Question

Remember a 100.0 gram sample was burned in the analysis. Post early to get your choice of compound as only one student may do each compound. Indicate the compound in the Subject Heading of your post.

CxHyOz Molar Mass g/mol 176.12

Grams CO2 150

Grams H2O. 40.88

Grams NO2 0.0

Determine its empirical formula from the combustion analysis given, showing all steps and explaining your work for your fellow classmates. Determine the molecular formula from approximate molar mass provided in the table.

balanced chemical reaction for the combustion of your compound.

Identify your compound, which should match one of the substances used in Discussion #1: Forensic Chemistry.

Answer 1

% of carbon = (12/44)*(mass of CO2)*100 / mass of sample = 12 * 150 * 100 / 44 * 100 = 40.91%

% of Hydrogen = (2/18)*(mass of H2O)*100 / mass of sample = 2*40.88*100/18*100 = 4.542 %

% of O = 100 - (40.91 + 4.542) = 54.55%

Moles of C = 40.91/12 = 3.41

Moles of H = 4.542/1 = 4.542

Moles of O = 54.55/16 = 3.41

Mole ratio;

C : H : O = 3.41 : 4.542 : 3.41

We will get Simplest ratio by dividing lowest mole i.e. 3.41

C : H : O = (3.41/3.41) : (4.542/3.41) : (3.41/3.41)

C : H : O = 1 : 1.332 : 1

Now, multiplying by number '3';

C : H : O = 3 : 4 : 3

Empirical formula = C3H4O3

Empirical formula mass = 3*12 + 4*1 + 3*16 = 88 g/mol

n = molar mass/empirical formula mass = 176.12/88 = 2

Hence,

Molecular formula = (empirical formula)n = (C3H4O3)2 = C6H8O6 .....Answer

Combustion reaction;

C6H8O6 + 5O2 -----> 6CO2 + 4H2O

It could be ascorbic acid i.e. vitamin C