The chemist investigated two mercury oxides. The mass of the first oxide was 6.74 grams, and from it the chemist

The chemist investigated two mercury oxides. The mass of the first oxide was 6.74 grams, and from it the chemist obtained 6.24 grams of mercury. The mass of the second oxide was 4.32 grams, from which 0.32 grams of oxygen was obtained. Did the chemist investigate oxides of the same composition? Confirm the answer by calculations and reaction equations.

Given:
m (HgxOy) = 6.74 g
m (Hg from HgxOy) = 6.24 g
m (HgaOb) = 4.32 g
m (O2 from HgaOb) = 0.32 g

To find:
HgxOy -?
HgaOb -?

Solution:
1) m (Hg from HgxOy) = m (Hg in HgxOy) = 6.24 g;
2) n (Hg in HgxOy) = m (Hg in HgxOy) / M (Hg) = 6.24 / 201 = 0.031 mol;
3) m (O in HgxOy) = m (HgxOy) – m (Hg in HgxOy) = 6.74 – 6.24 = 0.5 g;
4) n (O in HgxOy) = m (O in HgxOy) / M (O) = 0.5 / 16 = 0.031 mol;
5) x: y = n (Hg in HgxOy): n (O in HgxOy) = 0.031: 0.031 = 1: 1;
HgO;
6) m (Hg in HgaOb) = m (HgaOb) – m (O2 from HgaOb) = 4.32 – 0.32 = 4 g;
7) n (Hg in HgaOb) = m (Hg in HgaOb) / M (Hg) = 4/201 = 0.02 mol;
8) n (O2 from HgaOb) = m (O2 from HgaOb) / M (O2) = 0.32 / 32 = 0.01 mol;
9) n (O in HgaOb) = n (O2 from HgaOb) * 2 = 0.01 * 2 = 0.02 mol;
10) a: b = n (Hg in HgaOb): n (O in HgaOb) = 0.02: 0.02 = 1: 1;
HgO;
11) Oxides are the same in composition;
12) 2HgO => 2Hg + O2.

Answer: The oxides are the same in composition.