A compound was analyzed and found to contain 76.57% carbon, 6.43% hydrogen, and 17.00% oxygen by mass. Calculate the empirical formula of the compound. If the molar mass of the compound is 94.11 g/mol, what is the molecular formula of the compound? A second compound is composed of 53.30% Carbon 11.19% Hydrogen and 35.51% Oxygen by mass. Please Calculate the empirical formula of the compound of the molar mass of the compound is 90.12g/mol, what is the molecular formula for that compound?

See explanation

Explanation:

First we divide the percentage by mass of each element by it's relative atomic mass then we divide the quotients obtained by the lowest ratio obtained in the first step.

C- 76.57/12, H= 6.43/1, O = 17.00/16

C- 6.38/1.06, H= 6.43/1.06, O= 1.06/1.06

C- 6, H- 6, O- 1

Empirical formula: C6H6O

[(12 ×6) + (6 × 1) + (16 × 1)]n=94.11

[72 + 6 +16]n = 94.11

n = 94.11/94

n= 1

Molecular formula = C6H6O

2)

C- 53.30/12, H- 11.19/1, O- 35.51/16

C- 4.44/2.22, H- 11.19/2.22, O- 2.22/2.22

C- 2, H- 5, O- 1

Empirical formula: C2H5O

[(2×12) + (5× 1) + (1×16)]n = 90.12

[24 + 5 + 16] n = 90.12

n= 90.12/45

n= 2

Molecular formula = C4H10O2

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