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Cinnamaldehyde is a compound that is responsible for the characteristic aroma of cinnamon. It contains \(81.79 \% \mathrm{C}\), \(6.10 \% \mathrm{H},\) and the remaining is oxygen. Its molar mass is \(132 \mathrm{~g} / \mathrm{mol}\). Determine its molecular formula.

Short Answer

Expert verified
The molecular formula of cinnamaldehyde is \( \text{C}_9\text{H}_8\text{O} \).

Step by step solution

01

Determine the Mass of Oxygen

First, calculate the mass percentage of oxygen. Since the total percentage must equal 100%, the percentage of oxygen is given by:\[ 100 ext{ ext%} - 81.79 ext{ ext%} - 6.10 ext{ ext%} = 12.11 ext{ ext%} \]
02

Convert Percentages to Grams

Assume 100 g of cinnamaldehyde, where the percentages translate directly to grams: - Carbon: 81.79 g - Hydrogen: 6.10 g - Oxygen: 12.11 g
03

Convert Grams to Moles

Convert the mass of each element to moles using their respective molar masses:- Moles of C: \( \frac{81.79}{12.01} = 6.81 \text{ moles} \)- Moles of H: \( \frac{6.10}{1.008} = 6.05 \text{ moles} \)- Moles of O: \( \frac{12.11}{16.00} = 0.756 \text{ moles} \)
04

Find the Simplest Mole Ratio

Divide each number of moles by the smallest number of moles to find the simplest whole number ratio:- Carbon: \( \frac{6.81}{0.756} = 9.01 \approx 9 \)- Hydrogen: \( \frac{6.05}{0.756} = 8 \)- Oxygen: \( \frac{0.756}{0.756} = 1 \)
05

Determine the Empirical Formula

Using the mole ratios, the empirical formula is \( \text{C}_9\text{H}_8\text{O} \).
06

Calculate the Empirical Formula Mass

Calculate the molar mass of the empirical formula:\[ (9 imes 12.01) + (8 imes 1.008) + (1 imes 16.00) = 132.17 \text{ g/mol} \]
07

Determine the Molecular Formula

The empirical formula mass of 132.17 g/mol matches the given molar mass of 132 g/mol. Therefore, the empirical formula is also the molecular formula: \( \text{C}_9\text{H}_8\text{O} \).

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Key Concepts

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

Empirical Formula
An empirical formula provides the simplest whole-number ratio of atoms in a compound. This basic formula doesn't reflect the actual quantity of atoms but represents the proportion of each element in the molecule. Think of the empirical formula as the foundation, giving us the basic ratio of all elements involved. For instance, in the cinnamaldehyde example, when determining the empirical formula, we first ensure that the mass fractions of all the components (Carbon, Hydrogen, and Oxygen) add up to 100%, with Oxygen being calculated as the difference between 100% and the sum of Carbon and Hydrogen percentages.

Next, assume a mass of 100 grams of the compound to translate percentages directly into grams. Then, convert these masses into moles using the molar mass of each element. By dividing the number of moles for each element by the smallest number of moles calculated, you obtain the simplest ratio.

In the exercise, after setting the moles of each element in a simple ratio, we arrived at the formula \( \text{C}_9\text{H}_8\text{O} \), which represents the empirical—not yet confirmed as molecular—formula.
Molecule Composition
Molecule composition refers to the specific arrangement and types of atoms present in a single molecule of a substance. This composition is initially hinted at by the empirical formula, but the complete picture emerges with the molecular formula.

To determine the molecular formula, one needs to know the molar mass of the compound. In cinnamaldehyde's case, both empirical and molar masses are close enough to hint that the empirical formula \( \text{C}_9\text{H}_8\text{O} \) might actually be the molecular formula.

The molecular formula can sometimes be a simple multiple of the empirical formula. In case of differences, the molar mass will help scale up the empirical formula to match the true molecular composition. This is where stoichiometry becomes crucial, ensuring all atom counts and their respective atomic weights add up precisely to the known molar mass of the compound.
Stoichiometry
Stoichiometry is the mathematical heart of chemistry, focused on the quantitative relationships between substances as they undergo chemical reactions. With stoichiometry, you ensure that the law of conservation of mass holds true, meaning no atoms are lost or gained in the reactions.

In exercises involving empirical and molecular formulas, stoichiometry helps leverage known quantities to find unknown ones, using balanced equations and molar mass conversions. Consider the cinnamaldehyde example, where stoichiometry guided us through converting mass percentages to moles, calculating empirical formulas, and verifying molecular formulas.

By understanding stoichiometry, you're equipped to make accurate predictions about the outcomes of chemical reactions and the quantities of reactants needed. It is a fundamental tool that bridges experimental data with theoretical chemistry, making chemical calculations predictable and manageable across various conditions.

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Most popular questions from this chapter

(a) When the metallic element lithium combines with the nonmetallic element chlorine, \(\mathrm{Cl}_{2}(g),\) what is the chemical formula of the product? (b) Is the product a solid, liquid, or gas at room temperature? (c) In the balanced chemical equation for this reaction, what is the coefficient in front of the product if the coefficient in front of \(\mathrm{Cl}_{2}(g)\) is \(1 ?\)

Viridicatumtoxin B, \(\mathrm{C}_{30} \mathrm{H}_{31} \mathrm{NO}_{10}\), is a natural antibiotic compound. It requires a synthesis of 12 steps in the laboratory. Assuming all steps have equivalent yields of \(85 \%,\) which is the final percent yield of the total synthesis?

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Write balanced chemical equations to correspond to each of the following descriptions: (a) When sulfur trioxide gas reacts with water, a solution of sulfuric acid forms. (b) Boron sulfide, \(\mathrm{B}_{2} \mathrm{~S}_{3}(s),\) reacts violently with water to form dissolved boric acid, \(\mathrm{H}_{3} \mathrm{BO}_{3},\) and hydrogen sulfide gas. (c) Phosphine, \(\mathrm{PH}_{3}(g)\), combusts in oxygen gas to form water vapor and solid tetraphosphorus decaoxide. (d) When solid mercury(II) nitrate is heated, it decomposes to form solid mercury(II) oxide, gaseous nitrogen dioxide, and oxygen. (e) Copper metal reacts with hot concentrated sulfuric acid solution to form aqueous copper(II) sulfate, sulfur dioxide gas, and water.

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