Question

1. What is the empirical formula of a compound that is 58.80% barium, 13.75% sulfur and 27.45% oxygen by mass?

2. By chemical analysis, a molecular compound was found to consist of 80% carbon and 20% hydrogen by mass. By measuring the volume of a known mass of the compound in the gaseous phase, its molecular mass was found to be 30. Find the empirical and the molecular formulas of the compound.

Answer 1

Answer on Question #98203 – Chemistry – Inorganic Chemistry

Task:

1. What is the empirical formula of a compound that is 58.80% barium, 13.75% sulfur and 27.45% oxygen by mass?

2. By chemical analysis, a molecular compound was found to consist of 80% carbon and 20% hydrogen by mass. By measuring the volume of a known mass of the compound in the gaseous phase, its molecular mass was found to be 30. Find the empirical and the molecular formulas of the compound.

Solution (1):

Calculate per 100 grams of a compound:

m(Ba) = w(Ba) = 58.80 g;

m(S) = w(S) = 13.75 g;

m(O) = w(O) = 27.45 g.

Then,

58.8 g Ba * (1 mol Ba / 137.327) = 0.4282 mol;

13.75 g S * (1 mol S / 32.065) = 0.4288 mol;

27.45 g O * (1 mol O / 15.999) = 1.7157 mol.

n(Ba) : n(S) : (O) = 0.4282 : 0.4288 : 1.7157 = 1 : 1 : 4;

n(Ba) : n(S) : (O) = 1 : 1 : 4.

The empirical formula of a compound: BaSO₄ - barium sulfate.

Answer (1): BaSO4.

Solution (2):

Calculate per 100 grams of a compound:

m(C) = w(C) = 80 g;

m(H) = w(H) = 20 g.

Then,

80 g C * (1 mol C / 12.0096) = 6.6613 mol;

20 g H * (1 mol H / 1.0078) = 19.8452 mol.

n( C) : n(H) = 6.6613 : 19.8452 = 1 : 3.

The emperical formula of the compound: (CH₃)_n.

Molecular mass of the compound = Mr(C_xH_y) = (CH₃)_n = 30 g/mol;

(CH₃)*n = 30 g/mol;

(12.0096 + 3*1.0078)*n = 30;

15.033 *n = 30;

n = 2.

(CH₃)₂ = C₂H₆.

The molecular formula of the compound: C₂H₆.

Answer (2):

The emperical formula of the compound: (CH₃)_n.

The molecular formula of the compound: C₂H₆.