Question

Determine the simplest formulas of the following compounds: (a) the food enhancer monosodium glutamate (MSG), which has the composition \(35.51 \%\) C, \(4.77 \%\) H, \(37.85 \% \mathrm{O}, 8.29 \% \mathrm{~N},\) and \(13.60 \% \mathrm{Na}\) (b) zircon, a diamond-like mineral, which has the composition \(34.91 \% \mathrm{O}, 15.32 \% \mathrm{Si}\), and \(49.76 \% \mathrm{Zr}\) (c) nicotine, which has the composition \(74.0 \%\) C, \(8.65 \% \mathrm{H},\) and \(17.4 \% \mathrm{~N}\)

Short answer

Answer: C5H8O4NNa

Step-by-step solution

(a) Determine the simplest formula for monosodium glutamate (MSG)

1. Assume a 100 gram sample: C: 35.51 g, H: 4.77 g, O: 37.85 g, N: 8.29 g, Na: 13.60 g 2. Convert grams to moles: For C: \(35.51g \times \frac{1 \text{ mol}}{12.01 g} = 2.958 \text{ mol}\) , For H: \(4.77g \times \frac{1 \text{ mol}}{1.008 g} = 4.732 \text{ mol}\), For O: \(37.85g \times \frac{1 \text{ mol}}{16.00 g} = 2.366 \text{ mol}\), For N: \(8.29g \times \frac{1 \text{ mol}}{14.01 g} = 0.592 \text{ mol}\), For Na: \(13.60g \times \frac{1 \text{ mol}}{22.99 g} = 0.591 \text{ mol}\) 3. Divide by the smallest moles (0.591): For C: \(\frac{2.958}{0.591} = 5.00\) , For H: \(\frac{4.732}{0.591} = 8.00\), For O: \(\frac{2.366}{0.591} = 4.00\), For N: \(\frac{0.592}{0.591} = 1.00\), For Na: \(\frac{0.591}{0.591} = 1.00\) 4. Since the ratios are close to whole numbers, we can move on to the next step. 5. Simplest formula for monosodium glutamate (MSG): \(\text{C}_5 \text{H}_8 \text{O}_4 \text{N} \text{Na}\)

(b) Determine the simplest formula for zircon

1. Assume a 100 gram sample: O: 34.91 g, Si: 15.32 g, Zr: 49.76 g 2. Convert grams to moles: For O: \(34.91g \times \frac{1 \text{ mol}}{16.00 g} = 2.182 \text{ mol}\) , For Si: \(15.32g \times \frac{1 \text{ mol}}{28.09 g} = 0.545 \text{ mol}\), For Zr: \(49.76g \times \frac{1 \text{ mol}}{91.22 g} = 0.545 \text{ mol}\) 3. Divide by the smallest moles (0.545): For O: \(\frac{2.182}{0.545} = 4.00\) , For Si: \(\frac{0.545}{0.545} = 1.00\), For Zr: \(\frac{0.545}{0.545} = 1.00\) 4. Since the ratios are close to whole numbers, we can move on to the next step. 5. Simplest formula for zircon: \(\text{ZrSiO}_4\)

(c) Determine the simplest formula for nicotine

1. Assume a 100 gram sample: C: 74.0 g, H: 8.65 g, N: 17.4 g 2. Convert grams to moles: For C: \(74.0g \times \frac{1 \text{ mol}}{12.01 g} = 6.161 \text{ mol}\) , For H: \(8.65g \times \frac{1 \text{ mol}}{1.008 g} = 8.585 \text{ mol}\), For N: \(17.4g \times \frac{1 \text{ mol}}{14.01 g} = 1.242 \text{ mol}\) 3. Divide by the smallest moles (1.242): For C: \(\frac{6.161}{1.242} = 4.96\) , For H: \(\frac{8.585}{1.242} = 6.91\), For N: \(\frac{1.242}{1.242} = 1.00\) 4. Since the ratios are close to whole numbers, we can use 5:7:1 as our mole ratio. 5. Simplest formula for nicotine: \(\text{C}_{10} \text{H}_{14} \text{N}_2\)

Key concepts

Stoichiometry

Stoichiometry is like a recipe for chemists. It's the part of chemistry that deals with relationships between reactants and products in chemical reactions. By understanding stoichiometry, students can predict how much product will form in a reaction, or how much reactant is needed for a reaction to occur.

For instance, when we look at determining empirical formulas from percentages, we use stoichiometry to convert grams to moles. Since moles reflect the number of particles and not the mass, this conversion allows us to compare the actual number of each type of atom in the compound. By comparing these mole ratios, stoichiometry helps us determine the simplest whole number ratio of atoms in a compound, known as the empirical formula.

Molar Mass

Molar mass is a super helpful concept that connects the atomic world with the one we can see and measure. Think of it as the weight of one mole of a substance. The molar mass lets us convert grams to moles, which is a critical step in determining an empirical formula. For example, carbon has a molar mass of 12.01 grams per mole, so if you have 35.51 grams of carbon, you can convert this to moles by dividing by 12.01.

In the given solution for monosodium glutamate (MSG), different molar masses are used to convert the grams of carbon, hydrogen, oxygen, nitrogen, and sodium into moles. This conversion is fundamental for stoichiometry and gets us one step closer to finding the empirical formula.

Chemical Composition

Understanding the chemical composition of a substance means knowing what elements are present and in what proportions. It's like knowing the ingredients list for a dish. The composition is often given in percentages by mass, which assumes a certain amount of the compound, typically 100 grams, to make calculations straightforward.

For example, to find the empirical formula, you start with the percentage composition, convert that to grams (assuming a 100 gram sample), and then to moles. These mole amounts are used to find the simplest ratio of the atoms, revealing the basic building blocks of the substance's molecules. The chemical composition is the starting point to understanding the substance's molecular structure.

Mole-to-Mole Conversion

Mole-to-mole conversion is the heart of stoichiometry. It allows chemists to use a balanced chemical equation to determine the mole ratio between any two substances within the reaction.

In the context of finding an empirical formula, we use mole-to-mole conversion slightly differently. After converting the grams of each element to moles, we don’t look at the reactions but instead at the ratio between the moles of different elements in the compound. By dividing all of our mole quantities by the smallest number of moles present, we derive the mole-to-mole ratios that give us the simplest whole number ratios—the essence of the empirical formula.