Element, the higher oxide of which E2O5 creates a covalent compound with Hydrogen, the mass fraction

Element, the higher oxide of which E2O5 creates a covalent compound with Hydrogen, the mass fraction of Hydrogen in which is 3.85%. Determine the relative molecular weight of the higher oxide of element E.

Given:
E2O5 – higher oxide
H3E – compound with hydrogen
ω (H in H3E) = 3.85%

To find:
Mr (E2O5) -?

Decision:
1) ω (E in H3E) = 100% – ω (H in H3E) = 100% – 3.85% = 96.15%;
2) Mr (H3E) = N (H in H3E) * Ar (H) * 100% / ω (H in H3E) = 3 * 1 * 100% / 3.85 = 77.922;
3) Ar (E) = (ω (E in H3E) * Mr (H3E)) / (N (E in H3E) * 100%) = (96.15% * 77.922) / (1 * 100%) = 74 , 92 ≈ 75 – arsenic (As);
4) Mr (As2O5) = Ar (E) * N (E) + Ar (O) * N (O) = 75 * 2 + 16 * 5 = 230.

Answer: The relative molecular weight of As2O5 is 230.