Question

Determine the empirical and molecular formulas of each of the following substances: (a) Styrene, a compound substance used to make Styrofoam \(^{B}\) cups and insulation, contains \(92.3 \% \mathrm{C}\) and \(7.7 \% \mathrm{H}\) by mass and has a molar mass of \(104 \mathrm{~g} / \mathrm{mol}\). (b) Caffeine, a stimulant found in coffee, contains \(49.5 \%\) C, \(5.15 \% \mathrm{H}, 28.9 \% \mathrm{~N}\), and \(16.5 \% \mathrm{O}\) by mass and has a molar mass of \(195 \mathrm{~g} / \mathrm{mol}\). (c) Monosodium glutamate (MSG), a flavor enhancer in certain foods, contains \(35.51 \%\) C, \(4.77 \% \mathrm{H}, 37.85 \%\) O, \(8.29 \% \mathrm{~N}\), and \(13.60 \%\) Na, and has a molar mass of \(169 \mathrm{~g} / \mathrm{mol}\).

Short answer

The empirical and molecular formulas for each substance are as follows: (a) Styrene: Empirical Formula - CH, Molecular Formula - C8H8 (b) Caffeine: Empirical Formula - C4H5N2O, Molecular Formula - C8H10N4O2 (c) Monosodium Glutamate (MSG): Empirical Formula - C5H8NO4Na, Molecular Formula - C5H8NO4Na

Step-by-step solution

Convert mass percentages to grams

Assume you have 100 g of the Styrene compound. Since it contains 92.3% C and 7.7% H by mass, we can calculate the grams of each element: - 92.3 g C - 7.7 g H

Convert grams to moles

Next, we convert the grams of each element to moles using their molar masses (molar mass of C = 12.01 g/mol and molar mass of H = 1.008 g/mol): - Moles of C: \(92.3/12.01 \approx 7.68\) - Moles of H: \(7.7/1.008 \approx 7.64\)

Calculate the mole ratios

Now, divide the moles of each element by the smallest number of moles to find the mole ratio: - Mole ratio of C: \(7.68/7.64 \approx 1\) - Mole ratio of H: \(7.64/7.64 = 1\)

Determine the empirical formula

With the mole ratios, we can write the empirical formula as CH.

Calculate the molecular formula

The molar mass of CH is given by (12.01 + 1.008) = 13.018 g/mol. Then, we can divide the molar mass of Styrene (104 g/mol) by the empirical formula molar mass (13.018 g/mol) to find the molecular formula: Molecular formula factor: \(104/13.018 \approx 8\) So, multiplying the empirical formula by the factor gives the molecular formula: C8H8. The empirical formula for Styrene is CH, and the molecular formula is C8H8. (b) Caffeine: Follow the same steps as in part (a) with the appropriate mass percentages and molar masses. After performing the calculations, the empirical formula for Caffeine is found to be C4H5N2O, and the molecular formula is C8H10N4O2. (c) Monosodium Glutamate (MSG): Again, follow the same steps as in part (a) with the appropriate mass percentages and molar masses. After performing the calculations, the empirical formula for MSG is found to be C5H8NO4Na, and since the molar mass of MSG matches the molar mass of this empirical formula, the molecular formula is also C5H8NO4Na.

Key concepts

Empirical Formula Calculation

Understanding how to calculate an empirical formula is a fundamental skill in chemistry. Simply put, the empirical formula represents the simplest whole-number ratio of the elements in a compound. It is determined by converting the mass percentages of each element to moles, then reducing these values to the simplest whole-number ratio.

Let's consider how to calculate the empirical formula using the styrene example from the exercise. First, we assume a 100 g sample, which makes it easy to convert percentages to grams. With 92.3 g of carbon (C) and 7.7 g of hydrogen (H), the next step involves converting these amounts to moles using the molar masses of each element. To obtain the mole ratio, you divide the moles of each element by the smallest number of moles calculated. If necessary, the mole ratios are adjusted to the nearest whole number, which gives us the empirical formula. For Styrene, the empirical formula is CH, indicating a 1:1 ratio of carbon to hydrogen atoms.

Molar Mass

The molar mass plays a crucial role in relating the mass of a substance to the number of particles or moles. It is defined as the mass of one mole of a given substance and is expressed in grams per mole (g/mol). The molar mass of an element is numerically equivalent to its atomic weight from the periodic table and is critical for converting between grams and moles.

The concept of molar mass is applied when determining the molecular formula from the empirical formula. In the case of styrene, the empirical formula has a molar mass of 13.018 g/mol. By dividing the given molar mass of the compound (104 g/mol) by the empirical formula molar mass, we find how many times the empirical formula must be multiplied to get the molecular formula. This factor is then used to scale up the empirical formula to the molecular formula.

Chemical Composition

Chemical composition refers to the identity and ratio of the elements that make up a compound. The percentages of each element in a compound are important in finding both the empirical and molecular formulas. For styrene, we are provided with its composition by mass – 92.3% carbon and 7.7% hydrogen.

These percentages not only lead us to the empirical formula but are also used in stoichiometry calculations, where the understanding of the compound's chemical makeup is necessary for solving problems involving chemical reactions. The composition informs us about the relative amounts of elements and, subsequently, allows us to predict the amounts of reactants needed and products formed in chemical processes.

Stoichiometry

Stoichiometry is the quantitative study of reactants and products in a chemical reaction. It makes use of the conservation of mass and the stoichiometric coefficients from balanced chemical equations to calculate unknown quantities. The process often involves conversion factors such as molar mass, Avogadro’s number, and empirical and molecular formulas.

In the context of this exercise, stoichiometry is applied by first finding the empirical formula, which serves as a base to calculate the molecular formula. Knowing the quantity of one component allows for the calculation of others involved in the reaction using stoichiometric ratios. This ties back to the earlier steps involved in empirical formula calculation and the importance of understanding molar mass and chemical composition to solve stoichiometric problems.