Asample was decomposed in the laboratory and found to have 38.67g c, 16.22g h, and 45.11g n.

a)find the molecular formula of this compound if the formula mass is 62.12 g/mole.
b)determine how many h atoms would be in a 3.50g sample of this compound.

A) Molecular formula

Procedure

You need to pass the information in grams to moles.

Then you  need to find the proportion of moles, which will give you the subscript for each element in the formula, and then you will have the empirical formula.

With this you need to calculate the molar mass of the empirical formula.

Then divide the molar mass of the compound by the molar mass of the empirical formula to find the multiple that you have to use to establish the molercular formula.

Solution

Number of moles of an element = mass in grams / atomic mass

C: 38.67 g / 12 g/mol = 3.2225 moles
H: 16.22 g / 1 g/mol = 16.22 moles
N: 45.11 g / 14 g/mol = 3.2221 moles

Proportions: divide by the smaller number of moles

C: 3.2225 / 3.2221 = 1.0
H: 16.22 / 3.2221 = 5.0
N: 3.2221 / 3.2221 = 1.0

Empirical formula: C1H5N1

Empirical mass: 1mol*12g/mol + 5mol*1g/mol + 1 mol*14g/mol = 31 g/mol

Number of times the empirical formula is contained in the molecular formula: formula mass / empirical mass = 62.12 g/mole/31g/mole = 2

Molecular formula = 2* empirical formula = C2H10N2.

2) H atoms

Proportion of H: 10 moles of H / 62.12 g of C2H10N2 = x g of H / 35.0 g of C2H10N2

Solve for x:

x = 35.0*10 / 62.12 = 5.63 moles

Number of atoms = number of moles * Avogadro = 5.63 * 6.02 * 10^23 atoms

= 33. 89 * 10^23 atoms = 3.389 * 10^24 atoms

Both models are effective because they show the valence electrons, and these valence electrons indicate the reactivity of an element.