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Vanillin, the dominant flavoring in vanilla, contains C, H, and O. When \(1.05 \mathrm{~g}\) of this substance is completely combusted, \(2.43 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.50 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are produced. What is the empirical formula of vanillin?

Short Answer

Expert verified
The empirical formula of vanillin is C鈧侶O.

Step by step solution

01

Determine the moles of carbon, hydrogen, and oxygen in the CO鈧 and H鈧侽 produced

Given the masses of CO鈧 and H鈧侽 produced, we can calculate the moles of each element present. For Carbon: Mass of CO鈧 = 2.43 g Molar mass of CO鈧 = 12.01 g/mol (for carbon) + 2 * 16.00 g/mol (for oxygen) = 44.01 g/mol Moles of carbon = mass of CO鈧 / molar mass of CO鈧 Moles of carbon = \( \frac{2.43}{44.01} \) = 0.0552 moles For Hydrogen: Mass of H鈧侽 = 0.50 g Molar mass of H鈧侽 = 2 * 1.01 g/mol (for hydrogen) + 16.00 g/mol (for oxygen) = 18.02 g/mol Moles of hydrogen = mass of H鈧侽 / molar mass of H鈧侽 Moles of hydrogen = \( \frac{0.50}{18.02} \) = 0.0278 moles
02

Calculate the moles of oxygen in vanillin

Given the mass of vanillin combusted, we can now subtract the masses of carbon and hydrogen from it to find the mass of oxygen in vanillin. Mass of vanillin combusted = 1.05 g Mass of carbon = moles of carbon * molar mass of carbon = 0.0552 * 12.01 = 0.662 g Mass of hydrogen = moles of hydrogen * molar mass of hydrogen = 0.0278 * 1.01 = 0.028 g Mass of oxygen in vanillin = mass of vanillin - mass of carbon - mass of hydrogen Mass of oxygen = 1.05 - 0.662 - 0.028 = 0.360 g Now we can calculate the moles of oxygen in vanillin: Molar mass of oxygen = 16.00 g/mol Moles of oxygen = mass of oxygen / molar mass of oxygen Moles of oxygen = \( \frac{0.360}{16.00} \) = 0.0225 moles
03

Determine the empirical formula of vanillin

To find the empirical formula, we must find the mole ratio of each element. We will divide the moles of each element by the smallest number of moles among them to get the whole number ratio. Smallest number of moles = 0.0225 (oxygen) Mole ratio of carbon: \(\frac{0.0552}{0.0225} \approx 2.46\) 鈮 2 Mole ratio of hydrogen: \(\frac{0.0278}{0.0225} \approx 1.24\) 鈮 1 Mole ratio of oxygen: \(\frac{0.0225}{0.0225} \) = 1 Thus, the empirical formula of vanillin is C鈧侶鈧丱鈧, simplified as C鈧侶O.

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Key Concepts

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

Combustion Analysis
Combustion analysis is a widely used method in chemistry to determine the elemental composition of a substance, particularly those containing elements like carbon, hydrogen, and oxygen.
This technique involves combusting a sample in the presence of oxygen and measuring the masses of the resultant combustion products, usually carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)). For example, in the combustion of vanillin, measuring the \(\text{CO}_2\) and \(\text{H}_2\text{O}\) allows us to deduce the amounts of carbon and hydrogen.
  • The carbon present in the sample converts into \(\text{CO}_2\).
  • The hydrogen converts into \(\text{H}_2\text{O}\).
To determine the oxygen content, we use the conservation of mass principle. We know the total mass of the combusted sample and can subtract the masses of carbon and hydrogen to find the mass of oxygen. This process effectively gives the amounts of each element within the original compound, making it easier to deduce its empirical formula.
Mole Ratio
The mole ratio is a crucial concept when determining the empirical formula of a compound. It reflects the ratio of the number of moles of each element within a compound to each other. After combustion analysis calculates the moles of carbon, hydrogen, and oxygen, we utilize these figures to find the ratio.
  • Identify the smallest number of moles among the elements.
  • Divide all moles by this smallest number to derive a simple whole number ratio for each element.
This process is key in translating elemental mole values into a meaningful empirical formula. For vanillin, let's assume we calculated the following numbers of moles:
  • Carbon = 0.0552 moles
  • Hydrogen = 0.0278 moles
  • Oxygen = 0.0225 moles
In this case, 0.0225 is the smallest. Dividing each value by 0.0225 gives us approximately: - Carbon: 2 - Hydrogen: 1 - Oxygen: 1 These values correspond to the simplest whole-number mole ratio, and thus give us the empirical formula, C鈧侶O.
Chemical Composition
Chemical composition refers to the combination and proportion of different elements in a compound. Understanding it helps us predict the empirical formula, which indicates the simplest ratio of elements within a compound.
To determine a compound's chemical composition, combustion analysis provides a powerful pathway, while the mole ratio method simplifies these findings into an empirical formula. In our example of vanillin, we've delved into how carbon, hydrogen, and oxygen amounts are assessed following combustion.
The resulting ratios are simplified into an empirical formula. This formula is not just a representation of element types but reflects the relative proportions of each element in a molecule.
  • Empirical formulas do not indicate actual numbers of atoms, but only the simplest integer ratio.
  • The chemical composition tells us about the identity and nature of a compound's atoms.
By understanding these steps and concepts, chemists can recreate and understand the behavior and properties associated with a compound based on its composition.

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Most popular questions from this chapter

(a) One molecule of the antibiotic penicillin G has a mass of \(5.342 \times 10^{-21} \mathrm{~g}\). What is the molar mass of penicillin G? (b) Hemoglobin, the oxygen-carrying protein in red blood cells, has four iron atoms per molecule and contains \(0.340 \%\) iron by mass. Calculate the molar mass of hemoglobin.

(a) Combustion analysis of toluene, a common organic solvent, gives \(5.86 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(1.37 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O} .\) If the compound contains only carbon and hydrogen, what is its empirical formula? (b) Menthol, the substance we can smell in mentholated cough drops, is composed of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\). A \(0.1005-g\) sample of menthol is combusted, producing \(0.2829 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.1159 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula for menthol? If menthol has a molar mass of \(156 \mathrm{~g} / \mathrm{mol}\), what is its molecular formula?

A key step in balancing chemical equations is correctly identifying the formulas of the reactants and products. For example, consider the reaction between calcium oxide, \(\mathrm{CaO}(s)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) to form aqueous calcium hydroxide. (a) Write a balanced chemical equation for this combination reaction, having correctly identified the product as \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)\) (b) Is it possible to balance the equation if you incorrectly identify the product as \(\mathrm{CaOH}(a q)\), and if so, what is the equation?

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(74.0 \% \mathrm{C}, 8.7 \% \mathrm{H},\) and \(17.3 \% \mathrm{~N}\) (b) \(57.5 \% \mathrm{Na}, 40.0 \% \mathrm{O},\) and \(2.5 \% \mathrm{H}\) (c) \(41.1 \% \mathrm{~N}, 11.8 \% \mathrm{H},\) and the remainder \(\mathrm{S}\)

Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. If \(5.00 \mathrm{~g}\) of sulfuric acid and \(5.00 \mathrm{~g}\) of lead(II) acetate are mixed, calculate the number of grams of sulfuric acid, lead(II) acetate, lead(II) sulfate, and acetic acid present in the mixture after the reaction is complete.

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