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Chapter 3: Problem 147

Vitamin A has a molar mass of \(286.4 \mathrm{~g} / \mathrm{mol}\) and a general molecular formula of \(\mathrm{C}_{x} \mathrm{H}_{\mathrm{y}} \mathrm{E}\), where \(\mathrm{E}\) is an unknown element. If vitamin \(\mathrm{A}\) is \(83.86 \% \mathrm{C}\) and \(10.56 \% \mathrm{H}\) by mass, what is the molecular formula of vitamin \(\mathrm{A}\) ?

Short Answer

Expert verified
The short answer for the molecular formula of Vitamin A, given its molar mass and percentages of Carbon and Hydrogen, is \(\mathrm{C}_{20} \mathrm{H}_{30} \mathrm{O}\).

Step by step solution

01

Determine the grams of Carbon and Hydrogen

First, we will find the grams of Carbon and Hydrogen in one mole of Vitamin A. Using the given percentages, we can calculate the grams as follows: - Grams of Carbon: \(286.4 g/mol * 83.86\% = 240.12 g/mol\) - Grams of Hydrogen: \(286.4 g/mol * 10.56\% = 30.24 g/mol\)
02

Calculate the molar amount of Carbon and Hydrogen

Now, we will find the molar amounts of Carbon and Hydrogen by dividing the grams by their respective atomic masses: - Molar amount of Carbon: \(240.12 g/mol \div 12.01 g/mol = 20 mol\) - Molar amount of Hydrogen: \(30.24 g/mol \div 1.01 g/mol = 30 mol\)
03

Find the molar ratios of Carbon, Hydrogen to the unknown element (E)

As given, the molecular formula of Vitamin A is \(\mathrm{C}_{x} \mathrm{H}_{\mathrm{y}} \mathrm{E}\). We can write the following equations using the information we have gathered: - Grams of Carbon : \(12.01x\) - Grams of Hydrogen : \(1.01y\) - Grams of unknown element (E) : \(286.4 - (12.01x + 1.01y)\) We can now insert the molar amounts of Carbon and Hydrogen: - \(12.01 * 20 + 1.01 * 30 + M_E = 286.4\) Here, M_E is the molar mass of the unknown element (E) in g/mol.
04

Determine the molecular formula of Vitamin A

Solve the equation for M_E: \(M_E = 286.4 - (12.01 * 20 + 1.01 * 30) = 286.4 - (240.2 + 30.3) = 16 g/mol\) The molar mass of the unknown element (E) is 16 g/mol. This corresponds to the molar mass of Oxygen (O). Therefore, the molecular formula of Vitamin A is \(\mathrm{C}_{20} \mathrm{H}_{30} \mathrm{O}\).

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Key Concepts

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

Understanding Molar Mass
Molar mass is a fundamental concept in chemistry used to relate the mass of a substance to the number of particles it contains. It represents the mass of one mole of a substance, which is Avogadro's number of particles (approximately \(6.022 \times 10^{23}\)). For vitamin A, the molar mass is given as \(286.4 \text{ g/mol}\).
The molar mass helps us determine how much of each element is present in a molecule by mass. By understanding the molar mass, we can directly relate the mass percentages of elements to their actual quantities in the molecule. This understanding is crucial for calculating molar ratios and eventually determining the chemical formula of the compound.
Elemental Composition Insights
Elemental composition refers to the percentage by mass of each element in a compound. In the case of vitamin A, we know it consists of \(83.86\%\) carbon and \(10.56\%\) hydrogen by mass.
To find the actual mass of each element in one mole of vitamin A, we multiply these percentages by the molar mass. This gives us valuable insights into how much each element contributes to the total mass.
  • For Carbon: \(286.4 \text{ g/mol} \times 0.8386 = 240.12 \text{ g/mol}\)
  • For Hydrogen: \(286.4 \text{ g/mol} \times 0.1056 = 30.24 \text{ g/mol}\)
These figures reveal the mass contributions of each element, which is essential for further calculations.
Delving into Molar Ratios
Molar ratios indicate the number of moles of each element in a compound. They help specify the proportions of each element relative to one another, offering a direct way to construct the empirical or molecular formula.
By dividing the mass of each element by their respective atomic masses, we find the molar amounts:
  • Carbon: \(240.12 \text{ g/mol} \div 12.01 \text{ g/mol} = 20 \text{ mol}\)
  • Hydrogen: \(30.24 \text{ g/mol} \div 1.01 \text{ g/mol} = 30 \text{ mol}\)
This process uncovers the ratio of carbon to hydrogen, which informs us of the basic building blocks required to represent the compound accurately. These ratios lead us toward determining the formula.
Chemical Formula Determination Revealed
Determining the chemical formula involves understanding how the elements come together in specific ratios to form a compound. Once the molar ratios are known, we can set up equations to solve for the number of atoms in one molecule.
In our vitamin A example, the totals for carbon and hydrogen allow us to calculate the remaining mass attributable to the unknown element, \( E \). Using the given total molar mass of \(286.4 \text{ g/mol}\):
\(\text{Grams of } E = 286.4 -(240.2 + 30.3) = 16 \text{ g/mol}\)
The remaining mass corresponds to the element oxygen (O), as it has a molar mass of \(16 \text{ g/mol}\).
Therefore, the molecular formula of vitamin A is \(\text{C}_{20}\text{H}_{30}\text{O}\), representing the precise composition and arrangement of atoms in the molecule.

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Most popular questions from this chapter

Which of the following pairs of compounds have the same empirical formula? a. acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\), and benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\) b. ethane, \(\mathrm{C}_{2} \mathrm{H}_{6}\), and butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\) c. nitrogen dioxide, \(\mathrm{NO}_{2}\), and dinitrogen tetroxide, \(\mathrm{N}_{2} \mathrm{O}_{4}\) d. diphenyl ether, \(\mathrm{C}_{12} \mathrm{H}_{10} \mathrm{O}\), and phenol, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\)

Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane: $$ 2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g)+2 \mathrm{CH}_{4}(g) \longrightarrow 2 \mathrm{HCN}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ If \(5.00 \times 10^{3} \mathrm{~kg}\) each of \(\mathrm{NH}_{3}, \mathrm{O}_{2}\), and \(\mathrm{CH}_{4}\) are reacted, what mass of \(\mathrm{HCN}\) and of \(\mathrm{H}_{2} \mathrm{O}\) will be produced, assuming \(100 \%\) yield?

Several important compounds contain only nitrogen and oxygen. Place the following compounds in order of increasing mass percent of nitrogen. a. NO, a gas formed by the reaction of \(\mathrm{N}_{2}\) with \(\mathrm{O}_{2}\) in internal combustion engines b. \(\mathrm{NO}_{2}\), a brown gas mainly responsible for the brownish color of photochemical smog c. \(\mathrm{N}_{2} \mathrm{O}_{4}\), a colorless liquid used as fuel in space shuttles d. \(\mathrm{N}_{2} \mathrm{O}\), a colorless gas sometimes used as an anesthetic by dentists (known as laughing gas)

Consider the following data for three binary compounds of hydrogen and nitrogen: $$ \begin{array}{lcc} & \% \mathrm{H} \text { (by Mass) } & \text { \% N (by Mass) } \\ \hline \text { I } & 17.75 & 82.25 \\ \text { II } & 12.58 & 87.42 \\ \text { III } & 2.34 & 97.66 \end{array} $$ When \(1.00 \mathrm{~L}\) of each gaseous compound is decomposed to its elements, the following volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{N}_{2}(g)\) are obtained: $$ \begin{array}{lcc} & \mathrm{H}_{2} \text { (L) } & \mathrm{N}_{2} \text { (L) } \\ \hline \text { I } & 1.50 & 0.50 \\ \text { II } & 2.00 & 1.00 \\ \text { III } & 0.50 & 1.50 \end{array} $$ Use these data to determine the molecular formulas of compounds I, II, and III and to determine the relative values for the atomic masses of hydrogen and nitrogen.

The reaction between potassium chlorate and red phosphorus takes place when you strike a match on a matchbox. If you were to react \(52.9 \mathrm{~g}\) of potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\) with excess red phosphorus, what mass of tetraphosphorus decaoxide \(\left(\mathrm{P}_{4} \mathrm{O}_{10}\right)\) would be produced? $$ \mathrm{KClO}_{3}(s)+\mathrm{P}_{4}(s) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)+\mathrm{KCl}(s) \quad \text { (unbalanced) } $$
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