Question

HELP

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

my teacher hasnt graded yet, but i got the same answer as the guy above. i did my best to explain the process for y'all if you genuinely don't understand/know how to do it. THE PROCESS IS VERY LONG, but youll get it eventually if i didnt help lol GOODLUCK

and yes, the molecular formula is C6H12O6

Answer 2

I recently answered this question. The response I submitted is included below. I beleive my answer should be correct.

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