Question

3. Sulfur (S) and fluorine (F) combine to form three different compounds. Compound Ahas 1.118 g of F for every 1.000 9 S. Compound Bhas 2.375 g of F
for every 1.000 9 S. Compound Chas 3.563 of F for every 1.000 g S. How do these compounds support the law of multiple proportions?

Answer 1

Answer:

Compound A - $S_{2}F_{1}$

Compound B - $S_{3}F_{1}$

Compound C - $S_{4}F_{1}$

Different proportions of sulfur can be reacted with same mass of fluorine.

Explanation:

Law of multiple proportions:

When two elements form series of compounds.The ratio of the masses of the second element that combine with one gram of the first element can always be reduced to small whole numbers.

Compound -A

Mass of sulfur = 1.118 g

Mas of fluorine = 1.000 g

Divide the small whole number to each number.

$\frac{1.188}{1.000};\frac{1.000}{1.000}=2:1$

The formula of the compound will be $S_{2}F_{1}$

Compound -B

Mass of sulfur = 2.375 g

Mas of fluorine = 1.000 g

Divide the small whole number to each number.

$\frac{2.375}{1.000};\frac{1.000}{1.000}=3:1$

The formula of the compound will be $S_{3}F_{1}$

Compound -C

Mass of sulfur = 3.563 g

Mas of fluorine = 1.000 g

Divide the small whole number to each number.

$\frac{3.563}{1.000};\frac{1.000}{1.000}=3:1$

The formula of the compound will be $S_{4}F_{1}$

If we observe the formula of three compound have same mass of fluorine is reacted with the different masses of sulfur.

Hence, law of multiple proportion is proved.