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Darina [25.2K]
3 years ago
13

One method for determine the purity of a sample of titanium ( IV) oxide, an important industrial chemical, is to combine the sam

ple 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?
Chemistry
1 answer:
Tems11 [23]3 years ago
8 0

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)


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