A2.60 g sample of titanium metal chemically combines
withchlorine gas to form 10.31g of a titanium chloride.
a. what is the empirical formula of the titaniumchloride
b. what is the percent by mass of titanium and chlorine in
thesample?

For a: The empirical formula for the given compound is

For b: The percent by mass of titanium and chlorine in the sample is 25.55 % and 74.78 % respectively.

Explanation:

We are given:

Mass of Titanium = 2.60 g

Mass of sample = 10.31 g

Mass of Chlorine = 10.31 - 2.60 = 7.71 g

To formulate the empirical formula, we need to follow some steps:

Moles of titanium =

Moles of Chlorine =

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.054 moles.

For Titanium =

For Chlorine  =

Step 3: Taking the mole ratio as their subscripts.

The ratio of Ti : Cl = 1 : 4

Hence, the empirical formula for the given compound is

To calculate the percentage by mass of substance in sample, we use the equation:

      .......(1)

Mass of sample = 10.31 g

Mass of titanium = 2.60 g

Putting values in above equation, we get:

Mass of sample = 10.31 g

Mass of chlorine = 7.71 g

Putting values in above equation, we get:

Hence, the percent by mass of titanium and chlorine in the sample is 25.55 % and 74.78 % respectively.

your answer would be “dry”

gas

explanation:

smoke is the matter of a gas