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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.

#Divide given mass with their molar masses.

C=48.38÷12.01= 4.03mols

H= 6.74÷1.0079= 6.69mols

O= 53.5÷16.00= 3.34mols

#Divide the moles with the lowest mole.

C = 4.03÷3.34 = 1.21

H = 6.69÷3.34 = 2.00

O = 3.34÷3.34 = 1

∴Emperical Formula = CH₂O

#Molar mass of CH₂O = 12.01 + (1.0079 x 2) + 16.00 = 30.02 g/mol

#Divide both moler masses: 180.15g/mol÷30.02g/mol = 6

#Multiply the answer with the Emperical formula to get the Molecular formula: 6 x (CH₂O) = C₆H₁₂O₆

Empirical Formula: CH₂OMolecular Formula: C₆H₁₂O₆

Explanation:

Firstly, you have to divide the three masses of the elements with their respective molar mass to find their moles.  After you've found the moles use the lowest mole to divide the moles of each element again to find your empirical formula. Then you calculate the molar mass of the empirical formula and divide the molar mass with the given molar mass of  finally, multiply the answer with the empirical formula.