asked 26.3k views
5 votes
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.

asked
User Kiwiron
by
7.8k points

1 Answer

3 votes

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%
answered
User Prabhuram
by
8.1k points
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