Zinc is determined by precipitating and weighing as Zn2Fe(CN)6. a. What weight of zinc is contained in a sample that gives 0.348 g precipitate? b. What weight of precipitate would be formed from 0.500 g of Zinc?
MM = molar mass
AM = atomic mass
a.
0.348g Zn2Fe(CN)6 x [2*AM Zn/MM Zn2Fe(CN)6]
b.
0.5 g Zn x (MM Zn2Fe(CN)6/2*AM Zn)
(a) 0.26 g (b) 1.30 g
To determine the weight of zinc in a sample that gives 0.348 g precipitate, we need to use the formula weight of zinc = weight of precipitate / molar mass of Zn2Fe(CN)6.
a. To calculate the weight of zinc, we need the molar mass of Zn2Fe(CN)6. The molar mass of Zn2Fe(CN)6 is calculated by adding the atomic masses of zinc, iron, carbon, and nitrogen.
The atomic mass of zinc (Zn) is 65.38 g/mol.
The atomic mass of iron (Fe) is 55.85 g/mol.
The atomic mass of carbon (C) is 12.01 g/mol.
The atomic mass of nitrogen (N) is 14.01 g/mol.
The molar mass of Zn2Fe(CN)6 is:
(2 * atomic mass of zinc) + atomic mass of iron + (6 * (atomic mass of carbon + atomic mass of nitrogen))
= (2 * 65.38 g/mol) + 55.85 g/mol + (6 * (12.01 g/mol + 14.01 g/mol))
= 130.76 g/mol + 55.85 g/mol + (6 * 26.02 g/mol)
= 130.76 g/mol + 55.85 g/mol + 156.12 g/mol
= 342.73 g/mol
Now, we can calculate the weight of zinc:
weight of zinc = weight of precipitate / molar mass of Zn2Fe(CN)6
= 0.348 g / 342.73 g/mol
= 0.001 g/mol
Therefore, the weight of zinc contained in the sample that gives 0.348 g precipitate is 0.001 g.
b. To calculate the weight of precipitate formed from 0.500 g of zinc, we can use the same formula.
The molar mass of Zn2Fe(CN)6 is 342.73 g/mol (as calculated above).
weight of precipitate = weight of zinc * molar mass of Zn2Fe(CN)6
= 0.500 g * 342.73 g/mol
= 171.36 g/mol
Therefore, the weight of precipitate formed from 0.500 g of zinc is 171.36 g.
To determine the weight of zinc in a sample using the precipitate Zn2Fe(CN)6, we need to set up and solve a stoichiometry problem. Here's the step-by-step process to find the answers:
a. What weight of zinc is contained in a sample that gives 0.348 g precipitate?
1. Write down the balanced chemical equation for the precipitation reaction:
Zn(aq) + 2Fe(CN)6(aq) -> Zn2Fe(CN)6(s)
2. Determine the molar mass of Zn2Fe(CN)6:
Zn2Fe(CN)6: (2 × Atomic mass of Zn) + (1 × Atomic mass of Fe) + (6 × Atomic mass of C) + (6 × Atomic mass of N)
= 2(65.38 g/mol) + 1(55.85 g/mol) + 6(12.01 g/mol) + 6(14.01 g/mol)
= 271.84 g/mol
3. Determine the molar mass of Zn:
Atomic mass of Zn = 65.38 g/mol
4. Set up a proportion to relate the weight of the precipitate to the weight of Zn:
(Weight of Zn) / (Molar mass of Zn) = (Weight of precipitate) / (Molar mass of Zn2Fe(CN)6)
5. Substitute the given values and solve for the weight of Zn:
(Weight of Zn) / (65.38 g/mol) = 0.348 g / 271.84 g/mol
Weight of Zn = (0.348 g / 271.84 g/mol) × (65.38 g/mol)
Weight of Zn ≈ 0.0836 g
Therefore, the weight of zinc contained in the sample is approximately 0.0836 g.
b. What weight of precipitate would be formed from 0.500 g of Zinc?
1. Follow the same steps as above until step 4.
2. Set up a proportion to relate the weight of Zn to the weight of the precipitate:
(Weight of Zn) / (Molar mass of Zn) = (Weight of precipitate) / (Molar mass of Zn2Fe(CN)6)
3. Substitute the given values and solve for the weight of the precipitate:
(0.500 g) / (65.38 g/mol) = (Weight of precipitate) / (271.84 g/mol)
Weight of precipitate = (0.500 g / 65.38 g/mol) × (271.84 g/mol)
Weight of precipitate ≈ 2.085 g
Therefore, the weight of precipitate that would be formed from 0.500 g of zinc is approximately 2.085 g.