Problem 53 - Chemistry The Central Science In Si Units

Question

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

Step-by-Step Solution

Answer

(a) Styrene: Empirical = CH, Molecular = C8H8. (b) Caffeine: Empirical = C4H5N2O, Molecular = C8H10N4O2. (c) MSG: Empirical = Molecular = C5H8NO4Na.

Step 1: Calculate Moles for Styrene

Given mass percentages, for Styrene, which has 92.3% C and 7.7% H:- Assume 100 g of Styrene for easier calculation: Carbon: \(92.3\, \text{g} = \frac{92.3}{12.01} = 7.68\, \text{mol} \) of C, Hydrogen: \(7.7\, \text{g} = \frac{7.7}{1.008} = 7.64\, \text{mol} \) of H.

Step 2: Determine Simplest Ratio for Styrene

Divide the number of moles of each element by the smallest number of moles to find the simplest whole number ratio:- For Carbon: \(\frac{7.68}{7.64} \approx 1\), For Hydrogen: \(\frac{7.64}{7.64} = 1\).Thus, the empirical formula is \(CH\).

Step 3: Determine Molecular Formula for Styrene

Find the empirical formula mass: \(CH = 12.01 + 1.008 = 13.018\, \text{g/mol}\).Divide the molar mass by the empirical formula mass to find the factor: \( \frac{104}{13.018} \approx 8\).Multiply the empirical formula by this factor: \((CH)_8 = C_8H_8\).

Step 4: Calculate Moles for Caffeine

Assuming 100 g of Caffeine:- Carbon: \(49.5\, \text{g} = \frac{49.5}{12.01} = 4.12\, \text{mol} \)- Hydrogen: \(5.15\, \text{g} = \frac{5.15}{1.008} = 5.11\, \text{mol} \)- Nitrogen: \(28.9\, \text{g} = \frac{28.9}{14.01} = 2.06\, \text{mol} \)- Oxygen: \(16.5\, \text{g} = \frac{16.5}{16.00} = 1.03\, \text{mol} \).

Step 5: Determine Simplest Ratio for Caffeine

Divide by the smallest number of moles, which is Oxygen's 1.03:- For Carbon: \(\frac{4.12}{1.03} = 4\), For Hydrogen: \(\frac{5.11}{1.03} = 5\), For Nitrogen: \(\frac{2.06}{1.03} = 2\).Thus, the empirical formula is \(C_4H_5N_2O\).

Step 6: Determine Molecular Formula for Caffeine

Calculate the empirical formula mass: \(4(12.01) + 5(1.008) + 2(14.01) + 16.00 = 97.11\, \text{g/mol} \).Divide the molar mass by the empirical formula mass: \( \frac{195}{97.11} \approx 2 \).Thus, the molecular formula is \(C_8H_{10}N_4O_2\).

Step 7: Calculate Moles for MSG

Assuming 100 g of MSG:- Carbon: \(35.51\, \text{g} = \frac{35.51}{12.01} = 2.96\, \text{mol} \)- Hydrogen: \(4.77\, \text{g} = \frac{4.77}{1.008} = 4.73\, \text{mol} \)- Oxygen: \(37.85\, \text{g} = \frac{37.85}{16.00} = 2.37\, \text{mol} \)- Nitrogen: \(8.29\, \text{g} = \frac{8.29}{14.01} = 0.592\, \text{mol} \)- Sodium: \(13.60\, \text{g} = \frac{13.60}{22.99} = 0.592\, \text{mol} \).

Step 8: Determine Simplest Ratio for MSG

Divide all mole values by Nitrogen's 0.592 (minimum value):- For Carbon: \(\frac{2.96}{0.592} \approx 5\), For Hydrogen: \(\frac{4.73}{0.592} \approx 8\), For Oxygen: \(\frac{2.37}{0.592} \approx 4\), For Nitrogen: \(1\), For Sodium: \(1\).The empirical formula is \(C_5H_8NO_4Na\).

Step 9: Determine Molecular Formula for MSG

Calculate the empirical formula mass: \(5(12.01) + 8(1.008) + 14.01 + 4(16.00) + 22.99 = 169.11\, \text{g/mol} \).The molar mass matches the empirical formula mass; thus, the molecular formula is the same as the empirical formula: \(C_5H_8NO_4Na\).

Key Concepts

Chemical Composition Analysis

Chemical composition analysis is crucial in understanding what a substance is made from. It helps us identify the types and amounts of elements within a compound.
Imagine you have a cake, but you want to know how much flour, sugar, and butter is in it. You can do this by examining its "recipe," or in chemical terms, its composition.

To determine the composition of a substance, we start with mass percentages of each element present. For instance, in Styrene, we know it consists of 92.3% carbon (C) and 7.7% hydrogen (H). By analyzing these percentages, we can figure out the basic formula that represents the simplest ratio of atoms in the compound, which is termed the empirical formula.

• Mass percentages provide a straightforward starting point.
• Understanding these concepts is necessary for deeper chemical analysis.

Mole Concept

The mole concept is a foundational principle in chemistry that bridges the gap between the atomic world and the macroscopic world we observe. It allows us to count atoms by weighing them.

By definition, 1 mole of any substance contains exactly Avogadro's number of particles, which is approximately \( 6.022 \times 10^{23} \) particles. This could be atoms, molecules, ions, or electrons.

In practical terms, when we say there is a mole of salt, it means there are about \( 6.022 \times 10^{23} \) salt molecules present. When calculating the empirical formula of Styrene, for example, we use the mole concept by converting mass percentages into moles.
This step involves dividing the mass of each element by its atomic mass.

• Utilizing the mole concept allows chemists to work with reactants and products on a scale they can observe and measure directly.
• It is fundamental to linking masses observed in labs with quantities counted in atomic terms.

Mass Percentages

Mass percentages are used to express the proportions of different elements within a compound. They indicate how much of each element there is compared to the entire compound.

To find these percentages, take the mass of the specific element, divide it by the total mass of the compound, and then multiply by 100 to convert it into a percentage.

• This process helps us understand how much of each element is contained in a substance.
• Mass percentages serve as the foundation for finding empirical formulas.

For example, in the exercise, caffeine has a mass percentage composition of 49.5% carbon, 5.15% hydrogen, 28.9% nitrogen, and 16.5% oxygen. These percentages help us calculate the moles, which is the next step towards determining the empirical formula.

Step-by-Step Problem Solving

Solving chemical problems by breaking them down into smaller steps can make complex processes easier to manage and understand. Take the analysis of Monosodium Glutamate (MSG) in this exercise as an example.

First, start by assuming a 100 g sample to simplify calculations. Then, calculate the moles of each element using their respective mass and atomic mass.

• Breaking the problem down into steps makes it more understandable.
• Comparison of resulting moles helps in determining the simplest whole number ratio of atoms.

Once you have the smallest ratio, determine the empirical formula, and compare with the known molar mass to check if it fits, adjusting to find the molecular formula if necessary. This stepwise approach can be applied to almost any chemical calculation, making it a valuable skill in both education and practical chemistry work.