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

c2h3cl
c2h2cl
chcl4
chcl3

The right choice is CHCl₃

Explanation:

For Carbon :

no. of moles = (mass / molar mass) = (5.03 g / 12  g/mol )=  0.42 mol

For Hydrogen  :

no. of moles = (mass / molar mass) = (0.42 g / 1  g/mol )=  0.42 mol

For Chlorine :

no. of moles = (mass / molar mass) = (44.5 g / 35.5 g/mol )=  1.25 mol

The ratio for Carbon and Hydrogen is  1 : 1

0.42 mol / 042 mol =1.00

The ratio of Chlorine to both Carbon and Hydrogen is 3:1

1.25 mol / 0.42 mol = 2.98 ≅ 3

So the ratio is 1 C : 1 H : 3 Cl.

So, the right choice is CHCl₃

Carbon usually forms four bonds one to Hydrogen and 3 to Chlorine atoms.

Chlorine usually forms one bond to Carbon atom.

Hydrogen usually forms one bond Carbon atom.

CHCl₃

Explanation :

In order to deduce the empirical fomula of the compound, we first express the masses given as moles:

Elements C H Cl

Masses given(g) 5.030.4244.5

Then divide through by their respective molar masses:

Molar mass 12 1 35.5

(g/mol)

Number of

Moles(mol) 0.42 0.42 1.254

Dividing by

the smallest(0.42) 1 1 3

The empirical formula of the compound is

CHCl₃

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3