Question

ethylene glycol? Dioxane is a common solvent for plastics. If the molar mass is \(88 \mathrm{~g} / \mathrm{mol}\) and the percent composition is \(54.5 \% \mathrm{C}, 9.15 \% \mathrm{H}\), and \(36.3 \% \mathrm{O}\), what is the molecular formula of dioxane?

Short answer

The molecular formula of dioxane is \( \text{C}_4\text{H}_8\text{O}_2 \).

Step-by-step solution

Convert Percent to Mass

Assume we have 100 grams of the compound, so the mass of each element is equal to its percentage. - Carbon: 54.5 g - Hydrogen: 9.15 g - Oxygen: 36.3 g

Calculate Moles of Each Element

Use the molar mass of each element to convert these masses to moles.- Moles of Carbon: \( \frac{54.5 \, \text{g}}{12.01 \, \text{g/mol}} = 4.54 \, \text{mol} \)- Moles of Hydrogen: \( \frac{9.15 \, \text{g}}{1.01 \, \text{g/mol}} = 9.06 \, \text{mol} \)- Moles of Oxygen: \( \frac{36.3 \, \text{g}}{16.00 \, \text{g/mol}} = 2.27 \, \text{mol} \)

Determine the Simplest Mole Ratio

Divide each mole value by the smallest number of moles calculated to find the mole ratio.- Carbon: \( \frac{4.54}{2.27} = 2 \)- Hydrogen: \( \frac{9.06}{2.27} = 4 \)- Oxygen: \( \frac{2.27}{2.27} = 1 \)

Determine the Empirical Formula

With the simplest ratio, the empirical formula is \( \text{C}_2\text{H}_4\text{O} \).

Calculate Empirical Formula Molar Mass

Calculate the molar mass of the empirical formula (\( \text{C}_2\text{H}_4\text{O} \)).- \( (2 \times 12.01) + (4 \times 1.01) + (1 \times 16.00) = 44.08 \, \text{g/mol} \)

Determine the Molecular Formula

The molar mass of the compound is 88 g/mol. Divide this by the empirical molar mass.- \( \frac{88}{44.08} = 2 \)Multiply the empirical formula by this factor to get the molecular formula: \( \text{C}_4\text{H}_8\text{O}_2 \).

Key concepts

Percent Composition

Understanding percent composition is an essential step in finding the molecular formula of a compound. It tells us how much of each element is present in a compound, expressed as a percentage of the total mass. So when you see a percent composition, think of it as a recipe where each percentage represents the ingredient part of the whole.

Imagine you have a substance with a percent composition of 54.5% carbon, 9.15% hydrogen, and 36.3% oxygen. This means that in 100 grams of this compound, 54.5 grams are carbon, 9.15 grams are hydrogen, and 36.3 grams are oxygen. This conversion is very helpful because it allows us to work with actual masses to determine further properties such as mole ratios and, ultimately, molecular formulas.

Using these percent values as actual mass values is the first step in converting percent composition into something more tangible and calculable, like moles, which will lead us to the molecular details of the compound.

Empirical Formula

The empirical formula represents the simplest whole-number ratio of the elements in a compound. To find it, you need to convert the masses of the elements derived from the percent composition into moles. Once you have the number of moles, the next crucial step is to find the simplest ratio among them.

From the example, we calculated the moles of carbon, hydrogen, and oxygen and found them to be 4.54 mol, 9.06 mol, and 2.27 mol, respectively. The smallest number of these is 2.27 mol, corresponding to oxygen. By dividing all mole values by the smallest one, you will obtain the simplest mole ratio. In this case, the ratio comes out to be C: 2, H: 4, O: 1.

Thus, the empirical formula for the compound would be written as \(C_2H_4O\). This suggests that for every molecule of this substance, there are twice as many carbon atoms as oxygen atoms and four times as many hydrogen atoms as oxygen.

Molar Mass Calculation

Once you have the empirical formula, the next step is to calculate its molar mass. The molar mass is the sum of the masses of all the atoms in a molecule. For an empirical formula of \(C_2H_4O\), you add up the atomic masses: two carbons, four hydrogens, and one oxygen.

Here's how it breaks down:

• Carbon: 2 atoms \(\times\ 12.01\, \text{g/mol} = 24.02\, \text{g/mol}\)
• Hydrogen: 4 atoms \(\times\ 1.01\, \text{g/mol} = 4.04\, \text{g/mol}\)
• Oxygen: 1 atom \(\times\ 16.00\, \text{g/mol} = 16.00\, \text{g/mol}\)

The total is \(44.06\, \text{g/mol}\).

To identify the molecular formula, compare the empirical molar mass to the given molar mass of the compound, which is 88 g/mol in our example. Dividing the given molar mass by the empirical molar mass (88 / 44.06) gives approximately 2. This means the actual molecular formula is twice the empirical formula, resulting in \(C_4H_8O_2\). So, each molecule of the compound consists of four carbon atoms, eight hydrogen atoms, and two oxygen atoms.