Question 3.8: CALCULATING A PERCENT YIELD Methyl tert-butyl ether (MTBE, C...
CALCULATING A PERCENT YIELD
Methyl tert-butyl ether (MTBE, C_{5}H_{12}O), a gasoline additive now being phased out in many places because of health concerns, can be made by reaction of isobutylene (C_{4}H_{8}) with methanol (CH_{4}O). What is the percent yield of the reaction if 32.8 g of methyl tertbutyl ether is obtained from reaction of 26.3 g of isobutylene with sufficient methanol?
\underset{Isobutylene}{C_{4}H_{8}(g)} + CH_{4}O(l) → \underset{Methyl tert-butyl ether (MTBE)}{C_{5}H_{12}O(l)}STRATEGY
We need to calculate the amount of methyl tert-butyl ether that could be produced theoretically from 26.3 g of isobutylene and compare that theoretical amount to the actual amount (32.8 g). As always, stoichiometry problems begin by calculating the molar masses of reactants and products. Coefficients of the balanced equation then tell mole ratios, and molar masses act as conversion factors between moles and masses.
