What is the empirical formula of a compound containing 5.03 g carbon, 0.42 g hydrogen, and 44.5 g of chlorine

CHCl₃

Explanation:

We have the following data:

C = 5.03 g

H = 0.42 g

Cl= 44.5 g

First, we divide each mass by the molar mass (MM) of the chemical element to calculate the moles:

MM(C) = 12 g/mol

moles of C = mass/MM(C) = 5.03 g/(12 g/mol) = 0.42 mol C

MM(H) = 1 g/mol

moles of H = mass/MM(H) = 0.42 g/(1 g/mol) = 0.42 mol H

MM(Cl) = 35.4 g/mol

moles of Cl = mass/MM(Cl) = 44.5 g/(35.4 g/mol) = 1.26 mol Cl

Now, we divide the moles by the smallest number of moles (0.42):

0.42 mol C/0.42 = 1 C

0.42 mol H/0.42 = 1 H

1.26 mol Cl/0.42 = 3 Cl

Thus, the C:H:Cl ratio is 1:1:3.

Therefore, the empirical formula is CHCl₃

sodium (na), for example, has one electron in its valence shell. this is an unstable state because that valence shell needs eight electrons to be full. in order to fill its valence shell, find a way to add seven electrons to its valence shell