Problem 53 - Chemistry: The Central Science in SI Units, Global Edition

Question

Determine the empirical and molecular formulas of each of the following substances: (a) Styrene, a compound used to make Styrofoam \(^{*}\) 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 \(49.5 \% \mathrm{C}\), \(5.15 \% \mathrm{H}, 28.9 \% \mathrm{~N},\) 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 \(35.51 \% \mathrm{C}, 4.77 \% \mathrm{H}, 37.85 \% \mathrm{O},\) \(8.29 \% \mathrm{~N},\) and \(13.60 \% \mathrm{Na},\) and has a molar mass of \(169 \mathrm{~g} / \mathrm{mol} .\)

Step-by-Step Solution

Answer

The molecular formulas for the given substances are as follows: (a) Styrene: C8H8 (b) Caffeine: C8H10N4O2 (c) Monosodium glutamate (MSG): C5H8O4NNa

Step 1: Convert mass percentages to grams

Since the mass percentages are given, assume a 100g sample. This means we have: - 92.3g of Carbon (C) - 7.7g of Hydrogen (H)

Step 2: Convert grams to moles

Using the molar masses, convert the grams to moles: - Moles of C: \( \frac{92.3g}{12.01g/mol} = 7.69 \, mol \) - Moles of H: \( \frac{7.7g}{1.01g/mol} = 7.62 \, mol \)

Step 3: Find the mole ratio

Divide the moles by the smallest number of moles: - Mole ratio of C: \( \frac{7.69}{7.62} = 1.01\) - Mole ratio of H: \( \frac{7.62}{7.62} = 1\)

Step 4: Obtain whole-number subscripts

The mole ratios are already nearly whole numbers, so no need to multiply. The subscripts are: - C: 1 - H: 1

Step 5: Determine the empirical formula

The empirical formula of Styrene is CH.

Step 6: Calculate the empirical formula mass and find the molecular formula

The empirical formula mass for CH is (12.01 + 1.01) = 13.02 g/mol. Since the molar mass is given as 104 g/mol, we can divide it by the empirical formula mass: - \( \frac{104}{13.02} = 8 \) Thus, the molecular formula is 8(CH) which is C8H8. #(b) Caffeine# Following the same steps as above:

Step 1: Convert mass percentages to grams

Assume a 100g sample: - 49.5g of Carbon (C) - 5.15g of Hydrogen (H) - 28.9g of Nitrogen (N) - 16.5g of Oxygen (O)

Step 2: Convert grams to moles

Using the molar masses, convert the grams to moles: - Moles of C: \( \frac{49.5g}{12.01g/mol} = 4.12 \, mol \) - Moles of H: \( \frac{5.15g}{1.01g/mol} = 5.1 \, mol \) - Moles of N: \( \frac{28.9g}{14.01g/mol} = 2.06 \, mol \) - Moles of O: \( \frac{16.5g}{16.00g/mol} = 1.03 \, mol \)

Step 3: Find the mole ratio

Divide the moles by the smallest number of moles: - Mole ratio of C: \( \frac{4.12}{1.03} = 4 \) - Mole ratio of H: \( \frac{5.1}{1.03} = 5 \) - Mole ratio of N: \( \frac{2.06}{1.03} = 2 \) - Mole ratio of O: \( \frac{1.03}{1.03} = 1 \)

Step 5: Determine the empirical formula

The empirical formula of Caffeine is C4H5N2O.

Step 6: Calculate the empirical formula mass and find the molecular formula

The empirical formula mass for C4H5N2O is (4*12.01 + 5*1.01 + 2*14.01 + 1*16.00) = 97.09 g/mol. Since the given molar mass is 195 g/mol, we can divide it by the empirical formula mass: - \( \frac{195}{97.09} = 2 \) Thus, the molecular formula of Caffeine is 2(C4H5N2O) which is C8H10N4O2. #(c) Monosodium glutamate (MSG)# Following the same steps as above:

Step 1: Convert mass percentages to grams

Assume a 100g sample: - 35.51g of Carbon (C) - 4.77g of Hydrogen (H) - 37.85g of Oxygen (O) - 8.29g of Nitrogen (N) - 13.60g of Sodium (Na)

Step 2: Convert grams to moles

Using the molar masses, convert the grams to moles: - Moles of C: \( \frac{35.51g}{12.01g/mol} = 2.96 \, mol \) - Moles of H: \( \frac{4.77g}{1.01g/mol} = 4.72 \, mol \) - Moles of O: \( \frac{37.85g}{16.00g/mol} = 2.37 \, mol \) - Moles of N: \( \frac{8.29g}{14.01g/mol} = 0.592 \, mol \) - Moles of Na: \( \frac{13.60g}{22.99g/mol} = 0.592 \, mol \)

Step 3: Find the mole ratio

Divide the moles by the smallest number of moles: - Mole ratio of C: \( \frac{2.96}{0.592} = 5 \) - Mole ratio of H: \( \frac{4.72}{0.592} = 8 \) - Mole ratio of O: \( \frac{2.37}{0.592} = 4 \) - Mole ratio of N: \( \frac{0.592}{0.592} = 1 \) - Mole ratio of Na: \( \frac{0.592}{0.592} = 1 \)

Step 5: Determine the empirical formula

The empirical formula of MSG is C5H8O4N1Na1, or C5H8O4NNa.

Step 6: Calculate the empirical formula mass and find the molecular formula

The empirical formula mass for C5H8O4NNa is (5*12.01 + 8*1.01 + 4*16.00 + 14.01 + 22.99) = 169.00 g/mol. Since the given molar mass is 169 g/mol, the empirical formula mass and molar mass are equal. Thus, the molecular formula of Monosodium glutamate is C5H8O4NNa.

Key Concepts

Mole Ratio

The concept of mole ratio plays a fundamental role in determining empirical formulas. It all starts with calculating the moles of each element in a compound from their given mass or percentage composition. By converting the mass to moles using the element's molar mass, you obtain the number of moles for each element. To determine the mole ratio, these calculated moles are divided by the smallest number of moles present in the list. This approach scales down the values to a simple ratio. For example, if you find out you have 4 moles of carbon and 2 moles of hydrogen, you divide each by 2, providing a 2:1 ratio. This gives a clear picture of how each element relates to the others in terms of quantity. This ratio helps to establish the subscripts in the compound's empirical formula, translating lab findings into chemical language.

Molar Mass

Understanding molar mass is crucial when identifying molecular formulas. Molar mass, given in grams per mole (g/mol), represents the mass of a mole of a compound. By knowing this value, you can relate the empirical formula to the molecular formula. Take our Styrene example, with a molar mass of 104 g/mol. By first finding the empirical formula mass, which combines the atomic masses of all atoms in the empirical formula, you can see how many times the empirical formula units fit into the given molar mass. For instance, if the empirical formula mass is 13.02 g/mol, dividing 104 g/mol by 13.02 g/mol gives about 8. This means the actual molecule consists of eight times the empirical formula, leading to the molecular formula of C8H8.

Empirical Formula Mass

After determining the empirical formula, the next task is calculating its mass, known as the empirical formula mass. This process involves summing the atomic masses of the constituent elements in their empirical proportions. For example, consider a compound with an empirical formula of CH. We calculate its empirical formula mass by adding together the atomic mass of Carbon (12.01 g/mol) and Hydrogen (1.01 g/mol), resulting in 13.02 g/mol. This mass acts as a checkpoint when deducing the molecular formula. If the compound's molar mass is known, dividing it by the empirical formula mass reveals how many times the empirical formula repeats in the actual molecule, thereby guiding you to the true molecular formula.