Question #6bd7c

Question

1300 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-18T17:11:04

#sf(Cu_2O)#

Mass crucible + oxide #sf(= 27.128" " g)#
Mass crucible #sf(" " = 27.002" " g)#

#:.# Mass oxide = 27.128 - 27.002 = 0.126 g

Mass crucible + copper #sf(=27.114" "g)#
Mass crucible #sf(" "=27.002" "g)#

#:.# Mass copper = 27.114 - 27.002 = 0.112 g

#:.# Mass oxygen = 0.126 - 0.112 = 0.014 g

Ratio Cu : O in grams#sf(rArr)#

0.112 : 0.014

Ratio Cu : O in moles #rArr#

#sf(0.112/(63.546):0.014/15.9994)#

#sf(=0.0017625:0.00087503)#

#sf(=1:0.496)#

#sf(=2:0.9929)#

#sf(~=2:1)#

So the empirical formula is #sf(Cu_2O)# which is copper(I) oxide.