Problem 133 - Chemistry and Chemical Reactivity

Question

A compound called MMT was once used to boost the octane rating of gasoline. What is the empirical formula of MMT if it is \(49.5 \% \mathrm{C}, 3.2 \% \mathrm{H}, 22.0 \% \mathrm{O}\) and \(25.2 \%\) Mn?

Step-by-Step Solution

Answer

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

Step 1: Convert Percentages to Grams

Assume you have 100 grams of MMT, which makes calculations straightforward. Therefore, you have 49.5 grams of Carbon (C), 3.2 grams of Hydrogen (H), 22.0 grams of Oxygen (O), and 25.2 grams of Manganese (Mn).

Step 2: Convert Grams to Moles

Use the molar mass of each element to convert grams to moles: - Moles of C = \( \frac{49.5\,\text{g}}{12.01\,\text{g/mol}} = 4.12\,\text{mol} \)- Moles of H = \( \frac{3.2\,\text{g}}{1.008\,\text{g/mol}} = 3.17\,\text{mol} \)- Moles of O = \( \frac{22.0\,\text{g}}{16.00\,\text{g/mol}} = 1.375\,\text{mol} \)- Moles of Mn = \( \frac{25.2\,\text{g}}{54.94\,\text{g/mol}} = 0.459\,\text{mol} \)

Step 3: Determine the Mole Ratio

Divide the moles of each element by the smallest number of moles to find their ratio: - C: \( \frac{4.12}{0.459} = 8.98 \) - H: \( \frac{3.17}{0.459} = 6.91 \)- O: \( \frac{1.375}{0.459} = 3.00 \)- Mn: \( \frac{0.459}{0.459} = 1.00 \)

Step 4: Round to the Nearest Whole Number

Round the ratios to the nearest whole numbers to find the empirical formula: - C: 9 - H: 7 - O: 3 - Mn: 1

Step 5: Write the Empirical Formula

Based on the whole number mole ratios calculated, the empirical formula for MMT is \( \text{C}_9\text{H}_7\text{O}_3\text{Mn} \).

Key Concepts

Molar Mass

The molar mass is a crucial concept in chemistry that helps us convert between mass and amount of substance. It is defined as the mass of one mole of a given substance, usually expressed in grams per mole (g/mol). Each element has a unique molar mass based on its atomic weight. For instance, the molar mass of Carbon (C) is 12.01 g/mol, Hydrogen (H) is 1.008 g/mol, Oxygen (O) is 16.00 g/mol, and Manganese (Mn) is 54.94 g/mol.
Understanding molar mass allows chemists to determine how much of each substance is present in a sample. When we convert grams of an element to moles, we use its molar mass as a conversion factor. This conversion is fundamental when determining an empirical formula because it reveals the number of moles of each element in the compound.

Mole Ratio

The mole ratio is key in expressing the proportion of elements within a compound. By determining the number of moles of each element, we can compare them to understand how they combine to form compounds. The mole ratio shows the simplest integer ratio of moles of each element in a compound.
To find the mole ratio, you need to divide the moles of each element by the smallest number of moles from your calculations. This step simplifies the molar quantities to the smallest whole numbers, which are essential for constructing the compound's empirical formula.

Elemental Composition

Elemental composition refers to the percentage representation of each element within a compound. It highlights the specific contribution of each type of atom to the overall makeup of the compound. For example, in the compound MMT, which has an elemental composition of 49.5% Carbon (C), 3.2% Hydrogen (H), 22.0% Oxygen (O), and 25.2% Manganese (Mn), each percentage corresponds to the mass of that element in a 100 g sample of the compound.
These percentages are converted to grams when calculating empirical formulas, assuming a 100 g sample for simplicity, so that the percentage directly translates to mass in grams. This conversion helps in determining the number of moles of each element, which is pivotal in finding the empirical formula.

Octane Rating

Octane rating is a measure of a fuel's ability to resist knocking during combustion in an engine. It is a critical factor for gasoline quality and performance. A higher octane rating implies greater resistance to knocking, which can improve engine efficiency and performance.
Compounds like MMT were previously used to enhance the octane rating of gasoline. This is important because it allows engines to run smoothly at higher compression ratios without experiencing knocking, improving both power and fuel efficiency.

• An octane rating of 87 is considered regular, while ratings above 91 are considered premium-grade fuels.
• Compounds like MMT increased octane ratings but were subject to regulation due to environmental concerns.

Understanding octane ratings helps us see why certain additives were used in fuels and underscores the impact of chemistry on automotive technology.