Answer to Question #289567 in General Chemistry for Fish

Question

Mass of crucible+lid+Mg= 10.90

Crucible+lid= 10.03

Mass of MgO formed - Mass of crucible= 10.90-10.03= 0.87

Calculate:

Mass of MgO

Amount of Mg in the compound

Mass of O2 in the compound

Amount of oxygen in mole

Simple number ratio to oxygen to Mg

Empirical formula of the compound.

Answer 1

All information not given.

Suppose mass of crucible + lid + MgO$=11.47$

Mass of Mg $=10.90-10.03$

$=0.87$

Mass of MgO$=11.47-10.03$

$=1.44$

Mass of O$=1.44-0.87=0.57$

Number Of moles of Mg$=\frac{0.87}{24.305}=0.035795$

Number of moles of O$=\frac{0.57}{16}=0.035625$

Oxygen to Magnesium $=1:1$

Empirical formula $=$ MgO