Alright students, let's go through this step-by-step to understand how these data points illustrate the law of reciprocal proportions.
### Given Data:
1. Hydrogen Sulfide ([tex]\(\text{H}_2\text{S}\)[/tex]) contains [tex]\( 94.1\% \)[/tex] Sulfur (S).
2. Sulfur Dioxide ([tex]\(\text{SO}_2\)[/tex]) contains [tex]\( 50\% \)[/tex] Sulfur (S).
3. Water ([tex]\(\text{H}_2\text{O}\)[/tex]) contains [tex]\( 11.11\% \)[/tex] Hydrogen (H).
### Molecular Weights:
- Molecular weight of [tex]\(\text{H}_2\text{S}\)[/tex] is approximately [tex]\(34.08 \, \text{g/mol}\)[/tex].
- Molecular weight of [tex]\(\text{SO}_2\)[/tex] is approximately [tex]\(64.07 \, \text{g/mol}\)[/tex].
- Molecular weight of [tex]\(\text{H}_2\text{O}\)[/tex] is approximately [tex]\(18.015 \, \text{g/mol}\)[/tex].
### Step-by-Step Solution:
#### 1. Calculate the actual weight of S in 100g of each compound:
- Weight of S in 100g of [tex]\(\text{H}_2\text{S}\)[/tex]:
[tex]\[
\text{Weight of S in \(\text{H}_2\text{S}\)} = \frac{94.1}{100} \times 100 = 94.1 \, \text{g}
\][/tex]
- Weight of S in 100g of [tex]\(\text{SO}_2\)[/tex]:
[tex]\[
\text{Weight of S in \(\text{SO}_2\)} = \frac{50}{100} \times 100 = 50 \, \text{g}
\][/tex]
- Weight of H in 100g of [tex]\(\text{H}_2\text{O}\)[/tex]:
[tex]\[
\text{Weight of H in \(\text{H}_2\text{O}\)} = \frac{11.11}{100} \times 100 = 11.11 \, \text{g}
\][/tex]
#### 2. Calculate the amount of each element in 1 mole of the compound:
- Weight of S per mole of [tex]\(\text{H}_2\text{S}\)[/tex]:
[tex]\[
\text{Weight of S per mole of \(\text{H}_2\text{S}\)} = \frac{94.1}{100} \times 34.08 = 32.06928 \, \text{g}
\][/tex]
- Weight of S per mole of [tex]\(\text{SO}_2\)[/tex]:
[tex]\[
\text{Weight of S per mole of \(\text{SO}_2\)} = \frac{50}{100} \times 64.07 = 32.035 \, \text{g}
\][/tex]
- Weight of H per mole of [tex]\(\text{H}_2\text{O}\)[/tex]:
[tex]\[
\text{Weight of H per mole of \(\text{H}_2\text{O}\)} = \frac{11.11}{100} \times 18.015 = 2.0014665 \, \text{g}
\][/tex]
#### 3. Verify the law of reciprocal proportions:
The law of reciprocal proportion states that the ratios of the masses of one element that combine with a fixed mass of another element are the same as the ratios of the masses of these elements when they combine directly with each other.
We need to check the ratios:
- Ratio of masses of S in [tex]\(\text{H}_2\text{S}\)[/tex] and [tex]\(\text{SO}_2\)[/tex]:
[tex]\[
\text{Ratio} = \frac{\text{Weight of S per mole of \(\text{H}_2\text{S}\)}}{\text{Weight of S per mole of \(\text{SO}_2\)}} = \frac{32.06928}{32.035} \approx 1.001070079600437
\][/tex]
- Ratio of masses of H in [tex]\(\text{H}_2\text{O}\)[/tex] and S in [tex]\(\text{H}_2\text{S}\)[/tex]:
[tex]\[
\text{Ratio} = \frac{\text{Weight of H per mole of \(\text{H}_2\text{O}\)}}{\text{Weight of S per mole of \(\text{H}_2\text{S}\)}} = \frac{2.0014665}{32.06928} \approx 0.06241070894014458
\][/tex]
Since the calculated ratios are consistent:
- The ratio of S in [tex]\(\text{H}_2\text{S}\)[/tex] to S in [tex]\(\text{SO}_2\)[/tex] is approximately 1.
- The ratio of H in [tex]\(\text{H}_2\text{O}\)[/tex] to S in [tex]\(\text{H}_2\text{S}\)[/tex] is a fixed number.
These consistent ratios confirm that the initial data supports the law of reciprocal proportions.
