Question

Calcium tartrate is used as a preservative for certain foods and as an antacid. It contains \(25.5 \% \mathrm{C}, 2.1 \% \mathrm{H}, 21.3 \% \mathrm{Ca}\), and \(51.0 \% \mathrm{O}\). What is the emnirical formula for calcium tartrate?

Short answer

The empirical formula for calcium tartrate is \[ \text{C}_{4}\text{H}_{4}\text{Ca}\text{O}_{6} \]

Step-by-step solution

Convert Percentages to Masses

Assume you have a 100 g sample of calcium tartrate. This corresponds to having 25.5 g of C, 2.1 g of H, 21.3 g of Ca, and 51.0 g of O.

Convert Masses to Moles

Use the molar masses of each element to convert grams to moles using the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \] \ Molar masses are: Carbon (C) = 12.01 g/mol, Hydrogen (H) = 1.01 g/mol, Calcium (Ca) = 40.08 g/mol, Oxygen (O) = 16.00 g/mol. \ Therefore: \ \[ \text{Moles of C} = \frac{25.5}{12.01} \approx 2.12 \] \[ \text{Moles of H} = \frac{2.1}{1.01} \approx 2.08 \] \[ \text{Moles of Ca} = \frac{21.3}{40.08} \approx 0.532 \] \[ \text{Moles of O} = \frac{51.0}{16.00} \approx 3.19 \]

Divide by Smallest Number of Moles

To find the simplest ratio, divide each number of moles by the smallest number of moles (0.532 for Calcium): \[ \text{Ratio of C} = \frac{2.12}{0.532} \approx 3.99 \approx 4 \] \[ \text{Ratio of H} = \frac{2.08}{0.532} \approx 3.91 \approx 4 \] \[ \text{Ratio of Ca} = \frac{0.532}{0.532} = 1 \] \[ \text{Ratio of O} = \frac{3.19}{0.532} \approx 5.99 \approx 6 \]

Write the Empirical Formula

Using the smallest whole number ratios, the empirical formula is: \[ \text{C}_{4}\text{H}_{4}\text{Ca}\text{O}_{6} \]

Key concepts

Percentage Composition

Percentage composition helps us understand the proportion of each element in a compound. It is expressed as the percentage of the total mass of each element.
For instance, in calcium tartrate, we know:

• 25.5% is Carbon (C)
• 2.1% is Hydrogen (H)
• 21.3% is Calcium (Ca)
• 51.0% is Oxygen (O)

This tells us that in 100 grams of calcium tartrate, there are 25.5 grams of Carbon, 2.1 grams of Hydrogen, 21.3 grams of Calcium, and 51.0 grams of Oxygen. Understanding the percentage composition is the first step in determining the empirical formula.

Mole Conversion

To determine the empirical formula, converting the given mass of each element to moles is essential. Moles help us relate masses of substances to their quantities.
The conversion formula we use is:
\( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \)
Let's apply this to our example:

• For Carbon (C): \( \frac{25.5 \text{ grams}}{12.01 \text{ g/mol}} \approx 2.12 \text{ moles} \)


• For Hydrogen (H): \( \frac{2.1 \text{ grams}}{1.01 \text{ g/mol}} \approx 2.08 \text{ moles} \)


• For Calcium (Ca): \( \frac{21.3 \text{ grams}}{40.08 \text{ g/mol}} \approx 0.532 \text{ moles} \)


• For Oxygen (O): \( \frac{51.0 \text{ grams}}{16.00 \text{ g/mol}} \approx 3.19 \text{ moles} \)

Simplest Ratio

Once we have converted the masses to moles, the next step is to find the simplest ratio. This reveals the smallest whole number relationship between the elements.
To do this, divide the numbers of moles by the smallest number obtained.
Using our example:

• The smallest number of moles is 0.532 (Calcium).
• Ratio for Carbon: \( \frac{2.12}{0.532} \approx 3.99 \approx 4 \)
• Ratio for Hydrogen: \( \frac{2.08}{0.532} \approx 3.91 \approx 4 \)
• Ratio for Calcium: \( \frac{0.532}{0.532} \approx 1 \)
• Ratio for Oxygen: \( \frac{3.19}{0.532} \approx 5.99 \approx 6 \)

Simplifying these ratios leads to the smallest whole numbers, which will form the empirical formula.

Chemical Formula

The empirical formula represents the simplest whole number ratio of elements in a compound.
Using the ratios we found:

• Carbon: 4
• Hydrogen: 4
• Calcium: 1
• Oxygen: 6

Therefore, the empirical formula for calcium tartrate is \( \text{C}_{4}\text{H}_{4}\text{Ca}\text{O}_{6} \).
This empirical formula shows that these elements combine in the mentioned simplest ratio within the compound. Understanding how to calculate this from the percentage composition allows us to identify and verify chemical compounds analytically.