If given: A 1.4470g sample of an oxide of gold was reduced with ethane to 1.2900g of gold metal. What is the simple formula of the original gold oxide?
How many moles of gold is 1.29g? How many moles of O is in (1.447-1.29)g of oxygen?
divide the lower number of moles (gold, or O) into the larger, that is the mole ratio of the elements in the formula.
To determine the simple formula of the original gold oxide, we need to use the concept of chemical stoichiometry.
First, we need to calculate the mass of oxygen in the original gold oxide by subtracting the mass of gold metal produced from the initial mass of the sample:
Mass of oxygen = Initial mass of sample - Mass of gold metal produced
= 1.4470 g - 1.2900 g
= 0.1570 g
Next, we can convert the mass of oxygen to moles using the molar mass of oxygen, which is approximately 16.00 g/mol:
Moles of oxygen = Mass of oxygen / Molar mass of oxygen
= 0.1570 g / 16.00 g/mol
= 0.00981 mol
Now, we need to determine the mole ratio between gold and oxygen in the original gold oxide. Since the gold metal is produced from the reduction with ethane, we can assume that all the gold in the oxide is converted into gold metal. Therefore, the mole ratio between gold and oxygen in the original gold oxide is 1:1.
Finally, we write the simple formula of the original gold oxide using the mole ratio. The formula would be AuO.
Therefore, the simple formula of the original gold oxide is AuO.