A 1.42-g sample of a pure compound, with formula M2SO4, was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulphate ions as calcium sulphate. The precipitate was collected, dried, and found to weigh 1.36 g. Determine the atomic mass of M, and identify M.

Question

Please explain!

Answer 1

..........M2SO4 + CaCl2 --> CaSO4 + 2MCl
initial....1.42g....0........0........0
add................xs..........
change.......................1.36
equil......0..................1.36
moles CaSO4 = 136/molar mass CaSO4.
moles M2SO4 = moles CaSO4 (1 mole CaSO4 came from 1 mole M2SO4.)
moles M2SO4 = grams/molar mass. You know moles from above and grams from the problem, solve for molar mass.
Now that you know the molar mass of M2SO4, subtract 96 (due to the SO4) to leave 2 M^+, then divide by 2 to obtain M. You can look on the periodic table to identify M.

Answer 2

Atomic mass of the M is 23 g/mol
So M is actually Sodium (Na)