a. If a 1.303 g sample of aluminum metal is heated in a chlorine gas atmosphere,

Question

a. If a 1.303 g sample of aluminum metal is heated in a chlorine gas atmosphere, the mass of aluminum chloride produced is 6.438 g. Calculate the empirical formula of aluminum chloride.

b. When 6.538 g of zinc is heated in pure oxygen, the sample gains 1.600 g of oxygen in forming the oxide. Calculate the empirical formula of the compound.

c. A 1.7994 g sample of a new compound has been analyzed and found to contain the following masses of elements: carbon, 0.6966 g; hydrogen, 0.17544 g; oxygen, 0.9273 g. Calculate the empirical formula of the compound

d. A compound with empirical formula CH was found by experiment to have a molar mass of approximately 65 g. What is the molecular formula of the compound?

e. A compound has been analyzed and has been found to have the following composition: oxygen, 53.29%; carbon, 40.00%; hydrogen, 6.714%. Determine the empirical formula of the compound.

f. A compound was analyzed and was found to contain the following percentages of the elements by mass: calcium, 81.10%; nitrogen, 18.90%. Determine the empirical formula of the compound.

Explanation / Answer

There are multiple questions here. they are not related to each other. So then can't be part of 1 question. i am allowed to answer only 1 at a time

Let empirical formula be AlyClx
Where x can be fraction
Mass of Clx = Mass of AylClx - Mass of Aly
                         =6.438 - 1.303
                         =5.135 g
Molar mass of Cl = 35.5 g/mol
number of moles of Cl =mass /molar mass = 5.135/35.5 = 0.145

Molar mass of Al = 27 g/mol
mass of Al = 1.303 g
number of moles of Al = 1.303/27 =0.048

so,
AL : 0.048
Cl: 0.145
divide by simplest ratio:
AL : 0.048 / 0.048 =1 = y
Cl: 0.145 /0.048 = 3 = x
So,
empirical formula is AlCl3

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