Question
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
Please help I'm lost. I need step by step Please and thank you
Answers
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.
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.
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