Question

There are 0.0200 moles of B in 100 grams of the compound: AwBxCyDz (from Day 1 Pre-lab). AwBxCyDz (s) <---> wA(aq) + xB(aq) + yC(aq) + zD(aq)

Part A. You dissolve 0.250 g of your compound (AwBxCyDz) in exactly 10.00 mL of water producing 10.00 mL of solution. Using colorimetry, you find that the concentration of A is 2.00 x 10-2 M. The molar mass of A is 750.0 g/mol.

Calculate the following values: 1. Moles of A in the solution 2. Mass of A in the solution 3. Mass % of A in the compound 4. Moles of A in 100 g of the compound Part B. The compound, AwBxCyDz has an oxidation number of 0 (zero). The oxidation number of A is +1, the oxidation number of B is +2, the oxidation number of C is –3, and the oxidation number of D is 0. The molar mass of C is 500.0 g/mol. Previously determined: Moles of A in 100 grams of the compound: 0.0800 mol A Moles of B in 100 grams of the compound: 0.0200 mol B Calculate the following values: 5. Moles of C in 100 grams of the compound (use oxidation numbers) 6. Mass of C in 100 grams of the compound 7. Mass % of C in the compound Part C. The molar mass of D is 166.7 g/mol. Calculate the following values: 8. Mass % D in the compound 9. Mass of D in 100 grams of the compound 10. Moles of D in 100 grams of the compound 11. Formula of the compound

Answer 1

Solved the first 7 parts, the molar mass of B is missing which is leading to multiple unknowns for calculating the part D, please check the problem and revert in comments with the molar mass of B

1) Molarity of A (solution) = 2.00 * 10^(-2) M

Number of moles of A = Volume of solution (in L) * Molarity of A(solution) = 10/1000 * 2.00 * 10^(-2) = 2.00 * 10^(-4) moles

2) Mass of A in solution = Number of moles * Molar mass

=> 2.00 * 10^(-4) mol * 750.0 gm/mol

=> 0.150 grams

3) Mass %A in the compound = Mass of A/Total Mass * 100 = 0.15/0.25 * 100 = 60%

4) Mass of A in 100g of compound = 60% of 100g = 60g

Moles of A in 100g of compound = Mass of A/Molar mass = 60/750 = 0.08 moles

5) + charge provided by A = 0.08 * 1 (charge on A) = 0.08+ charge

+charge provided by B = 0.02 * 2 (charge on B) = 0.04+ charge

Hence all this + charge must be catered by C, which will have the number of moles equal to 0.12/3 (total charge on A+B/divided the valence state) = 0.04 moles

6) Mass of C = number of moles * molar mass

=> 0.04 mol * 500 gm/mol

=> 20 gms

7) Mass %C in compound = Mass of C/Total mass * 100 = 20%

8) Mass %D of compound = 100% - Mass % of A - Mass % of B - Mass % of C

(The molar mass of B is missing in the problem to compute this data)