A chemist 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%