Determining Empirical And Molecular Formulas From Combustion Analysis
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Determining Empirical And Molecular Formulas From Combustion Analysis

Empirical And Molecular Formulas From Combustion Analysis

(a) 5.86 mg of carbon dioxide (CO2) and 1.37 mg of water (H2O) are realized in the combustion analysis of toluene.
Determine the empirical formula of toluene, if it contains only carbon (C) and hydrogen (H).
(b) 0.1005 g of menthol is combusted. 0.2829 g of CO2 and 0.1159 g of H2O are realized.
Determine the empirical formula of menthol and its molecular formula if it is composed of carbon (C), hydrogen (H), oxygen (O) and its molar mass is 156 g/mol.

The strings: S7P2A22 (identity - chemical). S7P6A65 (grouping/interation-chemical).

The math:
Pj Problem of Interest is of type identity (chemical).

Empirical And Molecular Formulas From Combustion Analysis

(a) Let empirical formula of toluene be CxHy
Molar mass of CO2 = 12 + 32 = 44
1 mole CO2 weighs 44 g
So, 5.86 x 10-3 g is the weight of (5.86 x 10-3)/44 moles = 0.133 x 10-3 moles
1 mole of Carbon atom (C) in 1 mole of carbon dioxide CO2
So, (0.133 x 10-3) moles of C.
So, (0.133 x 10-3 x 12) g of C
So, (0.133 x 10-3 x 12)/12 moles of C.
Molar mass of H2O = 2 + 16 = 18
1 mole of H2O weighs 18 g
So, 1.37 x 10-3 gm is the weight of (1.37 x 10-3)/18 moles = 0.076 x 10-3 moles of H2O
2 moles of hydrogen atom (H) in 1 mole of H2O
So, 2(0.076 x 10-3) moles of H
So, 2(0.076 x 10-3 x 1)g of H
So, 2(0.076 x 10-3 x 1)/1 moles of H = 0.152 x 10-3

So, ratio of moles of C:H in toluene = 0.133 x 10-3:0.152 x 10-3
= 0.133:0.152 = 1: 1.14 = 1:8/7 = 7:8.
So, empirical formula of toluene = C7H8.

(b) Let empirical formula of menthol be CxHyOz
Molar mass of menthol = 156 g/mol
Molar mass of CO2 = 12 + 32 = 44
So, moles of CO2 produced = 0.2829/44 = 0.0064 moles
So, moles of C = 0.0064 (reasoning as in (a))
Molar mass of H2O = 2 + 16 = 18
So, moles of H2O = 0.1159/18 = 0.0064
So, moles of H = 2 x 0.0064 = 0.0128 moles
mass of oxygen = mass of sample -(mass of C + mass of H) = 0.1005 - 0.0192 = 0.0813 g
So, moles of oxygen = 0.0813/16 = 0.0050
So, O:C:H ratios = 1:1.28:2.56 = 1: 5/4:5/2 = 1: 10:20.
So, empirical formula = C10H20O.
empirical mass = molar mass
So molecular weight = C10H20O.

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