Determine the percentage by mass of sulfate in the fertilizer.
A sulfate is precipitated as barium sulfate from a solution containing a known mass of the fertiliser by adding an excess of barium chloride solution:
Ba2+ (aq) + SO42- (aq) ---- > BaSO4 (s)
The proportion of sulfate ions and therefore of sulfur in the fertiliser is determined by collecting and weighing the precipitate that is formed during the above reaction.
If the mass of precipitate = 1.02g, mass of plastic = 64.65g and mass of both precipitate and plastic = 65.67g,
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Since an excess of BaCl2 is added, we can be sure we precipitate all of the sulfate as the barium sulfate. We need the initial mass of fertilizer to determine the mass percent. Since the initial mass isn't given, let's assume an initial mass of 5.00 grams of fertilizer.
For mass percent of sulfate in the fertilizer, we need to find
mass of sulfate / mass of fertilizer * 100
The barium sulfate forms a solid which can then be dried and weighed. The mass of the sample will include both the barium and the sulfate so we have to convert to moles to determine the mass of just the sulfate. We don't want to include the barium in the mass percent because it's only asking for the sulfate.
1.02 g BaSO4 * 1 mol BaSO4/233.4 g * 1 mol SO4^2-/1 mol Ba^2+ * 96.06 g/mol SO4
= 0.4198 g SO4^2-
Now that we know the mass of sulfate, we can find the mass percent
0.4198 g SO4^2- / 5.00 g fertilizer * 100
= 8.40 % sulfate by mass in the fertilizer
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