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What is the empirical formula of a compound containing 24.56% potassium, 34.81% manganese, and 40.50% oxygen?

User McRist
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2 Answers

3 votes
the answer you are looking for is KMnO4
User ALUFTW
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2 votes
A(K)=39.098
A(Mn)=54.938
A(O)=15.994

w(K)=0.2456 (24.56%)
w(Mn)=0.3481 (34.81%)
w(O)=0.4050 (40.50%)

KxMnyOz

the molar mass is
M(KxMnyOz)=xA(K)+yA(Mn)+zA(O)


w(K)=xA(K)/[xA(K)+yA(Mn)+zA(O)]
w(Mn)=yA(Mn)/[xA(K)+yA(Mn)+zA(O)]
w(O)=zA(O)/[xA(K)+yA(Mn)+zA(O)]

substitute the numbers and solve the system of three equations
the calculated values for x, y and z
x=1
y=1
z=4
KMnO₄
User PhantomM
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