What is the molecular formula of this compound?

Question

Elemental ananalysis of a compound results in the following data

40.00 % C

6.71 % H

53.28 % O


1.80 g of the compound fills a 189.0 ml bottle with a pressure of 2931.6 mm Hg at a temperature of 23.0oC.

Answer 1

(1) Find the empirical formula. Assuming a 100 gram sample,

40.00 g C x (1 mol C/12.011 g C) = 3.33 mol C
6.71 g H x (1 mol H/1.0079 g H) = 6.66 mol H
53.28 g O x (1 mol O/16.00 g O) = 3.33 mol O

So the ratio of C:H:O is 1:2:1, and the Empirical Formula is C1H2O1.

(2) Now the molecular formula of something is some multiple of the empirical formula, or (C1H2O1) x z. We have to figure out what this z is.

If it's a gas, and the data given are:
P = 2931.6 mmHg x (1 atm/760 mmHg) = 3.86 atm
V = 189.0 mL = 0.1890 L
T = 23.0C = 296 K
R = 0.0821 atm-L/mol-K
so using the Ideal Gas Law, PV=nRT, n = 0.0300 mol

The molar mass of the compound is mass/moles = m/n = 1.80 g/0.0300 mol = 60.0 g/mol.

The Empirical Formula is C1H2O1; one empirical formula unit has a mass of (1 x 12.011) + (2 x 1.0079) + (1 x 16.00) = 30.03; if the molecule weighs 60.0 g/mol, then the z is 2 (in other words, it requires two empirical formula units to make up the molecule) nad the molecular formula is C2H4O2.

Answer 2

C2H4O2

Answer 3

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Answer 4

c2h4o2