Question

Carbohydrates are compounds containing carbon, hydrogen, and oxygen in which the hydrogen to oxygen ratio is 2: 1 . A certain carbohydrate contains 40.0 percent carbon by mass. Calculate the empirical and molecular formulas of the compound if the approximate molar mass is \(178 \mathrm{~g}\).

Short answer

The empirical and molecular formula is \(C_8H_{32}O_3\).

Step-by-step solution

Define mass percentages of elements

We know the compound is composed of carbon, hydrogen, and oxygen. The mass percentage of carbon is given as 40.0%. Since the hydrogen to oxygen ratio is 2:1 and the rest of the compound must be composed of hydrogen and oxygen, we express the remaining 60% mass as 2 parts of hydrogen and 1 part of oxygen.

Calculate mass percentages of hydrogen and oxygen

The remaining mass (60%) is distributed between hydrogen and oxygen such that the ratio is 2:1. Thus, hydrogen accounts for 2/3 of 60%, and oxygen accounts for 1/3 of 60%. \[\text{Hydrogen percentage} = \frac{2}{3} \times 60 = 40% \\text{Oxygen percentage} = \frac{1}{3} \times 60 = 20%\]

Calculate the moles of each element

Let's find the moles of each element by dividing their mass percentages by their respective molar masses:\[\text{Moles of Carbon} = \frac{40.0}{12.01} \approx 3.33 \, ext{mol}\\text{Moles of Hydrogen} = \frac{40.0}{1.008} \approx 39.68 \, ext{mol}\\text{Moles of Oxygen} = \frac{20.0}{16.00} \approx 1.25 \, ext{mol}\]

Determine the simplest mole ratio

Divide each mole value by the smallest number of moles calculated to find the simplest ratio:\[\text{Ratio for Carbon: } \frac{3.33}{1.25} \approx 2.66 \\text{Ratio for Hydrogen: } \frac{39.68}{1.25} \approx 31.74 \\text{Ratio for Oxygen: } \frac{1.25}{1.25} = 1\]Converting to the simplest whole-number ratio, multiply each by 3 (since 2.66 rounds to approximately 8 when multiplied by 3):\[\text{Empirical formula ratio (approx.): } \ C_8H_{32}O_3\]

Calculate empirical formula molar mass

Calculate the molar mass of the empirical formula:\[\text{Molar mass of } C_8H_{32}O_3 = 8(12.01) + 32(1.008) + 3(16.00) = 176.12 \ \text{g/mol}\]

Determine the molecular formula

Divide the given molar mass by the empirical formula molar mass to find the multiplier for the empirical formula:\[\frac{178}{176.12} \approx 1.011 \approx 1\]Since the multiplier is approximately 1, the empirical formula is also the molecular formula.

Key concepts

Empirical Formula

Carbohydrates, the energy source for all living cells, follow a specific chemical structure. An empirical formula represents the simplest whole-number ratio of atoms in a compound. It doesn't show the exact number of each atom but instead provides a reduced form.
For example, if you know a carbohydrate contains 40% carbon, you can deduce the rest is 60% divided between hydrogen and oxygen, maintaining a 2:1 ratio. The empirical formula of a carbohydrate with these percentages would be found by calculating the numerical proportions of the atoms. This involves determining the number of moles of each element based on their mass and then finding the simplest ratio of these mole values.
Through such calculations, you derive an empirical formula such as \(C_8H_{32}O_3\). It acts as a fingerprint for the compound's basic composition, even if it isn't the full picture of every molecule.

Molecular Formula

While the empirical formula gives us the simplest representation, the molecular formula provides the actual number of each type of atom in a molecule. For carbohydrates, this formula can often be a multiple of the empirical formula.
The molecular formula is calculated using the molar mass. Once the empirical formula is known, you find its molar mass. Dividing the given molar mass of the compound by the empirical molar mass gives you a ratio. This ratio helps determine whether the actual molecular formula is the same as the empirical or a multiple.
In our example, since the empirical formula is \(C_8H_{32}O_3\) and its molar mass closely matches the given molar mass of 178 g/mol (leading to a multiplier close to 1), the molecular formula here is the same as the empirical formula.

Molar Mass

The molar mass is a crucial part of solving such problems, acting as the bridge between empirical and molecular formulas. It is calculated as the sum of the masses of all atoms in one mole of a compound.
Understanding molar mass involves adding the atomic masses of all elements in the empirical formula. For \(C_8H_{32}O_3\), you multiply the atomic mass of carbon, hydrogen, and oxygen by the respective number of atoms, adding these values to get the empirical molar mass.

• Carbon: 8 atoms
• Hydrogen: 32 atoms
• Oxygen: 3 atoms

This calculation reveals whether the empirical formula matches the molar mass given for the molecular formula. In this case, the close match between 176.12 g/mol (empirical) and 178 g/mol (given) means they are indeed the same.

Chemical Composition

The chemical composition of carbohydrates directly influences their formulas. Knowing the percentage of each element enables insights into their composition, crucial for both empirical and molecular formula calculations.
In addition to carbon, hydrogen and oxygen play significant roles. Here, the chemical composition provided (40% carbon) helped compute the percentage of hydrogen and oxygen, respecting the hydrogen to oxygen ratio of 2:1. This allowed a full determination of the carbohydrates' composition, translating to real molecular structures.
Identifying chemical composition gives a more profound understanding of the stoichiometric relationships that make up complex carbohydrates, essential for both chemistry students and scientists.

Stoichiometry

Stoichiometry is the quantitative study of reactants and products in chemical reactions. It also applies to evaluating the ratios in compound formulas.
In the context of carbohydrates, stoichiometry helps us quantify each component's presence within the compound formation. Using stoichiometry, we calculated the ratio of carbon, hydrogen, and oxygen based on their mole calculations.

• Carbon's role: Ingredient that forms the backbone
• Hydrogen's role: Balances the oxygen to maintain hydration
• Oxygen's role: Connects atoms through covalent bonds

Simplifying these mole values into whole numbers gives us the stoichiometric formula, paying attention to the contributions and ratios of each element in the compound. Understanding such stoichiometric relationships demystifies chemical equations and interactions, making the subject fascinating and tangible.