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A 1.42-g sample of a pure compound with formula m2so4 was dissolved in water and treated with an ex- cess of aqueous calcium chloride, resulting in the precipi- tation of all the sulfate ions as calcium sulfate. the pre- cipitate was collected, dried, and found to weigh 1.36 g. determine the atomic mass of m and identify m.

Respuesta :

The reaction between [tex]m_{2}SO_{4}[/tex] with calcium chloride can be shown as-[tex]m_{2}SO_{4}[/tex]+[tex]CaCl_{2}[/tex]→[tex]CaSO_{4}[/tex]↓+2mCl. The molecular weight of [tex]CaSO_{4}[/tex] is 136.14g. The weight of sulfate ion is 96.06g. The molecular weight of [tex]m_{2}SO_{4}[/tex] = (2×m + 96.06). From the reaction we can see that 1 mole of calcium chloride reacts with 1 moles[tex]m_{2}SO_{4}[/tex] to produce 1 mole of calcium sulfate. Now 1.36g of calcium sulfate is equivalent to 1.36/136.14=9.989×[tex]10^{-3}[/tex] moles of calcium sulfate.

Thus, 9.989×[tex]10^{-3}[/tex] moles of [tex]m_{2}SO_{4}[/tex] reacts in this reaction.

Let assume the atomic mass of m is x thus the molecular weight of [tex]m_{2}SO_{4}[/tex] is 2x+96.

So we may write 9.989×[tex]10^{-3}[/tex]× (2x+96) =1.42

Or, 2x + 96 = 142.146

Or, 2x = 46.146

Or, x = 23.073

Thus the atomic mass of m is 23.073. The atom (m) is sodium (Na).  

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