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.

The answer is:

Empirical formula= C3H8O2

Molecular formula= C9H24O6

First, we need to calculate how many moles of each element are present in the chemical reaction by dividing the given information (grams) into the molecular mass of each element:

Carbon: 12,011 g/mol

Hydrogen: 1,008 g/mol

Oxygen: 15,999 g/mol

Calculating the moles of each element:

Moles of Carbon =

Moles of Hydrogen =

Moles of Oxygen =

Then, we need to determinate the mole ratio. We can do it by dividing each result into the smallest value of moles (2.62 moles (O))

To get entire numbers, we will use the 2 factor,

So,the empirical  formula will be:

C3H8O2

To know the molecular formula,

We need to divide the given molar mass of the compound by the empirical formula molecular mass to find the ratio:

So, the molecular formula will be:

Have a nice day!

Hello!

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?

Carbon (C) ≈ 12 a.m.u (g/mol)

Hydrogen (H) ≈ 1 a.m.u (g/mol)

Oxygen (O) ≈ 16 a.m.u (g/mol)

C: 47.37 g

H: 10.59 g

O: 42.04 g

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

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  

Answers:

The empirical formula: C3H8O2

The molecular formula: C9H24O6

your answer would be “dry”

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