Determine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxygen. In an experiment, the molar mass of the compound was determined to be 228.276 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. (8 points

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Determine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxygen.  

In an experiment, the molar mass of the compound was determined to be 228.276 g/mol. What is the molecular formula of the compound?

We have the following data:

Carbon (C) ≈ 12 a.m.u (g/mol)

Hydrogen (H) ≈ 1 a.m.u (g/mol)

Oxygen (O) ≈ 16 a.m.u (g/mol)

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100g solution)

C: 47.37 g

H: 10.59 g

O: 42.04 g

The values ​​(in g) will be converted into quantity of substance (number of mols), dividing by molecular weight (g / mol) each of the values, we will see:We realize that the values ​​found above are not integers, so we divide these values ​​by the smallest of them, so that the proportion does not change, let us see:

convert number of atomic radio into whole number

2 * (1.5 : 4 : 1)  

= 3 : 8 : 2  ← whole number of atomic radio

C = 3

H =  8

O = 2

Thus, the empirical formula found for the compound will be:We are going to find the Molar Mass (MM) of the Empirical Formula (EF), let's see:

if: C3H8O2

C = 3*(12 a.m.u) = 36 a.m.u  

H = 8*(1 a.m.u) = 8 a.m.u  

O = 2*(16 a.m.u) = 32 a.m.u  

Knowing that the Molar Mass of the Molecular Formula is 228.276 (in g/mol) and that the Molar Mass of the Empirical Formula is 76 (in g/mol), then we will find the number of terms (n) for the molecular formula of the compound, let us see:The Molecular Formula is the Empirical Formula times the number of terms (n), then, we have:

Answers:

The empirical formula: C3H8O2

The molecular formula: C9H24O6