Question

A certain vitamin extracted from plant sources has carbon and hydrogen in \(8: 1\) mass ratio. The percentage of oxygen is nearly 7.3. The compound gave no test for nitrogen or sulphur or any other element. What should be the empirical formula of the compound? (a) \(\mathrm{C}_{30} \mathrm{H}_{45} \mathrm{O}_{2}\) (b) \(\mathrm{C}_{15} \mathrm{H}_{23} \mathrm{O}\) (c) \(\mathrm{C}_{29} \mathrm{H}_{45} \mathrm{O}_{3}\) (d) \(\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{O}\)

Short answer

The empirical formula of the compound is (b) \(\mathrm{C}_{15} \mathrm{H}_{23} \mathrm{O}\).

Step-by-step solution

Determine the Percentage of Carbon and Hydrogen

Given the mass ratio of carbon to hydrogen as 8:1, the percentage of carbon and hydrogen in the compound can be calculated assuming the total mass of the compound is 100 g. Since the percentage of oxygen is 7.3%, the remaining percentage will be for carbon and hydrogen. The total percentage of carbon and hydrogen together will be 100% - 7.3% = 92.7%. Divide 92.7% in the ratio 8:1 to get the percentage of carbon and hydrogen respectively.

Calculate Percentages of Carbon and Hydrogen

The mass ratio of 8 parts carbon to 1 part hydrogen means for every 8 g of carbon, there is 1 g of hydrogen. To find the percentage of carbon, divide 8 by the total number of parts (8 + 1) and multiply by 92.7%. To find the percentage of hydrogen, divide 1 by the total number of parts (8 + 1) and multiply by 92.7%.

Convert Percentages to Masses

From the percentages, we can find the masses of carbon and hydrogen in a 100 g sample of the compound. The mass of carbon is the percentage of carbon times 100 g and the mass of hydrogen is the percentage of hydrogen times 100 g.

Determine the Mole Ratio

Using the atomic masses of carbon (approximately 12 g/mol) and hydrogen (approximately 1 g/mol), convert the masses of carbon and hydrogen to moles by dividing the mass by the respective atomic mass. This will give the mole ratio of carbon to hydrogen.

Simplify Mole Ratio to Find the Empirical Formula

Divide the mole ratio of carbon to hydrogen by the smallest value to simplify the ratio to the smallest whole numbers possible. This simplified mole ratio will give the subscripts in the empirical formula for carbon and hydrogen.

Include Oxygen in the Empirical Formula

Since the percentage of oxygen is 7.3%, convert this to moles by dividing by the atomic mass of oxygen (approximately 16 g/mol). The resulting mole number should be the subscript for oxygen in the empirical formula.

Choose the Correct Empirical Formula

Compare the mole ratio found for carbon, hydrogen, and oxygen with the options given. The correct empirical formula will have subscripts that match the simplified mole ratio that includes all three elements.

Key concepts

Molar Mass

Understanding molar mass is essential when determining an empirical formula. Molar mass is the weight of one mole of a substance, measured in grams per mole (g/mol). It corresponds to the relative atomic mass of an element or the sum of atomic masses in a compound. This information is critical in converting a mass percentage to moles, which is a key step in empirical formula calculation.

For instance, carbon has an approximate molar mass of 12 g/mol, hydrogen 1 g/mol, and oxygen 16 g/mol. By using the molar mass, we turn mass percentages into mole amounts, allowing for the determination of mole ratios. This conversion is the cornerstone of stoichiometry, which involves calculating the proportions of elements within compounds.

Stoichiometry

Stoichiometry lies at the heart of chemistry, involving the calculation of reactants and products in chemical reactions. In the context of determining empirical formulas, stoichiometry allows us to use the molar masses of elements and their mass percentages to find the mole ratio of elements in a compound.

Applying Stoichiometry to Empirical Formulas

When given mass percentages, we first convert these to actual masses, assuming a 100 g sample for mathematical simplicity. Next, we apply the concept of molar mass to convert these masses into moles. Finally, we can compare the mole amounts to establish the stoichiometric, or mole, ratios necessary to find the empirical formula. Mastery of stoichiometry is pivotal in making these calculations accurate.

Mole Ratio

The mole ratio is integral in articulating how elements combine to form a compound. When determining an empirical formula, we calculate the ratio of moles of each element in a sample. A correct mole ratio reflects the simplest whole-number ratio of atoms in the compound, which is the essence of the empirical formula.

Streamlining the Ratios

Once we have the number of moles for each element, we derive the mole ratio by dividing each by the smallest number of moles obtained in our calculations. The resulting simplified mole ratio is then used to designate the subscripts in the empirical formula that denote how many atoms of each element are present.

Chemical Composition

Chemical composition refers to the identity and proportion of elements within a chemical substance. The empirical formula is the simplest expression of a compound's chemical composition, showing the type of atoms present and their relative numbers in the smallest possible whole-number ratio.

Understanding a compound's chemical composition is essential for various applications in chemistry, including the synthesis of compounds, predicting reaction products, and analyzing material properties. In our exercise, the chemical composition is deduced from given mass ratios and percentages, leading us to determine the empirical formula and thus the fundamental makeup of the molecule in question.