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Chapter 2: Problem 42

Several compounds containing sulfur and fluorine are known. Three of them have the following compositions: i. 1.188 \(\mathrm{g}\) of \(\mathrm{F}\) for every 1.000 \(\mathrm{g}\) of \(\mathrm{S}\) ii. 2.375 \(\mathrm{g}\) of \(\mathrm{F}\) for every 1.000 \(\mathrm{g}\) of \(\mathrm{S}\) iii. 3.563 \(\mathrm{g}\) of \(\mathrm{F}\) for every 1.000 \(\mathrm{g}\) of \(\mathrm{S}\) How do these data illustrate the law of multiple proportions?

Short Answer

Expert verified
The given data illustrates the law of multiple proportions as the mass ratios of fluorine to sulfur in the three compounds have simple whole number ratios. For every 1 g of sulfur, compound ii has two times more fluorine than compound i, and compound iii has three times more fluorine than compound i. This supports the law of multiple proportions.

Step by step solution

01

Calculate the mass ratios of fluorine to sulfur in each compound.

For every 1 g of sulfur in each compound, we are given the mass of fluorine. i. 1.188 g of F for every 1 g of S ii. 2.375 g of F for every 1 g of S iii. 3.563 g of F for every 1 g of S The fluorine to sulfur mass ratios for the three compounds are already given, so we will use those in the next step.
02

Find the ratios between each mass ratio

Now we will find the ratio between these mass ratios to see if there is a simple whole number ratio. The easiest way to do this is to take the smallest value as the base and find the ratios with others. 1.188 g F/g S (compound i) 2.375 g F/g S (compound ii) ÷ 1.188 g F/g S (compound i) = 1.997 3.563 g F/g S (compound iii) ÷ 1.188 g F/g S (compound i) = 2.997
03

Round the ratios to the nearest whole numbers

Now, we will round the ratios found in step 2 to the nearest whole numbers. Compound ii : 1.997 ≈ 2 Compound iii: 2.997 ≈ 3
04

Conclude the relation to the law of multiple proportions

As we can see, the ratios found in step 3 are simple whole numbers. The compound ii has two times more fluorine in its composition than compound i, while compound iii has three times more fluorine in its composition than compound i. Since we found the small whole number ratios between the mass of fluorine that combines with a fixed mass (1 g) of sulfur in these compounds, the data given illustrates the law of multiple proportions.

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Most popular questions from this chapter

A sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) contains 2.02 \(\mathrm{g}\) of hydrogen, 32.07 \(\mathrm{g}\) of sulfur, and 64.00 \(\mathrm{g}\) of oxygen. How many grams of sulfur and grams of oxygen are present in a second sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) containing 7.27 \(\mathrm{g}\) of hydrogen?

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