Final answer:
To compare copper and chlorine compounds, the masses of copper per gram of chlorine are calculated for each, and then the ratio of these masses is determined, revealing a small whole-number ratio, indicating that Compound A has twice the amount of copper per chlorine compared to Compound B.
Step-by-step explanation:
The question involves applying the law of multiple proportions to two copper-chlorine compounds, Compound A and Compound B. To compare the masses of copper per gram of chlorine in the two samples, you must first find the mass of copper that combines with exactly 1.00 g of chlorine in each compound:
- For Compound A: (32.10g Cu) / (17.90g Cl) = 1.79g Cu per 1g Cl
- For Compound B: (7.53g Cu) / (8.40g Cl) = 0.896g Cu per 1g Cl
Then, to find the ratio of the masses of copper in the two compounds, divide the larger mass by the smaller mass:
(1.79g Cu in Compound A) / (0.896g Cu in Compound B) = 2
This result tells us that for a given mass of chlorine, Compound A contains twice the mass of copper as does Compound B, which suggests a small whole-number ratio in accordance with the atomic theory.