Asked by sh
I have the same question as Chopsticks
determine the emperical formula of the compound
25.3%copper, 12.9%sulphur, 25.7% oxygen, 36.1% water
Can the answer be CuSO4(H2O)5?
or does it have to be CuSO4 x 5H2O
thanks in advance.
determine the emperical formula of the compound
25.3%copper, 12.9%sulphur, 25.7% oxygen, 36.1% water
Can the answer be CuSO4(H2O)5?
or does it have to be CuSO4 x 5H2O
thanks in advance.
Answers
Answered by
bobpursley
Assume you have 100 g of the stuff.
25.3gCu=.398 mol Cu
12.9gS=.401 mol S
25.7gO=1.61 mol O
36.1gwater=2.01mol water.
Normalize, by dividing by the lowest number of moles
Cu 1
S 1
O 4
water 5
Yes, you are right.
25.3gCu=.398 mol Cu
12.9gS=.401 mol S
25.7gO=1.61 mol O
36.1gwater=2.01mol water.
Normalize, by dividing by the lowest number of moles
Cu 1
S 1
O 4
water 5
Yes, you are right.
Answered by
DrBob222
You may right the formula as
CuSO4(H2O)5 or CuSO4.5H2O or CuSO4*5H2O.
Most texts write it as CuSO4.5H2O BUT these boards don't let us put the decimal point in the middle of the line so we write a period (.) or an asterisk (*). Neither means TIMES. CuSO4*5H2O is read as copper II sulfate with 5 molecules of water of crystallization or as copper II sulfate pentahydrate.
CuSO4(H2O)5 or CuSO4.5H2O or CuSO4*5H2O.
Most texts write it as CuSO4.5H2O BUT these boards don't let us put the decimal point in the middle of the line so we write a period (.) or an asterisk (*). Neither means TIMES. CuSO4*5H2O is read as copper II sulfate with 5 molecules of water of crystallization or as copper II sulfate pentahydrate.
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