14.2 g of R2O5 contains 8 g of oxygen. Determine the molecular weight of RH3.

Given:
m (R2O5) = 14.2 g
m (O in R2O5) = 8 g

To find:
Mr (RH3) -?

Solution:
1) m (R in R2O5) = m (R2O5) – m (O in R2O5) = 14.2 – 8 = 6.2 g;
2) n (O in R2O5) = m (O in R2O5) / M (O) = 8/16 = 0.5 mol;
3) n (R2O5) = n (O in R2O5) / N (O) = 0.5 / 5 = 0.1 mol;
4) n (R in R2O5) = n (R2O5) * N (R) = 0.1 * 2 = 0.2 mol;
5) M (R) = m (R in R2O5) / n (R in R2O5) = 6.2 / 0.2 = 31 g / mol;
6) Ar (R) = M (R) = 31;
7) Mr (RH3) = Ar (R) * N (R) + Ar (H) * N (H) = 31 * 1 + 1 * 3 = 34.

Answer: The molecular weight of RH3 is 34.