Question

From the given empirical formula and molar mass, find the molecular formula of each compound. C6H7N, 186.24 g/mol C2HCl, 181.44 g/mol C5H10NS2,296.54 g/mol

Answer 1

Answer:C6H7N = C12H14N2

C2HCl = C6H3Cl3

C5H10NS2 = C10H20N2S4

Explanation:

There are certain steps you need to follow to find the molecular formula:

1. Determine the number of moles = subscript

So for the first one

C has 6 moles

H has 7 moles

N has 1 mole

2. Multiply moles by atomic mass = atomic mass found on periodic table

So...

6 × 12 = 72

7 × 1 = 7

1 × 14 = 14

Then we going to add all of them together to get the empirical weight:

72 + 7 + 14 = 93

3. Divide molar mass by the empirical weight

So

186.24 ÷ 93 = 2

4. Multiply each mole by molecular weight

So...

2(C6H7N) = C12H14N2 = Molecular formula

Answer 2

Answer:

(a) $C_{12}H_{14}N_2$

(b) $C_{6}H_{3}Cl_3$

(c) $C_{10}H_{20}N_2S_4$

Explanation:

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

(a) $C_6H_7N$

Mass from the Empirical formula = 6×12 + 7×1 + 1×14 = 93 g/mol

Molar mass = 186.24 g/mol

So,

Molecular mass = n × Empirical mass

186.24 = n × 93

⇒ n ≅ 2

Molecular formula = $C_{12}H_{14}N_2$

(b) $C_2HCl$

Mass from the Empirical formula = 2×12 + 1×1 + 1×35.5 = 60.5 g/mol

Molar mass = 181.44 g/mol

So,

Molecular mass = n × Empirical mass

181.44 = n × 60.5

⇒ n ≅ 3

Molecular formula = $C_{6}H_{3}Cl_3$

(c) $C_5H_{10}NS_2$

Mass from the Empirical formula = 5×12 + 10×1 + 1×14 + 2×32 = 148 g/mol

Molar mass = 296.54 g/mol

So,

Molecular mass = n × Empirical mass

296.54 = n × 148

⇒ n ≅ 2

Molecular formula = $C_{10}H_{20}N_2S_4$