Sulfur and oxygen form both sulfur dioxide and sulfur trioxide. when samples of these are decomposed, the sulfur dioxide produces 3.49 g oxygen and 3.50 g sulfur, while the sulfur trioxide produces 6.75 g oxygen and 4.50 g sulfur. calculate the mass of oxygen per gram of sulfur for each sample and show that these results are consistent with the law of multiple proportions.

The law of multiple proportions states that two elements when combined to form more than one compound, the mass of one element combines with a fixed mass of the other element in ratios of whole numbers. It is important to note that this only applies to compounds formed from the same elements, such as sulfur dioxide, SO2, and sulfur trioxide, SO3.

The atomic mass of sulfur is 32 while that of oxygen is 16 when rounded off to a whole number. Thus, we can say that sulfur dioxide is composed of 32 parts by mass of sulfur and 32 parts by mass of oxygen, thus, the ratio of sulfur to oxygen is 32:32. On the other hand, sulfur trioxide is composed of 32 parts by mass of sulfur and 48 parts by mass of oxygen, thereby, the ratio of sulfur to oxygen is 32:48. We then divide both ratios by the smallest number which is 32. For sulfur dioxide, for every 1 g of sulfur, there is also 1 g of oxygen. For sulfur trioxide, for every 1 g of sulfur, there is 1.5 g of oxygen. With this, we take the ratio of the oxygen for the two compounds. The ratio of the oxygen of SO2 to oxygen of SO3 is 1:1.5. To prove that the results from the experiment is consistent with the law of multiple proportions, we take the ratio of the mass of oxygen per gram of sulfur for each sample which was given in the problem.

For SO2,

Mass of O2/ g sulfur = 3.49/3.50 = 0.9971  1

For SO3,  

Mass of O2/ g sulfur = 6.75/4.50 = 1.5

Thus, we have proved that the results are consistent with the law of multiple proportions.

pure water does not conduct electricity very well. however, when certain substances are dissolved in water, the solution does conduct electricity. you can make a simple device that shows how well a solution conducts electricity. this device uses a flashlight bulb to indicate how well the solution conducts electricity. the better the solution conducts electricity, the brighter the bulb will glow.

explanation: