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
