Question

Determine the empirical formula of a compound containing 48. 38 grams of carbon, 6. 74 grams of hydrogen, and 53. 5 grams of oxygen.

In an experiment, the molar mass of the compound was determined to be 180. 15 g/mol. What is the molecular formula of the compound?

For both questions, show your work or explain how you determined the formulas by giving specific values used in calculations

Answer 1

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Step-by-step explanation:

Question 1:

C: 48.38g(1mol/12g) = 4.0317

H: 8.12g(1mol/1.01g) = 8.12

O: 53.5g(1mol/16g) = 3.34375

Divide by the smallest amount (3.34375)

C = 4.0317/3.34375 = 1.206 = 1

H = 8.12/3.34375 = 2.42 = 2

O = 3.34375/3.34375 = 1

Empirical formula = CH2O

Question 2:

Molecular formula = n(empirical formula)

n = molar mass (compound)/molar mass (empirical)

Empirical formula: CH2O

Molar mass of CH2O = 12 + 2x1 + 16 = 30 g/mol

Molar mass of compound: 180.15 g/mol

`n=(180.15g/mol)/(30g/mol)= 6`

Molecular formula = C6H12O6