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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.