Answer:
AlCl₃
Explanation:
Data Given:
Mass of aluminum metal = 1.271 g
Mass of aluminum chloride = 6.280 g
Empirical formula of aluminum chloride = ?
Solution:
First find mass of Chlorine
As 6.280 g of aluminum chloride produced by 1.271 g so
the mass of chlorine in 6.280 g will be 6.280 g -1.271 g)
Mass of chlorine = 5.009 g
Now
Find the number of moles of Al and chlorine in aluminum chloride.
Molar Mass of Al = 26.98 g/mole
Molar mass of Cl = 35.5 g/mol
Mole of Al
Formula Used
no. of moles = mass in grams / molar mass . . . . . . . . (1)
Put values in equation 1
no. of moles of Al = 1.271 g / 26.98 g/ mole
no. of moles of Al = 0.0471
Mole of Chlorine
Formula used
no. of moles = mass in grams / molar mass . . . . . . . . (1)
Put values in equation 1
no. of moles of Cl = 5.009 g / 35.5 g/ mole
no. of moles of Cl = 0.1411
Now we have
Al = 0.0471 moles
Cl = 0.1411 moles
As we Know
Empirical formula shows the simplest ratio of atoms in the molecule but not whole numbers of atoms in a compound.
So,
The ratio of moles of Al to chlorine is
Al : Cl
0.0471 0.1411
Divide the ratio by smallest number to get simplest whole number ratio
Al : Cl
0.0471 / 0.0471 0.1411/ 0.0471
Al : Cl
1 : 3
The simplest ratio of Al to cl is 1 to 3, so the formula will be
Emperical formula of aluminum chloride = AlCl₃
Answer: AlCl3
Explanation:
Al: 1.271 g /29.48 g= 0.0471/0.0471=1
Cl: 6.280 g-1.271=5.009 g / 35.45/0.0411/0.0471= 3