4· One of the additives in unleaded gasoline that replaced tetraethyl lead in le

Question

4· One of the additives in unleaded gasoline that replaced tetraethyl lead in leaded gasoline is called MTBE burned completely, 37.640 g CO2 and 18.489 g H20 form. In a sepa & O atoms) is found to be 88.150 g. What is the molecular formula for MTBE? . When 15.078 g MTBE is Co 2-H 20 rate experiment the molecular mass of MTBE (composed of C, H c x Hyo z· 2 mo Co2. 31.lHogco2 l4.ol gco2 31.DAd 1s.4eng 815012.3442484 150 520.17 H2o imolco 2 4.73,185948 mo 955200 12.015 C imolc 2-o5205329 2.4x5=12, H 4,312 3948 96520017 5.5 " 2.07257% H

Explanation / Answer

Answer

Molecular formula of MTBE is C5H12O

Explanation

In complete burning carbon completely converted into CO2 and Hydrogen completely converted into H2O

i)mass of CO2 obtained = 37.640g

conversion factor of CO2 to C = 12.01g/44.01g = 0.27289

mass of C in MTBE = 0.27289 × 37.640g = 10.2715g

mass fraction of C in MTBE = 10.2715g/15.078g = 0.6812

ii) mass of H2O obtained = 18.489g

conversion factor of H2O to H = 2.02g/18.02 = 0.11210

mass of H in MTBE = 0.11210× 18.489g = 2.0726g

mass fraction of H in MTBE = 2.0726g/ 15.078g = 0.1375

iii) mass of Oxygen = mass of MTBE - (mass of C+ mass of H)

= 15.078g - (10.2715g + 2.0726g)

= 2.7339g

mass fraction of O in MTBE = 2.7339g/15.078g = 0.1813

iv) Molar mass of MTBE = 88.150g

mass of C in one mole MTBE = mass fraction of C × Molar mass of MTBE

= 0.6812×88.150g

= 60.0478g

No of moles of C in one mole of MTBE = 60.0478g/12.01g/mol = 5mol

v) mass of H in one mole of MTBE = 0.1375 × 88.150g = 12.1206g

No of moles of H in one mole of MTBE = 12.1206g/1.01g/mol = 12mol

vi) mass of H in one mole of MTBE = 0.1813 × 88.150g = 15.9815g

No of moles of O in one mole of MTBE = 15.9815g/16.00g/mol = 1mol

Therefore

molecular formula of MTBE is C5H12O

The error in your calculation are 1. the calculation of mass of oxygen in MTBE sample and 2. you confused with empirical formula.

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