To analyze for barium in an unknown white powder, an excess of sodium sulfate solution was added to a 0.576 g sample of the powder dissolved in water. A white precipitate of BaSO4 was isolated, dried, and found to weigh 0.438 g. What is the mass percent of Ba in the white powder
I got 44.5 for the answer i can show u my work if needed
I tried it and obtained 44.7%. I used 233.39 for molar mass BaSO4 and 137.33 for atomic mass Ba.
137.33/233.43*.438=.258
.258/567*100= 44.8
got it now thanks
Zxff
Sure, I'd be happy to help verify your work. Please go ahead and show me your calculations so I can assist you in determining the correct answer.
To find the mass percent of Ba in the white powder, we need to calculate the percentage of Ba in the BaSO4 precipitate.
First, we need to calculate the moles of BaSO4 precipitate formed. The molar mass of BaSO4 is 233.39 g/mol.
0.438 g BaSO4 * (1 mol BaSO4 / 233.39 g BaSO4) = 0.001875 mol BaSO4
Since the reaction is 1:1 between BaSO4 and Ba, the moles of Ba in the BaSO4 precipitate is also 0.001875 mol.
Next, we need to calculate the moles of Ba in the original sample. The molar mass of Ba is 137.33 g/mol.
0.001875 mol Ba * (137.33 g Ba / 1 mol Ba) = 0.257 grams Ba
Finally, we can calculate the mass percent of Ba in the white powder:
Mass percent of Ba = (mass of Ba / mass of powder) * 100
Mass of powder = 0.576 g (given)
Mass percent of Ba = (0.257 g Ba / 0.576 g powder) * 100 = 44.6%
So, based on your calculations, the correct answer is 44.6%, not 44.5%.