1.

A mineral contained MgO= 31.88%, SiO_(2)=63.37% and H_(2)O=4.75%. Show that the simplest formula for the mineral is H_(2)Mg_(3)Si_(4)O_(12). (H=1, Mg=24, Si=28, O=16)

Answer»

Solution :Suppose the weight of the mineral is 100g. Then weight of MGO= 31.88g
weight of `SO_(4)=` 63.37g,
weight of `H_(2)O`=4.75g.
Moles of Mg in MgO= `1 XX` moles of MgO
`=(31.88)/(40)=0.797`
Moles of Si in `SiO_(2)=1 xx` moles of `SiO_(2)`
`=(63.37)/(60)=1.0561`
Moles of H in `H_(2)O=2 xx` moles of `H_(2)O`
`=(2xx4.75)/(18)= 0.5278`
moles of O = moles of O in MgO+ moles of O in `SiO_(2)` + moles of O in `H_(2)O`
`=1 xx` moles of MgO +2 `xx` moles of `SiO_(2)+1 xx` moles of `H_(2)O`
`=(31.88)/(40) + (2 xx 63.37)/(60)+ (4.75)/(18)=3.172`
Moles of O= 3.172
Now, by inspection, we have moles of H= 0.5278 `=0.2639 xx 2`
moles of Mg= 0.797 `~~0.2639 xx 3`
moles of Si=1.0561 `~~0.2639 xx 4`
moles of O=3.172 `~~0.2639 xx 12`
`therefore H: Mg: Si: O= 2: 3: 4: 12`
The FORMULA is `H_(2)Mg_(3)Si_(4)O_(12)`


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