Problem 53 - Chemistry The Central Science

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

(a) Styrene: Molecular formula = Empirical formula = CH (b) Caffeine: Molecular formula = C₈H₁₀N₄O₂ (c) Monosodium Glutamate (MSG): Molecular formula = Empirical formula = C₅H₈O₄NNa

Step 1: Convert percentage composition to grams

Assuming a 100g sample, we have 92.3g of carbon (C) and 7.7g of hydrogen (H).

Step 2: Convert mass to moles

Divide the mass of each element by its molar mass to find the moles: Moles of C: \( \frac{92.3g}{12.01g/mol} = 7.68 \, mol \) Moles of H: \( \frac{7.7g}{1.008g/mol} = 7.64 \, mol \)

Step 3: Find the empirical formula

Divide each mole values with the smallest mole value to find the mole ratio: Mole ratio of C:H: \( \frac{7.68}{7.64}:1 = 1:1 \) So, the empirical formula is CH.

Step 4: Determine the molecular formula

Find the molar mass of the empirical formula. In this case, it's the same as the given molar mass of 104 g/mol as the empirical formula already represents the substance. Molecular formula = Empirical formula = CH (b) Caffeine

Step 1: Convert percentage composition to grams

Assuming a 100g sample, we have 49.5g of carbon (C), 5.15g of hydrogen (H), 28.9g of nitrogen (N), and 16.5g of oxygen (O).

Step 2: Convert mass to moles

Moles of C: \( \frac{49.5g}{12.01g/mol} = 4.12 \, mol \) Moles of H: \( \frac{5.15g}{1.008g/mol} = 5.11 \, mol \) 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 empirical formula

Divide each mole values with the smallest mole value to find the mole ratio: Mole ratio of C:H:N:O: \( 4:5:2:1 \) So, the empirical formula is C₄H₅N₂O.

Step 4: Determine the molecular formula

Find the molar mass of the empirical formula: \( (4 \times 12.01) + (5 \times 1.008) + (2 \times 14.01) + (1 \times 16.00) = 97 \, g/mol \) Divide the given molar mass by the molar mass of the empirical formula to find the multiplier, which in this case is approximately 2. Molecular formula = (Empirical formula) x 2 = C₈H₁₀N₄O₂ (c) Monosodium Glutamate (MSG)

Step 1: Convert percentage composition to grams

Assuming a 100g sample, we have 35.51g of carbon (C), 4.77g of hydrogen (H), 37.85g of oxygen (O), 8.29g of nitrogen (N), and 13.60g of sodium (Na).

Step 2: Convert mass to moles

Moles of C: \( \frac{35.51g}{12.01g/mol} = 2.96 \, mol \) Moles of H: \( \frac{4.77g}{1.008g/mol} = 4.73 \, mol \) 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 empirical formula

Divide each mole values with the smallest mole value to find the mole ratio: Mole ratio of C:H:O:N:Na: \( 5:8:4:1:1 \) So, the empirical formula is C₅H₈O₄NNa.

Step 4: Determine the molecular formula

Find the molar mass of the empirical formula: \( (5 \times 12.01) + (8 \times 1.008) + (4 \times 16.00) + (1 \times 14.01) + (1 \times 22.99) = 169 \, g/mol \) In this case, the molar mass of the empirical formula is equal to the given molar mass. Molecular formula = Empirical formula = C₅H₈O₄NNa

Key Concepts

Percentage Composition

Understanding the percentage composition of a compound is crucial in chemistry as it indicates the relative amount of each element within a substance. This information allows scientists and students alike to determine the empirical formula of a compound, which displays the simplest whole-number ratio of atoms in the molecule.

Let's simplify it with an analogy: Imagine you have a recipe that tells you the percentages of each ingredient needed. If you know the total quantity of the recipe, you can calculate the exact amount of each ingredient required. In chemistry, percentage composition serves a similar purpose, but instead of ingredients for a cake, we're dealing with elements in a compound.

In analytical chemistry, often a sample is assumed to be 100 grams for ease of calculation. This way, the percentage can be directly converted into grams — just like we see in our example involving Styrene, where 92.3% carbon translates to 92.3 grams of carbon.

Molar Mass

Molar mass is like the 'chemical weight' of a substance. It refers to the weight of one mole (6.022 × 10²³ particles) of a compound or element and is expressed in grams per mole (g/mol).

Here’s an easy comparison: Every fruit has its average weight, like an apple might weigh around 150 grams. In the same manner, each element on the periodic table has its own molar mass. Carbon’s molar mass, for instance, is 12.01 g/mol. It is a critical concept when converting between the mass of a substance and the number of moles, laying the groundwork for further stoichiometric calculations.

Mole Ratio

The mole ratio is the bridge between the elements in a reaction. It tells us how the atoms of different elements compare to each other in a compound. Imagine beads on a string, where the pattern is the formula, and the mole ratio tells you how many beads of each color you need. The simpler the pattern, the easier it is to replicate, this is how empirical formulas work.

By dividing the number of moles of each element by the smallest number of moles present, the empirical formula can be determined, showing the simplest ratio among the atoms. This is evident in our Styrene example, where the mole ratio of carbon to hydrogen ended up being 1:1, leading to the very straightforward empirical formula CH.

Stoichiometry

Stoichiometry is the backbone of chemical reactions. It involves calculations that relate to the quantities of substances involved in reactions. Think of it as the math behind a chemical recipe: It tells you how much of each reactant you need and how much product you'll get.

For instance, in cooking, if you know you need twice as much flour as sugar for a cake, stoichiometry would be the equivalent of calculating how much flour and sugar is required for 10 cakes. In our exercises, stoichiometry is applied to find the molecular formula, often seen as an integral step in the determination of the compound’s exact structure based on its empirical formula and molecular weight.