Atomic Weight Calculation for Elements X and Y in Compounds XY2 and X3Y2

Given Information:
- Compounds formed: XY2 and X3Y2
- 0.1 mole of XY2 weighs 10g
- 0.05 mole of X3Y2 weighs 9g

Step 1: Determine the Molar Mass of XY2 and X3Y2
- Molar mass is the mass of one mole of a substance.
- We can calculate the molar mass of XY2 and X3Y2 using the information provided.

Calculation for XY2:
- Given: 0.1 mole of XY2 weighs 10g
- Molar mass of XY2 = Mass / Moles = 10g / 0.1 mol = 100g/mol

Calculation for X3Y2:
- Given: 0.05 mole of X3Y2 weighs 9g
- Molar mass of X3Y2 = Mass / Moles = 9g / 0.05 mol = 180g/mol

Step 2: Identify the Atomic Weight of X and Y
- Let the atomic weight of X be 'a' and the atomic weight of Y be 'b'.
- In XY2, we have one atom of X and two atoms of Y.
- In X3Y2, we have three atoms of X and two atoms of Y.

Calculation for XY2:
- Molar mass of XY2 = (Atomic weight of X) + 2 * (Atomic weight of Y) = a + 2b
- Substituting the molar mass value calculated earlier: 100g/mol = a + 2b

Calculation for X3Y2:
- Molar mass of X3Y2 = 3 * (Atomic weight of X) + 2 * (Atomic weight of Y) = 3a + 2b
- Substituting the molar mass value calculated earlier: 180g/mol = 3a + 2b

Step 3: Solve the Equations
- We have two equations with two variables (a and b).
- We can solve these equations simultaneously to find the values of a and b.

Equation 1: 100g/mol = a + 2b
Equation 2: 180g/mol = 3a + 2b

- Multiply Equation 1 by 3 and subtract from Equation 2:
(3a + 2b) - 3(a + 2b) = 180 - 300
3a + 2b - 3a - 6b = -120
-4b = -120
b = -120 / -4
b = 30

- Substituting the value of b in Equation 1:
100g/mol = a + 2(30)
100g/mol = a + 60
a = 100 - 60
a = 40

Step 4: Interpretation and Conclusion
- The atomic weight of X is 40 and the

Given 0.1 mole of XY2 weight 10g
x+2y=10/0.1=100 (equation 1)
0.05 mole of X3Y2 weight 9g
3x+2y=9/0.05=180(equation 2)
x=100-2y
substituting for x
3(100-2y)+2y=180
300-6y+2y=180
y=30
so X =100-2(30)=40