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

=
`(Mass)/(Molar mass)=(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=(mass titanium ( IV) oxideX100)/(mass impure sample)`

`Masspercent=(0.357X100)/(2.367)=15.08`

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