Question

The compound \(\mathrm{X}_{2}\left(\mathrm{YZ}_{3}\right)_{3}\) has a molar mass of \(282.23 \mathrm{~g}\) and a percent composition (by mass) of \(19.12 \% \mathrm{X}, 29.86 \% \mathrm{Y}\), and \(51.02 \% \mathrm{Z}\). What is the formula of the compound?

Short answer

The empirical formula is \text{X}_{6}\text{Y}_{3}\text{Z}_9\.

Step-by-step solution

- Convert Percentages to Masses

Assume you have 100 grams of the compound. Therefore, you have 19.12 grams of X, 29.86 grams of Y, and 51.02 grams of Z.

- Convert Masses to Moles

Use the molar masses of X, Y, and Z to convert the masses to moles. Let the molar masses be as follows: X = 12 g/mol, Y = 24 g/mol, and Z = 16 g/mol (these are just assumed for calculation purposes).\[\text{Moles of X} = \frac{19.12 \text{ g}}{12 \text{ g/mol}} = 1.593 \text{ mol} \]\[\text{Moles of Y} = \frac{29.86 \text{ g}}{24 \text{ g/mol}} = 1.244 \text{ mol} \]\[\text{Moles of Z} = \frac{51.02 \text{ g}}{16 \text{ g/mol}} = 3.189 \text{ mol} \]

- Determine Mole Ratios

Divide the moles of each element by the smallest number of moles calculated in the previous step to determine the simplest whole number ratio.\[\text{Ratio of X} = \frac{1.593}{1.244} = 1.281 \approx 1.29 \]\[\text{Ratio of Y} = \frac{1.244}{1.244} = 1 \]\[\text{Ratio of Z} = \frac{3.189}{1.244} = 2.563 \approx 2.56 \]

- Simplify the Ratios

Multiply each ratio by the appropriate factor to convert to the nearest whole number. Here, multiply by approximately 2.\[\text{X Ratio} = 1.29 \times 2 = 2.58 \approx 3 \]\[\text{Y Ratio} = 1 \times 2 = 2 \]\[\text{Z Ratio} = 2.56 \times 2 = 5.12 \approx 5 \]

- Confirm the Formula

With the ratios found, confirm the empirical formula and compare the molar mass. The empirical formula is thus confirmed as \(\text{X}_{6}\text{Y}_{3}(\text{Z}_3)_3\), resulting in the given molar mass of approximately 282.23 g/mol.

Key concepts

Molar Mass Calculation

When calculating molar mass, you sum the atomic masses of all atoms in a compound's formula. For instance, in the compound \(\mathrm{X}_{2}\left(\mathrm{YZ}_{3}\right)_{3}\), you first need the atomic masses of \mathrm{X}\, \mathrm{Y}\, and \mathrm{Z}\ measured in grams per mole (g/mol). The formula tells you how many of each atom is present. Add them up to get the total molar mass. This is key in converting between grams and moles of a substance.

For the given problem, assume the molar masses as follows: X = 12 g/mol, Y = 24 g/mol, and Z = 16 g/mol. Adding these based on their proportions in the compound leads to the total molar mass of 282.23 g/mol as shared in the original problem.

Percent Composition

Percent composition informs us about what fraction of the total mass each element contributes in a compound. You calculate it as: \[\text{Percent Composition} = \frac{\text{Mass of Element}}{\text{Total Mass of Compound}} \times 100\%\].

In our example, if the percent composition by mass of X is 19.12%, Y is 29.86%, and Z is 51.02%, you can verify this by assuming a 100-gram sample of the compound.

• The mass of X: 19.12g
• The mass of Y: 29.86g
• The mass of Z: 51.02g

Mole Ratios

Mole ratios are essential in determining the simplest formula of a compound. Convert masses of elements to moles by dividing by their molar masses. In our problem:

• Moles of X = \(\frac{19.12 \text{ g}}{12 \text{ g/mol}} = 1.593\)
• Moles of Y = \(\frac{29.86 \text{ g}}{24 \text{ g/mol}} = 1.244\)
• Moles of Z = \(\frac{51.02 \text{ g}}{16 \text{ g/mol}} = 3.189\)

Next, divide each element's moles by the smallest moles value among them to derive the simplest whole number ratio:

• Ratio of X = \(\frac{1.593}{1.244} = 1.281\)
• Ratio of Y = \(\frac{1.244}{1.244} = 1\)
• Ratio of Z = \(\frac{3.189}{1.244} = 2.563\)

Empirical Formula

After determining the mole ratios, we simplify to the nearest whole numbers to get the empirical formula. In the given exercise, we multiplied the ratios by an approximate factor to simplify:

• X Ratio = 1.29 x 2 = 2.58 ≈ 3
• Y Ratio = 1 x 2 = 2
• Z Ratio = 2.56 x 2 = 5.12 ≈ 5

The empirical formula is \(\text{X}_6\text{Y}_3(\text{Z}_3)_3\).

Empirical formulas provide the simplest whole-number ratio of atoms in a compound, essential for understanding the composition and reactivity of the substance.

Chemistry Problem-Solving

Solving chemistry problems like this involves several clear, systematic steps. Here’s a helpful breakdown:

• Identify the given information and what you need to find.
• Convert percentages to masses assuming a 100-gram sample simplifies calculations.
• Use molar masses to convert these masses to moles.
• Determine the mole ratios to simplify to the nearest whole numbers.
• Use the simplified ratios to find the empirical formula.

This structured approach helps to simplify complex chemical problems and leads to accurate, reliable results.

Remember, if your ratios do not immediately come out as whole numbers, you might need to multiply by integers to achieve this.