A 7.96 gram sample of silver reacts with oxygen to form 8.55 gram of the metal oxide.What is the formula of the oxide?

Question

Quizplus · Accepted Answer

First, we need to find the mass of oxygen that reacted with the silver. Mass of oxygen = 8.55 g - 7.96 g = 0.59 g Next, we need to find the mole ratio of silver to oxygen in the compound. Moles of silver = 7.96 g / 107.87 g/mol = 0.074 moles Moles of oxygen = 0.59 g / 15.999 g/mol = 0.037 moles The simplest ratio of moles of silver to oxygen is 2:1. This means that the formula of the oxide is Ag2O. However, since the question is asking for the full formula, we need to check if there are any subscripts needed for the elements. The only possible element that could have a subscript is sulfur (S), as it is often found with silver. To check if sulfur is present, we need to calculate the expected mass of the compound assuming it is Ag2O and compare it to the actual mass given in the question. Expected mass of Ag2O = 2(107.87 g/mol) + 15.999 g/mol = 231.74 g/mol Actual mass given in question = 8.55 g Since the actual mass is less than the expected mass, this suggests that there is another element present with a lower atomic mass. This could be sulfur. To account for sulfur, we need to adjust the formula to Ag2₂O. However, we need to make sure that the subscripts are in their simplest ratio. Moles of sulfur = (8.55 g - 7.96 g) / 32.06 g/mol = 0.0186 moles The mole ratio of sulfur to silver is 0.0186 moles / 0.074 moles = 0.25 or 1/4 To get whole number subscripts, we can multiply both subscripts by 4, giving us the final formula of Ag₂O.

A 796 Gram Sample of Silver Reacts with Oxygen to Form