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Chapter 3: Problem 95

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 C₂HO.

Step by step solution

01

Determine the moles of carbon, hydrogen, and oxygen in the CO₂ and H₂O produced

Given the masses of CO₂ and H₂O produced, we can calculate the moles of each element present. For Carbon: Mass of CO₂ = 2.43 g Molar mass of CO₂ = 12.01 g/mol (for carbon) + 2 * 16.00 g/mol (for oxygen) = 44.01 g/mol Moles of carbon = mass of CO₂ / molar mass of CO₂ Moles of carbon = \( \frac{2.43}{44.01} \) = 0.0552 moles For Hydrogen: Mass of H₂O = 0.50 g Molar mass of H₂O = 2 * 1.01 g/mol (for hydrogen) + 16.00 g/mol (for oxygen) = 18.02 g/mol Moles of hydrogen = mass of H₂O / molar mass of H₂O 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: \(\frac{0.0552}{0.0225} \approx 2.46\) ≈ 2 Mole ratio of hydrogen: \(\frac{0.0278}{0.0225} \approx 1.24\) ≈ 1 Mole ratio of oxygen: \(\frac{0.0225}{0.0225} \) = 1 Thus, the empirical formula of vanillin is C₂H₁O₁, simplified as C₂HO.

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Most popular questions from this chapter

(a) The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of \(2.78 \mathrm{mg}\) of ethyl butyrate produces \(6.32 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and $2.58 \mathrm{mg}\( of \)\mathrm{H}_{2} \mathrm{O}$. What is the empirical formula of the compound? (b) Nicotine, a component of tobacco, is composed of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{N}\). A \(5.250-\mathrm{mg}\) sample of nicotine was combusted, producing \(14.242 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(4.083 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula for nicotine? If nicotine has a molar mass of $160 \pm 5 \mathrm{~g} / \mathrm{mol},$ what is its molecular formula?

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?

Write a balanced chemical equation for the reaction that occurs when (a) titanium metal reacts with \(\mathrm{O}_{2}(g) ;(\mathbf{b})\) silver(I) oxide decomposes into silver metal and oxygen gas when heated; (c) propanol, \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(l)\) burns in air; \((\mathbf{d})\) methyl tert-butyl ether, \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}(l),\) burns in air.

The reaction between potassium superoxide, \(\mathrm{KO}_{2}\), and \(\mathrm{CO}_{2}\), $$ 4 \mathrm{KO}_{2}+2 \mathrm{CO}_{2} \longrightarrow 2 \mathrm{~K}_{2} \mathrm{CO}_{3}+3 \mathrm{O}_{2} $$ is used as a source of \(\mathrm{O}_{2}\) and absorber of \(\mathrm{CO}_{2}\) in selfcontained breathing equipment used by rescue workers. (a) How many moles of \(\mathrm{O}_{2}\) are produced when \(0.400 \mathrm{~mol}\) of \(\mathrm{KO}_{2}\) reacts in this fashion? (b) How many grams of \(\mathrm{KO}_{2}\) are needed to form \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\) ? (c) How many grams of \(\mathrm{CO}_{2}\) are used when \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\) are produced?

A chemical plant uses electrical energy to decompose aqueous solutions of \(\mathrm{NaCl}\) to give \(\mathrm{Cl}_{2}, \mathrm{H}_{2},\) and \(\mathrm{NaOH}\) : $2 \mathrm{NaCl}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g)$ If the plant produces \(1.5 \times 10^{6} \mathrm{~kg}\) ( 1500 metric tons) of \(\mathrm{Cl}_{2}\) daily, estimate the quantities of \(\mathrm{H}_{2}\) and \(\mathrm{NaOH}\) produced.
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