An unknown compound contains 75.69% carbon, 8.80% hydrogen, and 15.51% oxygen by mass. a mass spectrometry analysis reveals that the molecular weight is 412 g/mol. what is the molecular formula of this compound?

The molecular formula of the compound is, C28H36O4.

We have to divide the mass percent of each atom by the atomic number;

C - 75.69%/12         H - 8.80%/1       O - 15.51%/16

C - 6.3                     H - 8.80             O - 0.97

Next we have to divide each ratio by the lowest ratio;

C -                    H -            O -

C - 7                                H - 9                             O - 1

The empirical formula is C7H9O and the molecular mass is 412 g/mol.

Hence;

[7(12) + 9(1) + 16]n = 412

[84 + 9 + 16]n = 412

109n = 412

n = 412/109

n = 4

The molecular formula is, C28H36O4        

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Molecular formula is C₂₆H₃₆O₄

Explanation:

The compound is 75.69 % C, 8.80 % H and 15.51 % O. This data means, that in 100 g of compound we have 75.69 g, 15.51 g and 8.80 g of, C, O and H, respectively. We know the molar mass of the compound, so we can work to solve the moles of each element.

In 100 g of compound we have 75.69 g C, 15.51 g O and 8.80 g H

In 412 g of compound we would have:

(412 . 75.69) / 100 = 311.8 of C

(412 . 15.51) / 100 = 63.9 g of O

(412 . 8.80) / 100 = 36.2 g of H

Now, we can determine the moles of each, that are contained in 1 mol of compound.

312 g / 12 g/mol 26 C

64 g / 16 g/mol = 4 O

36 g / 1 g/mol = 36 H

Molecular formula is C₂₆H₃₆O₄

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3