Question

Determine the empirical and molecular formulas of each of the following substances: (a) Styrene, a compound substance used to make Styrofoam \(^{8}\) cups and insulation, contains \(92.3 \% \mathrm{C}\) and \(7.7 \%\) H by mass and has a molar mass of \(104 \mathrm{~g} / \mathrm{mol}\). (b) Caffeine, a stimulant found in coffee, contains \(49.5 \%\) \(\mathrm{C}, 5.15 \% \mathrm{H}, 28.9 \% \mathrm{~N}\), 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, \(4.77 \%\) H, \(37.85 \%\) O, \(8.29 \% \mathrm{~N}\), and \(13.60 \% \mathrm{Na}\), and has a molar mass of \(169 \mathrm{~g} / \mathrm{mol}\).

Short answer

The empirical and molecular formulas for the given substances are as follows: (a) Styrene: Empirical Formula - CH, Molecular Formula - C8H8 (b) Caffeine: Empirical Formula - C4H5N2O, Molecular Formula - C4H5N2O (c) Monosodium glutamate (MSG): Empirical Formula - C5H8O4N1Na1, Molecular Formula - C5H8O4N1Na1

Step-by-step solution

Convert the percentage composition into grams

Assuming 100 grams of the substance, we can directly use the given percentages as the mass of each element in the compound: Carbon (C) = 92.3 g Hydrogen (H) = 7.7 g

Convert grams to moles

Divide the mass of each element by its molar mass to get the number of moles: Moles of Carbon (C) = \( \frac{92.3}{12.01} \) ≈ 7.69 moles Moles of Hydrogen (H) = \( \frac{7.7}{1.008} \) ≈ 7.63 moles

Find the mole ratio

Divide the moles of each element by the smallest number of moles to obtain the whole number ratio: Mole ratio (C:H) = \( \frac{7.69}{7.63} : \frac{7.63}{7.63} \) ≈ 1:1 The empirical formula for Styrene is CH.

Calculate the molecular formula

Determine the molecular formula for Styrene by multiplying the empirical formula by the whole number factor (n) obtained from the molar mass: Empirical formula mass of Styrene (CH) = 12.01 + 1.008 = 13.018 g/mol n = \( \frac{104}{13.018} \) ≈ 8 The molecular formula for Styrene is C8H8. **For substance (b): Caffeine**

Convert percentage composition into grams

Assuming 100 grams of the substance, we have: Carbon (C) = 49.5 g Hydrogen (H) = 5.15 g Nitrogen (N) = 28.9 g Oxygen (O) = 16.5 g

Convert grams to moles

Moles of each element: Carbon (C) = \( \frac{49.5}{12.01} \) ≈ 4.12 moles Hydrogen (H) = \( \frac{5.15}{1.008} \) ≈ 5.11 moles Nitrogen (N) = \( \frac{28.9}{14.01} \) ≈ 2.06 moles Oxygen (O) = \( \frac{16.5}{16.00} \) ≈ 1.03 moles

Find the mole ratio

Mole ratio (C:H:N:O) = \( \frac{4.12}{1.03} : \frac{5.11}{1.03} : \frac{2.06}{1.03} : \frac{1.03}{1.03} \) ≈ 4:5:2:1 The empirical formula for Caffeine is C4H5N2O.

Calculate the molecular formula

The empirical formula mass of C4H5N2O is 97.11 g/mol. Since the molar mass of Caffeine is 195 g/mol, the molecular formula is the same as the empirical formula: The molecular formula for Caffeine is C4H5N2O. **For substance (c): Monosodium glutamate (MSG)**

Convert percentage composition into grams

Assuming 100 grams of the substance, we have: Carbon (C) = 35.51 g Hydrogen (H) = 4.77 g Oxygen (O) = 37.85 g Nitrogen (N) = 8.29 g Sodium (Na) = 13.60 g

Convert grams to moles

Moles of each element: Carbon (C) = \( \frac{35.51}{12.01} \) ≈ 2.96 moles Hydrogen (H) = \( \frac{4.77}{1.008} \) ≈ 4.74 moles Oxygen (O) = \( \frac{37.85}{16.00} \) ≈ 2.37 moles Nitrogen (N) = \( \frac{8.29}{14.01} \) ≈ 0.59 moles Sodium (Na) = \( \frac{13.60}{22.99} \) ≈ 0.59 moles

Find the mole ratio

Mole ratio (C:H:O:N:Na) = \( \frac{2.96}{0.59} : \frac{4.74}{0.59} : \frac{2.37}{0.59} : \frac{0.59}{0.59} : \frac{0.59}{0.59} \) ≈ 5:8:4:1:1 The empirical formula for Monosodium glutamate (MSG) is C5H8O4N1Na1.

Calculate the molecular formula

The empirical formula mass of MSG (C5H8O4N1Na1) is 169 g/mol, which is the same as the given molar mass. Thus, the molecular formula is the same as the empirical formula: The molecular formula for Monosodium glutamate (MSG) is C5H8O4N1Na1.

Key concepts

Molar Mass Calculation

Molar mass is a fundamental concept in chemistry that refers to the mass of one mole of a given substance. The unit for molar mass is grams per mole (g/mol). To calculate the molar mass, we sum the masses of all the atoms in a molecule based on the atomic masses found on the periodic table.

For instance, to find the molar mass of water (H₂O), you would add the molar masses of two hydrogen atoms (2 x 1.008 g/mol) with one oxygen atom (16.00 g/mol), giving a molar mass of approximately 18.02 g/mol. In the context of the exercise where we determine the molar mass for various substances, this step is vital as it allows for the conversion of the mass of each element into moles, enabling the further determination of the empirical and molecular formulas.

Mole Ratio Determination

Determining the mole ratio is essential when decoding the empirical formula of a compound. After the conversion of each element's mass (in grams) to the number of moles, the next step is to derive the simplest whole number ratio of these elements in the compound.

This process begins by dividing all of the elements' mole quantities by the smallest mole value obtained, leading to the simplest whole number ratio. In our earlier example for Styrene, the moles of carbon and hydrogen were divided by the smallest mole quantity to get a mole ratio of 1:1. This basic mole ratio reflects the simplest ratio of atoms within the molecule, which is CH for Styrene. Moreover, when solving empirical and molecular formulas, it's sometimes necessary to multiply these ratios by a common factor to obtain whole numbers, ensuring the accuracy of the empirical formula.

Chemical Composition Analysis

Chemical composition analysis involves determining the percentage by mass of each element within a compound. Knowledge of chemical composition is critical for calculating empirical and molecular formulas. In chemistry, it is customary to assume a 100-gram sample of a substance when assessing percentage composition. This simplification allows the percentage values to be directly treated as mass in grams.

In the given examples, the substances are analyzed for their composition by mass: Styrene with 92.3% C and 7.7% H, Caffeine with a distribution among C, H, N, and O, and MSG with its more complex mixture of C, H, O, N, and Na. Understanding the chemical composition is the starting point for molar mass and mole ratio calculations that ultimately lead to the empirical and molecular formulas vital for the identification of substances.