A 1.250 g sample of copper wire was heated in air and reacted with oxygen to give 1.565 g of copper oxide product. Calculate the empirical formula of the copper oxide.
Please help I'm lost. I need step by step Please and thank you
When I worked the question I got cu = 0.0196 and o= 0.0196 so the empirical formula is CUO
If you had shown your work I could have found the error. As It is I have nothing to go on.
1.565g = mass Cu + oxygen
-1.250 = mass Cu
-----------------------
0.315g = mass oxygen
Convert g Cu and g oxygen to moles.
moles = grams/molar mass
moles Cu = 1.250/63.54 = about 0.0196 which agrees with your number BUT you gave no units to tell me it was # moles Cu.
moles oxygen = 0.315/16 = 0.0197 so the ratio is indeed 1:1 and the formula is CuO and you made no errors.
When Cu is heated in air the product formed is Cu2O, not CuO.
Sure! I can help you step by step to determine the empirical formula of the copper oxide. Here's how you can approach it:
1. Determine the mass of copper (Cu) in the copper wire:
Since the sample is 1.250 g and it contains only copper wire, the mass of copper is also 1.250 g.
2. Determine the mass of oxygen (O) that combined with the copper:
The mass of oxygen can be calculated by subtracting the mass of copper from the mass of the copper oxide product.
Mass of oxygen = Mass of copper oxide product - Mass of copper
Mass of oxygen = 1.565 g - 1.250 g
Mass of oxygen = 0.315 g
3. Convert the masses of copper and oxygen to moles:
To convert the masses to moles, you need the molar mass of copper and oxygen.
The molar mass of copper is 63.55 g/mol, and the molar mass of oxygen is 16.00 g/mol.
Moles of copper = Mass of copper / Molar mass of copper
Moles of copper = 1.250 g / 63.55 g/mol
Moles of copper = 0.01967 mol
Moles of oxygen = Mass of oxygen / Molar mass of oxygen
Moles of oxygen = 0.315 g / 16.00 g/mol
Moles of oxygen = 0.01969 mol
4. Determine the empirical ratio:
Divide the moles of each element by the smallest number of moles to get the empirical ratio.
Empirical ratio (Cu:O) = Moles of copper / Moles of oxygen
Empirical ratio (Cu:O) = 0.01967 mol / 0.01969 mol
Empirical ratio (Cu:O) ≈ 0.998 : 1
Since the ratio is approximately 1:1, it means the empirical formula of the copper oxide is CuO.
Therefore, the empirical formula of the copper oxide is CuO.