Question

8. One compound of mercury with a formulas weight of 519 contains 77.26% Hg, 9.25% C, 1.17% H, and the remainder is oxygen. What are the empirical and molecular formulas of this compound?

Answer 1

answer and explanation

Hg = 77.26 %

C = 9.25 %

H = 1.17 %

O = 100 – 87.68 = 12.32 %

To answer this question, we assume a compound sample mass of 100 grams. We use the given mass percent values to find the mass of each element in the sample:

mC=0.0925×100 g=9.25 g

mH=0.0117×100 g=1.17 g

mHg=0.7726×100 g=77.26 g

mO=(1−0.0925−0.0117−0.7726)×100 g=12.32 g

Using the literature molar masses of the elements, the molar quantities of these elements are:

C= 9.25 g x 12.01 g/mol =0.770 mol

H= 1.17 g x 1.008 g/mol =1.16 mol

Hg= 77.26 g x 200.59 g/mol =0.385 mol

O= 12.32 g x 16.0 g/mol =0.770 mol

The molar ratio of the elements is:

Hg 0.385

C 0.770

H 1.16

O 0.770

If we divide each subscript by the smallest value, then we get the empirical formula of:

HgC₂H₃O₂

and the empirical formula mass is 259 g/mol

therefor the molecular formula can be determined to be:

519/259 = 2.00

the molecular formula is

Hg₂C₄H₆O₄