Question

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.

Will give brainlist but have to explain a lot

Answer 1

Okay so the trick is to find the number of moles of each compound you have:

Carbon: 47.37 g / (12.01 g/mol)  = 3.944

Hydrogen: 10.59/1.008 = 10.55

Oxygen: 42.04/16  = 2.6275

Now divide each my the smallest mole number which is oxygen.

3.944/2.6275 = 1.5

10.55/2.6275 = 4

2.6275/2.6275 = 1

So the integer ratio between C:H:O is 3:8:2.

That means the empirical fomrula of the compound is C3H8O2.

However, for the molecular formula, we have to look at the Molar Mass value given. since C3H8O2 has a molar mass of 12.01*3 + 1.008*8 + 16*2 = 76.094,

we can do 228.276/76.084 = 3. So that means the molecular formula of the compound is C9H24O6

Consider giving brainliest

Answer 2

Answer:

any more info?

Explanation: