percent composition?: 85.05 g of Mg combines completely with 112.21 g of S. empirical formula?: 78.8% C, 20.2% H. 67.6% Hg, 10.8% S, 21.6% O

To find the percent composition of a compound, you’ll need the mass of each element in a given sample and the total molar mass of the compound. You then divide the mass of the element by the total mass, and multiply by 100 to get the percentage.

1. For the Mg and S compound, first determine the moles of each element: - Moles of Mg = mass of Mg / atomic mass of Mg - Moles of S = mass of S / atomic mass of S (Atomic mass of Mg is approximately 24.305 g/mol and that of S is approximately 32.06 g/mol.)

Moles of Mg = 85.05 g / 24.305 g/mol = 3.50 moles Moles of S = 112.21 g / 32.06 g/mol = 3.50 moles

Since the molar ratio is 1:1, the empirical formula is MgS.

2. For the empirical formula from percentages, the assumption is that there is a 100 g sample so that the percentages can be directly converted to grams.

For the compound with C, H: - C: 78.8 g of C is present. (Atomic mass is about 12.01 g/mol.) - H: 20.2 g of H is present. (Atomic mass is about 1.008 g/mol.)

Moles of C = 78.8 g / 12.01 g/mol ≈ 6.56 moles Moles of H = 20.2 g / 1.008 g/mol ≈ 20.04 moles

We now find the simplest whole number ratio between C and H: Dividing by the smallest number of moles (6.56 in this case): - C: 6.56 / 6.56 = 1 - H: 20.04 / 6.56 ≈ 3

The empirical formula is CH_3.

3. For the compound with Hg, S, O: - Hg: 67.6 g (Atomic mass is about 200.59 g/mol.) - S: 10.8 g (Atomic mass is about 32.06 g/mol.) - O: 21.6 g (Atomic mass is about 16.00 g/mol.)

Determine the moles of each: Moles of Hg = 67.6 g / 200.59 g/mol ≈ 0.337 moles Moles of S = 10.8 g / 32.06 g/mol ≈ 0.337 moles Moles of O = 21.6 g / 16.00 g/mol = 1.35 moles

Find the ratio by dividing by the smallest number of moles (0.337 in this case): - Hg: 0.337 / 0.337 = 1 - S: 0.337 / 0.337 = 1 - O: 1.35 / 0.337 ≈ 4

The empirical formula is HgSO_4.