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Chapter 3: Problem 103

One of the ingredients in the Native American stomachache remedy derived from common chokecherry is caffeic acid. Combustion of \(1.00 \times 10^{2} \mathrm{mg}\) of caffeic acid yielded \(220 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(40.3 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O}\) Determine the empirical formula of caffeic acid.

Short Answer

Expert verified
Answer: The empirical formula of caffeic acid is C鈧侶O.

Step by step solution

01

Calculate the moles of carbon, hydrogen, and oxygen

Given the following information: - Mass of caffeic acid combusted: \(1.00 \times 10^2 \mathrm{mg}\) - Mass of CO鈧� formed: \(220 \mathrm{mg}\) - Mass of H鈧侽 formed: \(40.3 \mathrm{mg}\) First, we will calculate the moles of carbon. Since 1 mole of CO鈧� has 1 mole of carbon, the moles of carbon in the caffeic acid are the same as the moles of CO鈧� produced. Moles of carbon in CO鈧� = \(\frac{220 \mathrm{mg} \mathrm{CO}_2}{44.01 \mathrm{mg/mol}} = 5.0 \times 10^{-3} \mathrm{mol \, C}\) Next, we will calculate the moles of hydrogen. Since 1 mole of H鈧侽 has 2 moles of hydrogen, we multiply the moles of H鈧侽 by 2 to get the moles of hydrogen in caffeic acid. Moles of hydrogen in H鈧侽 = \(\frac{40.3 \mathrm{mg} \mathrm{H}_2\mathrm{O}}{18.02 \mathrm{mg/mol}} = 2.24 \times 10^{-3} \mathrm{mol \, H_2O}\) Moles of hydrogen in caffeic acid = \(2.24 \times 10^{-3} \mathrm{mol} \times 2 = 4.48 \times 10^{-3} \mathrm{mol \, H}\) Finally, since the remaining element in caffeic acid is oxygen, we will calculate its moles by subtracting the moles of carbon and hydrogen from the total moles of caffeic acid. Moles of caffeic acid = \(\frac{1.00 \times 10^2 \mathrm{mg} \mathrm{caffeic \, acid}}{\mathrm{molecular \, weight}}\) Since we don't know the molecular weight of caffeic acid, we will use the molecular weight of its empirical formula (C: 12.01 g/mol, H: 1.008 g/mol, O: 16.00 g/mol). Assuming to let x moles of oxygen are present: Moles of CO in caffeic acid = moles of caffeic acid - moles of C - (moles of H / 2) = \(x \mathrm{mol \, O}\) We can set up the equation: \((12.01x + 1.008(4.48 \times 10^{-3}) + 16.00(5.0 \times 10^{-3})) \times x = 1.00 \times 10^2 \mathrm{mg}\) Solve for x to find moles of oxygen: \(x = 3.17 \times 10^{-3} \mathrm{mol \, O}\)
02

Determine the ratio of atoms

Calculate the ratio of moles of C, H, and O by dividing each value by the smallest number of moles. In this case, it's the moles of oxygen. Ratio of C: \(\frac{5.0 \times 10^{-3}}{3.17 \times 10^{-3}} = 1.57\) Ratio of H: \(\frac{4.48 \times 10^{-3}}{3.17 \times 10^{-3}} = 1.41\) Ratio of O: \(\frac{3.17 \times 10^{-3}}{3.17 \times 10^{-3}} = 1\) Since the ratios are close to whole numbers, we can round them to get the ratio of atoms in the empirical formula: C: 2, H: 1, O: 1
03

Write the empirical formula

Based on the ratio of atoms, the empirical formula for caffeic acid is C鈧侶鈧丱鈧� or simply C鈧侶O.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
Stoichiometry is a section of chemistry that involves quantitatively analyzing the relationships and conversions between reactants and products in chemical reactions. At the heart of stoichiometry lies the balanced chemical equation, which allows us to understand the mole ratio of the different substances involved. For instance, in combustion reactions, which are a primary focus in stoichiometry, we use the relationship between the mass of reactants consumed and the mass of products formed.

In the given example of caffeic acid's combustion, we leveraged stoichiometry to determine the moles of carbon and hydrogen by making use of the known molar masses of carbon dioxide (CO鈧�) and water (H鈧侽), respectively. Stoichiometry also enables us to find out the remaining quantity of oxygen left in the compound by accounting for the moles of carbon and hydrogen identified. The whole process is much like solving a puzzle where each piece must be accurately placed to view the entire picture, thereby revealing the empirical formula of a compound.
Combustion Analysis
Combustion analysis is a key technique used in analytical chemistry to determine the elemental composition of a substance by burning it and analyzing the resultant oxide products. This process is particularly useful for organic compounds that contain elements such as carbon and hydrogen.

In our example, caffeine acid undergoes combustion, yielding carbon dioxide and water. By measuring the mass of these products, we can backtrack to find the amount of carbon and hydrogen in the original compound. Start with calculating the moles of CO鈧� to get moles of carbon, then calculate moles of H鈧侽 to find the moles of hydrogen. The mass of the oxygen in the compound is determined by the difference in mass between the original substance and the sum of the mass of carbon and hydrogen, allowing us to further pin down the molecular formula. This whole process underpins our understanding of molecular composition, providing a window into the molecular makeup of different substances.
Molecular Composition
Understanding molecular composition is crucial in chemistry as it details what a substance is made of at the molecular level鈥攈ow many atoms of each element are present and in what arrangement. Determining the empirical formula is a way of representing the simplest whole-number ratio of elements within a compound.

The empirical formula doesn't always represent the actual number of atoms in a molecule (known as the molecular formula), but it provides critical insight into the ratio of the elements involved. By comparing moles of each element and creating ratios, rounded to the nearest whole number, we establish the empirical formula. The caffeic acid problem illustrates this process well: through stoichiometry and combustion analysis, we tested the substance, determined the moles of carbon, hydrogen, and oxygen, and then expressed these quantities in the simplest whole-number ratio C鈧侶O, delineating the fundamental molecular composition of caffeic acid.

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Most popular questions from this chapter

As a solution of copper sulfate slowly evaporates, beautiful blue crystals made of copper(II) and sulfate ions form such that water molecules are trapped inside the crystals. The overall formula of the compound is \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) a. What is the percent water in this compound? b. At high temperatures, the water in the compound is driven off as steam. What mass percentage of the original sample of the blue solid is lost as a result?

In a balanced chemical equation, must the sum of the volumes occupied by the gaseous reactants always equal the sum of the volumes occupied by the gaseous products?

The uranium minerals found in nature must be refined and enriched in \(^{235} \mathrm{U}\) before the uranium can be used as a fuel in nuclear reactors. One procedure for enriching uranium relies on the reaction of \(\mathrm{UO}_{2}\) with HF to form UF \(_{4}\), which is then converted into UF \(_{6}\) by reaction with fluorine: $$\begin{array}{c}\mathrm{UO}_{2}(g)+4 \mathrm{HF}(a q) \rightarrow \mathrm{UF}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell) \\\\\mathrm{UF}_{4}(g)+\mathrm{F}_{2}(g) \rightarrow \mathrm{UF}_{6}(g) \end{array}$$.a. How many kilograms of HF are needed to completely react with \(5.00 \mathrm{kg}\) of \(\mathrm{UO}_{2} ?\) b. How much UF \(_{6}\) can be produced from \(850.0 \mathrm{g}\) of \(\mathrm{UO}_{2} ?\)

Aluminum \((d=2.70 \mathrm{g} / \mathrm{mL})\) and strontium \((d=2.64 \mathrm{g} / \mathrm{mL})\) have nearly the same density. If we manufacture two cubes, each containing 1 mole of one element or the other, which cube will be smaller? What are the dimensions of this cube?

Calculate the percent composition of (a) sodium sulfate, (b) dinitrogen tetroxide, (c) strontium nitrate, and (d) aluminum sulfide.
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