The mass fraction of metal sulfide (I) in a mixture with sulfate of this metal

The mass fraction of metal sulfide (I) in a mixture with sulfate of this metal is 70%, and the mass fraction of sulfur atoms in this mixture of salts is 25.88%.

Given:
ω (Me2S) = 70%
ω (Me2SO4) = 30%
ω (S) = 25.88%

To find:
Me -?

Decision:
1) Find the mass of Me2S and Me2SO4:
Take the mass of the mixture as 100 g, then
m (Me2S) = ω (Me2S) * m mixture / 100% = 70% * 100/100% = 70 g;
m (Me2SO4) = m mixture – m (Me2S) = 100 – 70 = 30 g;

2) Find the amount of substance Me2S and Me2SO4:
Let the atomic mass Me be (x) g / mol, then
n (Me2S) = m (Me2S) / Mr (Me2S) = 70 / (2x + 32) mol;
n (Me2SO4) = m (Me2SO4) / Mr (Me2SO4) = 30 / (2x + 96) mol;

3) Find the amount of substance S in Me2S and Me2SO4:
n (S in Me2S) = n (Me2S) = 70 / (2x + 32) mol;
n (S in Me2SO4) = n (Me2SO4) = 30 / (2x + 96) mol;
n total (S) = n (S in Me2S) + n (S in Me2SO4) = (70 / (2x + 32)) + (30 / (2x + 96)) mol;

4) Find the mass S in Me2S and Me2SO4:
m (S) = n (S) * Mr (S) = (70 / (2x + 32)) + (30 / (2x + 96)) * 32 g;

5) Make and solve the equation according to the formula for finding the mass fraction of S in Me2S and Me2SO4:
ω (S) = m (S) * 100% / m mixture;
25.88% = ((70 / (2x + 32)) + (30 / (2x + 96)) * 32) * 100% / 100;
25.88 = (70 * 32 / (2x + 32)) + (30 * 32 / (2x + 96));
103.52x ^ 2 + 225.88x – 166256.64 = 0;

We solve the quadratic equation by the formulas:
ax ^ 2 + bx + c = 0;
D = b ^ 2 – 4ac;
x1,2 = (-b + – √D) / 2a;
x1 = (-225.88 + 8300.28) / (2 * 103.25) = 39;
x2 = (-225.88 – 8300.28) / (2 * 103.25) = -41;

Because the atomic mass of an element cannot be negative, then the answer is x = 39.

6) Find in the table D.I. Mendeleev’s element with an atomic mass equal to 39. This is potassium (K).

Answer: The unknown metal is potassium (K).