Question

A compound subjected to analysis was found to have the following composition by weight: \(69.96 \%\) carbon (atomic weight \(=12.0 \mathrm{~g} / \mathrm{mole}), 7.83 \%\) hydrogen (atomic weight \(=1.01 \% \mathrm{~g} /\) mole \()\), and \(22.21 \%\) oxygen (atomic weight \(=\) \(16.0 \mathrm{~g} / \mathrm{mole})\). If the molecular weight of this compound is 360 \(\mathrm{g} /\) mole, what is its molecular formula?

Short answer

The molecular formula for the given compound is \(C_{21}H_{28}O_{5}\).

Step-by-step solution

Convert weight percentages into moles

First, we need to convert the weight percentages of each element into moles. We can do this by using the atomic weights given in the problem statement. - Carbon: \(\frac{69.96 \text{ g}}{12.0 \text{ g/mol}} = 5.83 \text{ moles}\) - Hydrogen: \(\frac{7.83 \text{ g}}{1.01 \text{ g/mol}} = 7.75 \text{ moles}\) - Oxygen: \(\frac{22.21 \text{ g}}{16.0 \text{ g/mol}} = 1.39 \text{ moles}\)

Divide moles by the smallest number of moles

Now, we need to find the mole ratios by dividing the moles by the smallest number of moles. In this case, the smallest number of moles is 1.39 moles from Oxygen. - Carbon: \(\frac{5.83}{1.39} = 4.2\) - Hydrogen: \(\frac{7.75}{1.39} = 5.6\) - Oxygen: \(\frac{1.39}{1.39} = 1\)

Multiply the ratios to find the whole-number ratio

As we can see, the mole ratios are not integers. In order to make them whole numbers, we need to find the least common multiplier. In this case, the least common multiplier is 5. - Carbon: \(4.2 \times 5 = 21\) - Hydrogen: \(5.6 \times 5 = 28\) - Oxygen: \(1 \times 5 = 5\) Now we have a whole-number ratio of 21:28:5 for Carbon:Hydrogen:Oxygen.

Determine the molecular formula

Finally, we can use the whole-number ratio to determine the molecular formula of the compound. The molecular formula for the compound is \(C_{21}H_{28}O_{5}\).

Key concepts

Percent Composition

Understanding percent composition is essential when analyzing the makeup of a compound. It describes the percentage by mass of each element in a substance. To calculate it, you divide the mass of a specific element by the total mass of the compound and multiply the result by 100. This gives you the percentage of that element in the compound.

For example, in the given exercise, the compound has a percent composition of 69.96% carbon, 7.83% hydrogen, and 22.21% oxygen. These percentages are crucial for determining the compound's empirical and molecular formulas, as they provide a starting point to convert mass into moles, which leads to the mole ratio of the elements within the compound.

Atomic Weight

Atomic weight, also known as atomic mass, is the weight of a single atom of an element. It's usually expressed in atomic mass units (amu) or grams per mole (g/mol). This value is the average mass of all isotopes of an element, weighted by their natural abundance.

In the exercise, atomic weights are given: for carbon, it is 12.0 g/mol; for hydrogen, it is 1.01 g/mol; and for oxygen, it is 16.0 g/mol. These values are used to convert the percent composition of each element into moles, which is the first step in determining the empirical formula of the compound.

Molecular Weight

Molecular weight, or molecular mass, is the sum of the atomic weights of all the atoms in a molecule. It's also typically expressed in grams per mole (g/mol). This measurement is fundamental when converting between moles and grams, which is necessary for stoichiometric calculations and determining the molecular formula of a compound.

In the provided exercise, the molecular weight of the compound is given as 360 g/mol. This information, combined with the empirical formula of the compound, can help in determining the molecular formula by comparing the weight of the empirical formula unit to the actual molecular weight.

Mole Concept

The mole concept is a bridge between the microscopic world of atoms and molecules and the macroscopic world we experience. One mole represents Avogadro's number, which is approximately 6.022 x 10^23 particles. This concept allows chemists to count atoms by weighing them.

In the exercise solution, the percent composition by mass is first converted to moles, thereby translating mass ratios into mole ratios which correspond to the ratios of atoms in the compound. Dividing these mole ratios by the smallest number of moles among them gives the simplest whole-number atomic ratio, which is the basis of the empirical formula.

Empirical and Molecular Formulas

The empirical formula represents the simplest whole-number ratio of elements in a compound. In contrast, the molecular formula shows the actual number of atoms of each element in a molecule of the compound. The empirical formula may be the same as or a simple multiple of the molecular formula.

In the step-by-step solution of the exercise, the empirical formula is initially determined by converting the percent composition to moles, and then to the simplest whole-number ratio. Multiplying these numbers by an integer, if necessary, gives a whole-number molecular formula. In some cases, like in the given exercise, the molecular formula can be directly derived by adjusting the ratios using the molecular weight of the compound.