brady problem 2.83
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BRADY PROBLEM 2.83. Tutorial on the procedure of solving for empirical and molecular formulas from a combustion reaction. H 2 O 0.1874 g. CO 2 0.6871 g. O 2. (C + H + O) 0.5000 g. +. +. Problem. - PowerPoint PPT PresentationTRANSCRIPT
BRADY PROBLEM 2.83
Tutorial on the procedure of solving for empirical and molecular formulas
from a combustion reaction.
ProblemCitric Acid, the substance that makes lemon juice sour, is composed of only carbon , hydrogen, and oxygen. When a 0.5000 g sample of citric acid was burned, it produced 0.6871 g CO2 and 0.1874 g H2O.
The molecular mass of the compound is 192. What are the empirical and molecular formulas for citric acid?
(C + H + O)0.5000 g
CO2
0.6871 g+ H2O
0.1874 g+ O2
(C + H + O)0.5000 g
CO2
0.6871 g+ H2O
0.1874 g+ O2
All the hydrogen has gone to the product of H2O
All the carbon has gone to the product of CO2
From these gram quantities, convert to moles of compound, then to moles of elements…(you may wish to subtotal as you
will use the number of moles later)…then to grams of elements.
Starting with CO2….
Convert grams of CO2 to moles of CO2
to moles of C then to grams of carbon.
0.6871 g CO2 1 mol CO2 1 mol C = 0.01562 mol C
44.0 g 1 mol CO2
= 0.01562 mol C 12.011 g = 0.1876 grams C
1 mol C
Convert grams of H2O to moles of H2O
to moles of H then to grams of hydrogen.
0.1874 g H2O 1 mol H2O 2 mol H = 0.02082 mol H
18.0 g 1 mol H2O
= 0.02082 mol H 1.008 g = 0.02099 grams H
1 mol H
Find the number of grams of oxygenin the original sample
0.5000 g original sample size
<0.02099 g> subtract grams hydrogen
<0.1876 g> subtract grams carbon
0.2914 g remaining grams are the grams of oxygen in the original sample
0.2914 g O 1 mol O = 0.01821 mol O
15.994 g
Convert these grams to moles of oxygen
Find the lowest mole ratio for the empirical formula
0.1562 mol C/ 0.1562 = 1 x 6 = 6
0.02082 mol H/ 0.1562 = 1.333 x 6 = 8
0.01821 mol O/ 0.1562 = 1.166 x 6 = 7
C6H8O7
Find the Molecular Formulafrom the given Molecular Weight
Given molecular weight = 192 g/molEmpirical Formula determined: C6H7O7
Calculate the mass of the empirical formula:
C = 12.011 X 6 = 72.061H = 1.008 x 8 = 8.064O = 15.9994 x 7 = 111.99
192.12 g/mol
Divide the mass of the empirical into the molecular weight:
192/192.12 = 0.99938 = 1
therefore the empirical formula is the same as the molecular formula