A salt contains only barium and one of halides ions. A 0.1480g sample of the salt was dissolved in water and an water and an excess of sulfuric acid was added to form barium sulfate which was filtered, dried and weighed its mass was found to be 0.1660g.what is the formula for for the barium Halide?
BaCl2
To determine the formula for the barium halide, we need to analyze the given information and perform some calculations.
1. Calculate the mass of barium sulfate formed:
Mass of barium sulfate = Mass of filter paper + Barium sulfate - Mass of filter paper
= 0.1660g - 0g (assuming the filter paper has negligible mass)
= 0.1660g
2. Convert the mass of barium sulfate to moles:
Molar mass of barium sulfate (BaSO4) = Molar mass of Ba + 4 × Molar mass of S + 4 × Molar mass of O
= 137.33 g/mol + 4 × 32.07 g/mol + 4 × 16.00 g/mol
= 233.40 g/mol
Moles of barium sulfate = Mass of barium sulfate / Molar mass of barium sulfate
= 0.1660g / 233.40 g/mol
≈ 0.000711 mol
3. The ratio between the moles of barium sulfate and the moles of the barium halide will give us the subscripts for the formula. Since barium sulfate has a 1:1 ratio with barium, the same ratio applies to the barium halide.
4. Calculate the moles of barium in the barium sulfate:
Moles of barium = Moles of barium sulfate
≈ 0.000711 mol
5. The molar mass of barium (Ba) is 137.33 g/mol. We can compare this value with the mass of the barium in grams to determine the number of moles, which will correspond to the subscript in the formula for the barium halide.
Mass of barium = Moles of barium × Molar mass of barium
= 0.000711 mol × 137.33 g/mol
≈ 0.0976 g
6. Calculate the mass of the halide component of the salt by subtracting the mass of barium from the total mass of the sample:
Mass of halide = Mass of sample - Mass of barium
= 0.1480 g - 0.0976 g
= 0.0504 g
7. The molar mass of the halide can be used to determine the number of moles, which will correspond to the subscript in the formula for the barium halide. You will need to look up the molar masses of the different halides: chloride (Cl-) = 35.45 g/mol, bromide (Br-) = 79.90 g/mol, and iodide (I-) = 126.90 g/mol.
Now, divide the mass of the halide by its respective molar mass to obtain the number of moles of the halide:
- Chloride (Cl-): Moles of chloride = Mass of chloride / Molar mass of chloride
= 0.0504 g / 35.45 g/mol
≈ 0.001423 mol
- Bromide (Br-): Moles of bromide = Mass of bromide / Molar mass of bromide
= 0.0504 g / 79.90 g/mol
≈ 0.000631 mol
- Iodide (I-): Moles of iodide = Mass of iodide / Molar mass of iodide
= 0.0504 g / 126.90 g/mol
≈ 0.000397 mol
Comparing the moles of barium with the moles of the halide, we find that the ratio is approximately 1:2 for chloride, 1:1 for bromide, and 1:3 for iodide. This ratio tells us the formula of the barium halide.
Thus, the possible formulas for the barium halide are BaCl2 (barium chloride), BaBr2 (barium bromide), and BaI3 (barium iodide).