical properties.

Overall, elemental analysis is the foundation for understanding the complete chemical identity of a compound and supports further exploratory research in chemical synthesis and application.