Achemist is asked to find the percentage of mgso4 in a commercial sample of epsom salts. a 1.2500 g sample of an epsom salts preparation is dissolved and treated with excess barium chloride solution. the dried barium sulfate precipitate weighs 0.9165 g. calculate the percent by mass of mgso4 in the original mixture.

37.81%  
The balanced equation for the reaction is
 BaCl2 + MgSO4 ==> MgCl2 + BaSO4 
 So for every mole of MgSO4 consumed, 1 mole of BaSO4 should be produced. Let's calculate the molar mass of MgSO4 and BaSO4.
 Atomic weight magnesium = 24.305
 Atomic weight oxygen = 15.999
 Atomic weight sulfur = 32.065 
 Atomic weight barium = 137.327 
  Molar mass MgSO4 = 24.305 + 32.065 + 4 * 15.999 = 120.366 g/mol
 Molar mass BaSO4 = 137.327 + 32.065 + 4 * 15.999 = 233.388 g/mol 
 Moles BaSO4 = 0.9165 g / 233.388 g/mol = 0.003926937 mol
 Mass MgSO4 = 0.003926937 mol * 120.366 g/mol = 0.472669699 g 
 % by mass of MgSO4 = 0.472669699 g / 1.2500 g = 0.378135759 =
37.8135759% 
 And since we only have 4 significant digits in our data, the result rounds to
37.81%

It could possibly be beneficial if the experiment is successful.

Answer trigonal pyramidal;