Question

Chemical analysis shows that citric acid contains 37.51% C, 4.20% H, and 58.29% O. What is the empirical formula for citric acid?

Short answer

The empirical formula for citric acid is \( C_3H_4O_4 \).

Step-by-step solution

- Convert Percentages to Grams

Assume a 100 g sample of citric acid. Thus, you have 37.51 g of C, 4.20 g of H, and 58.29 g of O.

- Convert Grams to Moles

Use the molar mass of each element to convert grams to moles.\[ \text{Moles of C} = \frac{37.51}{12.01} = 3.12 \]\[ \text{Moles of H} = \frac{4.20}{1.008} = 4.17 \]\[ \text{Moles of O} = \frac{58.29}{16.00} = 3.64 \]

- Determine the Simplest Ratio

Divide each mole value by the smallest number of moles calculated.\[ \text{Ratio of C} = \frac{3.12}{3.12} = 1 \]\[ \text{Ratio of H} = \frac{4.17}{3.12} = 1.34 \]\[ \text{Ratio of O} = \frac{3.64}{3.12} = 1.17 \]

- Adjust Ratios to Whole Numbers

Adjust the ratios to the nearest whole numbers. The ratios of H and O suggest a small error or need to multiply each ratio by the same factor. Multiply by 3 to convert them to whole numbers.\[ \text{New ratio of C} = 1 \times 3 = 3 \]\[ \text{New ratio of H} = 1.34 \times 3 = 4.02 \approx 4 \]\[ \text{New ratio of O} = 1.17 \times 3 = 3.51 \approx 4 \]

- Write the Empirical Formula

Using the whole number ratios, the empirical formula is\[ C_3H_4O_4 \].

Key concepts

Understanding Chemical Composition

Chemical composition refers to the types and amounts of elements that make up a substance. In the context of citric acid, we're interested in the percentages of carbon (C), hydrogen (H), and oxygen (O) present. These proportions can tell us much about the molecular make-up of the compound. To find the empirical formula, we need to convert these percentages into actual moles of each element. This process helps to understand how atoms combine in fixed ratios to form compounds.

Mole Calculations

A mole is a unit used to measure the amount of a substance. It's similar to counting things in dozens but on a much larger scale since one mole equals about 6.022 x 10^23 particles. For mole calculations, we use the molar mass of elements to convert grams to moles. For example, with citric acid:

• For carbon (C), the calculation is: \[ \frac{37.51 \text{ grams}}{12.01 \text{ g/mol}} = 3.12 \text{ moles} \]
• For hydrogen (H): \[ \frac{4.20 \text{ grams}}{1.008 \text{ g/mol}} = 4.17 \text{ moles} \]
• For oxygen (O): \[ \frac{58.29 \text{ grams}}{16.00 \text{ g/mol}} = 3.64 \text{ moles} \]

These calculations convert mass into an easily comparable quantity, moles, helping us to better analyze the substance's makeup.

Determining the Empirical Formula

The empirical formula represents the simplest whole-number ratio of elements in a compound. After converting to moles, we simplify these values to get the lowest possible ratio. Following this, we divide each mole value by the smallest number of moles calculated:

• Ratio for C: \[ \frac{3.12}{3.12} = 1 \]
• Ratio for H: \[ \frac{4.17}{3.12} = 1.34 \]
• Ratio for O: \[ \frac{3.64}{3.12} = 1.17 \]

Since the ratios for H and O are not whole numbers, we multiply all of them by the same factor to find the simplest whole number ratio. Multiplying by 3, we get:

• New ratio for C: 3
• New ratio for H: 4.02 ≈ 4
• New ratio for O: 3.51 ≈ 4

Therefore, the empirical formula for citric acid is \[ C_3H_4O_4 \].

Percent Composition

Percent composition helps us understand how much of each element is in a compound by mass. This is crucial for determining formulas like the empirical formula of citric acid. To illustrate:

• Citric acid is found to contain 37.51% C, 4.20% H, and 58.29% O.

These percentages are used as a starting point to convert to moles and then ratios. Using percent composition aligns the calculations with real-world data, making the results practical and accurate. This approach is useful not only for academic exercises but also in various industries where precise chemical compositions are critical.