Question

A compound that contains only carbon, hydrogen, and oxygen is \(48.64 \% \mathrm{C}\) and \(8.16 \% \mathrm{H}\) by mass. What is the empirical formula of this substance?

Short answer

The empirical formula for the given compound is \(C_3H_6O_2\).

Step-by-step solution

Convert percentages to grams

Since we are given the percentage composition by mass, let's assume we have a 100 g sample of the substance. This way, the percentages directly correspond to grams: - Carbon: 48.64% of 100 g = 48.64 g C - Hydrogen: 8.16% of 100 g = 8.16 g H - Oxygen: 100% - (48.64% + 8.16%) = 43.20% of 100 g = 43.20 g O

Convert grams to moles

Now, we'll convert the grams of each element into moles using their respective molar masses: Carbon: \( \frac{48.64 \, \text{g} }{12.01 \, \frac{\text{g}}{\text{mol}} } = 4.05 \, \text{mol} \) Hydrogen: \( \frac{8.16 \, \text{g} }{1.008 \, \frac{\text{g}}{\text{mol}} } = 8.10 \, \text{mol} \) Oxygen: \( \frac{43.20 \, \text{g} }{16.00 \, \frac{\text{g}}{\text{mol}} } = 2.70 \, \text{mol} \)

Find the simplest whole-number ratio

We'll divide each of the calculated values in moles by the smallest value (which is 2.70 mol), and round the result to the nearest whole number: - Carbon: \( \frac{4.05 \, \text{mol}}{2.70 \, \text{mol}} \approx 1.50 \rightarrow 1.5 \approx 1.5 \) - Hydrogen: \( \frac{8.10 \, \text{mol} }{2.70 \, \text{mol}} \approx 3.00 \rightarrow 3.0 \approx 3 \) - Oxygen: \( \frac{2.70 \, \text{mol} }{2.70 \, \text{mol}} = 1.00 \rightarrow 1.0 \approx 1 \) Since the ratio of carbon is approximately 1.5, we'll multiply all the ratios by 2 to get whole numbers: - Carbon: 1.5 × 2 = 3 - Hydrogen: 3 × 2 = 6 - Oxygen: 1 × 2 = 2

Write the empirical formula

Based on the whole-number ratios found in step 3, the empirical formula for this compound is: \(C_3H_6O_2\)

Key concepts

Percentage Composition

When we analyze a compound's makeup, we often start with its percentage composition. This indicates how much of each element is present in the compound, by mass. For example, if a compound is 48.64% carbon, this means that out of a hypothetical 100 g of the compound, 48.64 g would be pure carbon.
To solve our problem, where we determine the empirical formula, it is efficient to assume a 100 g sample of the compound. This helps in simplifying the conversion of percentages to grams—making them equal. So, in such a sample:

• Carbon (C): 48.64 g
• Hydrogen (H): 8.16 g
• Oxygen (O): Remaining percentage, calculated as 100% - (C% + H%)

Following these steps prepares us for further conversions and calculations.

Moles Calculation

Once we have the mass of each element in grams from the percentage composition, the next step is to convert these into moles. Moles are a basic unit in chemistry that represent a specific amount of particles (like atoms or molecules). Using the formula \[\text{Moles} = \frac{\text{mass in grams}}{\text{molar mass in g/mol}}\]For carbon, hydrogen, and oxygen:

• Carbon: \(\frac{48.64 \text{ g}}{12.01 \text{ g/mol}} \approx 4.05 \text{ mol}\)
• Hydrogen: \(\frac{8.16 \text{ g}}{1.008 \text{ g/mol}} \approx 8.10 \text{ mol}\)
• Oxygen: \(\frac{43.20 \text{ g}}{16.00 \text{ g/mol}} \approx 2.70 \text{ mol}\)

Calculating in moles allows comparable measurements across different elements, who naturally have different atomic weights.

Whole Number Ratio

Converting moles into a whole number ratio is crucial for determining the empirical formula. We do this by taking the moles calculated for each element and dividing them by the smallest value in the set. This equalizes the comparison among different elements’ amounts.
In our problem:

• Divide each mole count by 2.70 (the smallest number):
• For Carbon: \( \frac{4.05}{2.70} \approx 1.5 \)
• For Hydrogen: \( \frac{8.10}{2.70} \approx 3.00 \)
• For Oxygen: \( \frac{2.70}{2.70} = 1.00 \)

But, we want whole numbers, because atoms exist in whole units in a chemical formula. Therefore, multiplying all by 2 gives us integers: 3:6:2.

Chemical Formula Determination

The empirical formula is the simplest whole-number ratio of elements in a compound and shows the type and relative number of atoms. For our compound:

• Carbon (C): 3 parts
• Hydrogen (H): 6 parts
• Oxygen (O): 2 parts

This ratio gives us the empirical formula \(C_3H_6O_2\). It is essential to note that the empirical formula does not always provide the exact number of atoms in a molecule, but rather the simplest ratio of the different types of atoms present.