胆矾:CuSO4·5H2O
n(CuSO4·5H2O)=m/M=2.5/250=0.01mol
CuSO4·5H2O===(加热)CuSO4·xH2O+(5-x)H2O
失去的质量就是失去的水的质量为2.5-2.14=0.36g
n(H2O)失=m/M=0.36/18=0.02mol
n(H2O)总=5*n(CuSO4·5H2O)=0.05mol
就是说水还剩0.05-0.02=0.03mol
那么n(CuSO4):n(H2O)=0.01:0.03=1:3
剩余固体的化学式CuSO4·3H2O