Question

One method for determine the purity of a sample of titanium ( IV) oxide, an important industrial chemical, is to combine the sample with bromine trifluoride to produce titanium ( IV) fluoride, liquid bromine , and oxygen has. Suppose 2.367g of an impure sample( impure meaning that the sample has titanium (IV) oxide as well as other “ stuff” in it ) evolves 0.143 g of oxygen gas. What is the mass percent of titanium ( IV) oxide in the impure sample?

Answer 1

The reaction between titanium ( IV) oxide and bromine trifluoride to give  liquid bromine , and oxygen is:

$3TiO_{2}+4BrF_{3} -->3TiF_{4}+2Br_{2}+3O_{2}$

Thus from each three moles of  titanium ( IV) oxide we will get three moles of oxygen molecule

The mass of oxygen obtained = 0.143g

The moles of oxygen obtained

      = $\frac{Mass}{Molar mass}=\frac{0.143}{32}= 0.00447mol$

So moles of  titanium ( IV) oxide required will be 0.00447mol

the molar mass of  titanium ( IV) oxide is 79.87g/mol

mass of titanium ( IV) oxide used will be = $molesXmolarmass=0.00447X79.87=0.357g$

This mass of  titanium ( IV) oxide is present in 2.367g of impure sample

the mass percent will be

$Mass=\frac{mass titanium ( IV) oxideX100}{mass impure sample}$

$Masspercent=\frac{0.357X100}{2.367}=15.08$

The percentage of titanium ( IV) oxide in impure sample is 15.08% (w/w)