Consider a compound that is 31.17% c, 6.54% h, and 62.29% o by mass. assume that we have a 100 g sample of this compound. also consider that the molecular formula mass of this compound is 231.2 amu. what are the subscripts in the actual molecular formula for this compound?

 The molecular formula will be

Explanation:

If percentage are given and we are given total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 31.17 g

Mass of H = 6.54 g

Mass of O = 62.29 g

Step 1 : convert given masses into moles.

Moles of C =

Moles of H=

Moles of O =

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C=

For H =

For O =

The ratio of C: H: O= 1: 2.5: 1.5

Converting them into whole number ratios by multiplying by 2.

Hence the empirical formula is

The empirical weight of = 2(12)+5(1)+3(16)= 77 amu

The molecular weight = 231.2 amu

Now we have to calculate the molecular formula.

The molecular formula will be=

The molecular formula will be

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explanation:

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the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be