6. A chemist analyzes the sulfate salt of an unknown alkaline earth metal. The c

Question

6. A chemist analyzes the sulfate salt of an unknown alkaline earth metal. The chemist adds 1.273 g of th salt to excess barium chloride solution. After filtering and drying the mass of precipitate is found to be 2.468 g (a) Use the formula MS04 to represent the unknown salt. Write the molecular and net ionic equations for the reaction. (2) (b) Calculate the amount (in mol) of MS04 used in the reaction. (3) (c) Determine the molar mass of the unknown salt. (2) (d) What is the likely identity of the unknown metal cation? What test might the chemist perform to help confirm this conclusion? (3)

Explanation / Answer

1. the reaction is

MSO4 + BaCl2 === BaSO4 (s)+ MCl2, the barium sulphate will precipitate

the ionic equation is

M+2 + SO4-2 + Ba+2 + 2Cl*-1 ====== BaSO4 (s) + M+2 + 2Cl-1

remove the ions that appears on both sides

  SO4-2 + Ba+2 ==== BaSO4 (s)

calculate the moles of BaSO4, molar mass of baso4 = 233.38 g/mol

moles = 2.468 / 233.38 = 0.010575 moles of precipitate

so if we see the reaction 1 mole of salt produces 1 mole of precipitate so there are 0.010575 moles of unknown salt

the molar mass of the SO4 ion is 96 g/gmol then

moles = mass / molar mass

moles = 1.273 / (96 + X) = 0.010575

1.273 = 1.0152 + 0.010575x

1.273 - 1.0152 = 0.010575 x

x = 0.2578 / 0.010575 = 24.37 g/gmol

so we must look at the periodic table, since the form is MSO4, the unknown must be an element from group two so the metal that matches this description is Magnesium (Mg)

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