Problem 112 - Introductory Chemistry Atoms First

Question

What is the empirical formula of a compound that is \(17.552 \% \mathrm{Na}, 39.696 \% \mathrm{Cr}\), and \(42.752 \% \mathrm{O} ?\)

Step-by-Step Solution

Answer

The empirical formula of the compound is \( \mathrm{Na}_2\mathrm{Cr}_2\mathrm{O}_7 \).

Step 1: Convert percentage composition to grams

Let's assume we have 100 grams of the compound - this makes it easier to convert the percentage composition directly to grams. - Na: 17.552% of 100 g = 17.552 g - Cr: 39.696% of 100 g = 39.696 g - O: 42.752% of 100 g = 42.752 g

Step 2: Convert grams to moles

Next, we'll convert the grams of each element to moles using their respective molar masses: - Na: Molar mass = 22.99 g/mol - Cr: Molar mass = 51.996 g/mol - O: Molar mass = 16.00 g/mol Moles of each element: - Moles of Na: \( \frac{17.552 \text{ g}}{22.99 \text{ g/mol}} \) = 0.7632 mol - Moles of Cr: \( \frac{39.696 \text{ g}}{51.996 \text{ g/mol}} \) = 0.7632 mol - Moles of O: \( \frac{42.752 \text{ g}}{16.00 \text{ g/mol}} \) = 2.672 mol

Step 3: Calculate mole ratios

Now, we'll find the mole ratios of each element by dividing the moles of each element by the lowest number of moles: - Mole ratio of Na: \( \frac{0.7632}{0.7632} \) = 1 - Mole ratio of Cr: \( \frac{0.7632}{0.7632} \) = 1 - Mole ratio of O: \( \frac{2.672}{0.7632} \) ≈ 3.5

Step 4: Simplify to the lowest whole numbers

Since we have a ratio of about 3.5 for O, we need to find the lowest whole number ratios. To do this, we'll multiply each mole ratio by a small whole number (e.g., 2) to obtain: - Mole ratio of Na: 1 × 2 = 2 - Mole ratio of Cr: 1 × 2 = 2 - Mole ratio of O: 3.5 × 2 = 7 So, the empirical formula of the compound is: \( \mathrm{Na}_2\mathrm{Cr}_2\mathrm{O}_7 \)