Respuesta :
Answer:
For a: The empirical formula for the given compound is [tex]TiCl_4[/tex]
For b: The percent by mass of titanium and chlorine in the sample is 25.55 % and 74.78 % respectively.
Explanation:
- For a:
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:
- Step 1: Converting the given masses into moles.
Moles of titanium =[tex]\frac{\text{Given mass of Titanium}}{\text{Molar mass of Titanium}}=\frac{2.60g}{47.867g/mole}=0.054moles[/tex]
Moles of Chlorine = [tex]\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{7.71g}{35.5g/mole}=0.217moles[/tex]
- Step 2: Calculating the mole ratio of the given elements.
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 = [tex]\frac{0.054}{0.054}=1[/tex]
For Chlorine = [tex]\frac{0.217}{0.054}=4.01\approx 4[/tex]
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 [tex]TiCl_4[/tex]
- For b:
To calculate the percentage by mass of substance in sample, we use the equation:
[tex]\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of sample}}\times 100[/tex] .......(1)
- For Titanium:
Mass of sample = 10.31 g
Mass of titanium = 2.60 g
Putting values in above equation, we get:
[tex]\%\text{ composition of titanium}=\frac{2.60g}{10.31g}\times 100=25.22\%[/tex]
- For Chlorine:
Mass of sample = 10.31 g
Mass of chlorine = 7.71 g
Putting values in above equation, we get:
[tex]\%\text{ composition of chlorine}=\frac{7.71g}{10.31g}\times 100=74.78\%[/tex]
Hence, the percent by mass of titanium and chlorine in the sample is 25.55 % and 74.78 % respectively.
