Question

Find empirical formula of 4 gram CA 6.022 x 10 22 atom of c and 4.8 gram of oxygen

Answer 1

one atom of carbon,

1

×

C

two atoms of oxygen,

2

×

O

This means that the given number of atoms of oxygen would correspond to

4.8

⋅

10

22

atoms O

⋅

1 molecule CO

2

2

atoms O

=

2.4

⋅

10

22

molecules CO

2

Now, one mole of any molecular substance contains exactly

6.022

⋅

10

22

molecules of that substance -- this is known as Avogadro's constant.

In your case, the sample of carbon dioxide molecules contains

2.4

⋅

10

22

molecules CO

2

⋅

1 mole CO

2

6.022

⋅

10

23

molecules CO

2

=

0.03985 moles CO

2

Finally, carbon dioxide has a molar mass of

44.01 g mol

−

1

, which means that your sample will have a mass of

0.03985

moles CO

2

⋅

44.01 g

1

mole CO

2

=

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

∣

∣

a

a

1.8 g

a

a

∣

∣

−−−−−−−−−

Answer 2

Explanation:

plz observe the photo. Answer in the photo