Question: Show that these results are consistent with the law of multiple propor

Question

Question: Show that these results are consistent with the law of multiple proportions (fluorine in sulfur hexafluoride : fluorine in sulfur tetrafluoride = ). Express your answers as integers separated by a comma.

Sulfur and fluorine form several different compounds including sulfur hexafluoride and sulfur tetrafluoride. Decomposition of a sample of sulfur hexafluoride produced 4.48 g of fluorine and 1 g of sulfur, while decomposition of a sample of sulfur tetrafluoride produced 4.45 g of fluorine and 1.88 g of sulfur

Explanation / Answer

decomposition of sulfur hexafluoride produces 4.48 g of fluorine

number of moles of fluorine =mass/ molecular weight = 4.48 / 18.99 = 0.235 M

mass of sulfur hexafluoride taken = 4.48 + 1.26 = 5.74g

moles of sulfur hexafluoride = 5.74 / 146.06 = 0.039

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decomposition of sulfur tetrafluoride produces 4.45 g of fluorine

number of moles of fluorine =mass/ molecular weight = 4.45 / 18.99 = 0.234 M

mass of sulfur tetrafluoride taken = 4.45 + 1.88 = 6.33 g

moles of sulfur tetrafluoride = 6.33 / 108.07 = 0.0586

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0.0586 moles of  tetrafluoride produced 4.45 g of fluorine

0.039 moles of  tetrafluoride produce = (0.039 X 4.45)/0.0586 = 2.96

Therefore 0.039 moles of sulfur hexafluoride produce 4.48 g of fluorine whereas same moles of sulfur tetrafluoride produce 2.96 g of sulfur

mass of sulfur in hexafluoride / mass of sulfur in tetrfluorid =4.48 / 2.96 = 1.5

which is in the ratio of 3 : 2

Hence it follows law of multiple proportion ( 6F/4F = 6/4 = 3/2 =1.5)

fluorine is in the ratio of multiples

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