Determining Empirical And Molecular Formulas From Combustion Analysis

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Determining Empirical And Molecular Formulas From Combustion Analysis

(a) 5.86 mg of carbon dioxide (CO_{2}) and 1.37 mg of water (H_{2}O) 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 CO_{2} and 0.1159 g of H_{2}O 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**:
S_{7}P_{2}A_{22} (identity - chemical).
S_{7}P_{6}A_{65} (grouping/interation-chemical).
**The math**:

Pj Problem of Interest is of type *identity* (chemical).

**(a)** Let empirical formula of toluene be C_{x}H_{y}

Molar mass of CO_{2} = 12 + 32 = 44

1 mole CO_{2} 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 CO_{2}

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 H_{2}O = 2 + 16 = 18

1 mole of H_{2}O 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 H_{2}O

2 moles of hydrogen atom (H) in 1 mole of H_{2}O

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 = C_{7}H_{8}.
**(b)** Let empirical formula of menthol be C_{x}H_{y}O_{z}

Molar mass of menthol = 156 g/mol

Molar mass of CO_{2} = 12 + 32 = 44

So, moles of CO_{2} produced = 0.2829/44 = 0.0064 moles

So, moles of C = 0.0064 (reasoning as in (a))

Molar mass of H_{2}O = 2 + 16 = 18

So, moles of H_{2}O = 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 = C_{10}H_{20}O.

empirical mass = molar mass

So molecular weight = C_{10}H_{20}O.

Math

The *point* **.** is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.

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