Three pure compounds form when 1.00 gg samples of element X combine with 0.472 gg, 0.210 gg , and 0.157 gg of element ZZ, respectively. The first compound has the formula X2Z3. Find the empirical formulas of the other two compounds.

We are tasked with finding the empirical formulas of two compounds formed by elements X and Z when 1.00 g of element X combines with different masses of element Z. Let's break down the information and solve step by step.

### Step 1: Determine the molar ratio for the first compound (X₂Z₃)
We know that for the first compound, the formula is X₂Z₃, and 1.00 g of X combines with 0.472 g of Z. 

To find the empirical formula for the other two compounds, we will first calculate the mass ratio of Z to X for the first compound (X₂Z₃). The mass ratio will serve as a reference for the other two compounds.

1. Assume the atomic mass of X is \( M_X \) and the atomic mass of Z is \( M_Z \).
2. In the compound \( \text{X}_2\text{Z}_3 \), the mass of X is \( 2M_X \) and the mass of Z is \( 3M_Z \).
3. Given that 1.00 g of X combines with 0.472 g of Z, the mass ratio of Z to X is:
   \[
   \frac{0.472}{1.00} = \frac{3M_Z}{2M_X}
   \]
4. Therefore, we have the equation:
   \[
   \frac{3M_Z}{2M_X} = 0.472
   \]
   Solving for \( \frac{M_Z}{M_X} \), we get:
   \[
   \frac{M_Z}{M_X} = \frac{2}{3} \times 0.472 = 0.315
   \]
This ratio of the atomic masses of Z to X is \( \frac{M_Z}{M_X} = 0.315 \).

### Step 2: Analyze the second compound
For the second compound, 1.00 g of X combines with 0.210 g of Z. We can use the same method as for the first compound to find the empirical formula.

1. The mass ratio of Z to X in this compound is:
   \[
   \frac{0.210}{1.00} = \frac{nM_Z}{mM_X}
   \]
   where \( m \) and \( n \) represent the subscript values in the empirical formula.
2. Using \( \frac{M_Z}{M_X} = 0.315 \), the ratio becomes:
   \[
   \frac{0.210}{1.00} = \frac{n(0.315)}{m}
   \]
   Simplifying:
   \[
   0.210 = \frac{0.315n}{m}
   \]
   \[
   \frac{n}{m} = \frac{0.210}{0.315} = 0.667
   \]
   This ratio \( \frac{n}{m} \approx \frac{2}{3} \), so the empirical formula for the second compound is likely \( X_3Z_2 \).

### Step 3: Analyze the third compound
For the third compound, 1.00 g of X combines with 0.157 g of Z. Following the same method:

1. The mass ratio of Z to X is:
   \[
   \frac{0.157}{1.00} = \frac{nM_Z}{mM_X}
   \]
2. Using \( \frac{M_Z}{M_X} = 0.315 \), we get:
   \[
   0.157 = \frac{0.315n}{m}
   \]
   \[
   \frac{n}{m} = \frac{0.157}{0.315} = 0.498
   \]
   This ratio \( \frac{n}{m} \approx \frac{1}{2} \), so the empirical formula for the third compound is likely \( X_2Z \).

### Final Empirical Formulas:
- **Second compound**: \( X_3Z_2 \)
- **Third compound**: \( X_2Z \)

Would you like more details or have any questions?

Here are 5 related questions to explore:
1. How do you determine the molecular formula from the empirical formula?
2. What are the differences between empirical and molecular formulas?
3. How can you calculate the molar mass of a compound using its empirical formula?
4. How is the law of multiple proportions applied in this context?
5. How do you experimentally determine the mass of elements in a compound?

**Tip**: Always ensure that you simplify the atomic ratio to its smallest whole number to get the correct empirical formula.

Three pure compounds form when 1.00 g samples of element X combine with 0.472 g, 0.210 g, and 0.157 g of element Z, respectively. The first compound has the formula X2Z3. Find the empirical formulas of the other two compounds.

Discover how to find the empirical formulas of compounds formed by elements X and Z, given mass ratios. The problem demonstrates the calculation of mass ratios and application of the Law of Multiple Proportions to find the formulas X3Z2 and X2Z. Suitable for Grades 10-12.

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