Question

An organic compound contains 49.2% carbon, 9.68% hydrogen, 19.19% nitrogen. If molecular mass of given compound is 146g, calculate the molecular formula of compound. please explain the process.
:3

Answer 1

The percentage is taken in terms of mass (to calculate moles).
1. Calculate the total percentage or mass. It should equal 100.
49.2+9.68+19.19= 78.07 i.e. the remaining 21.93% is of oxygen.

2. Find the number of moles:
Carbon- 49.2/12= 4.1
Hydrogen - 9.68/1 =9.68
Nitrogen - 19.19/14 = 1.37
Oxygen - 21.93/16 =1.37

3. Check for the lowest number and divide the above answers with the same.
So, Carbon- 4.1/1.37 = 2.99 = 3
Hydrogen- 9.68/1.37 = 7
Nitrogen - 1.37/1.37 = 1
Oxygen - 1.37/1.37 = 1

4. The simplest whole number ratio is
C:H:N:O = 3:7:1:1

5. The emperical formula = C3H7NO

6. Calculate emperical mass of the formula.
(12x3)+7+14+16 = 73g

7. Emperical mass x n= Molecular formula. Calculate n.
73 x 2= 146
n = 146/73 = 2.

8. Multiply the emperical formula by n to get the molecular formula.
(C3H7NO) x 2= C6H14N2O2.
This is lysine which is an amino acid.

Answer 2

Hello!

An organic compound contains 49.2% carbon, 9.68% hydrogen, 19.19% nitrogen. If molecular mass of given compound is 146 g, calculate the molecular formula of compound.

data:

Carbon (C) ≈ 12 a.m.u (g/mol)

Hydrogen (H) ≈ 1 a.m.u (g/mol)

Nitrogen (N) ≈ 14 a.m.u (g/mol)

Oxygen (O) ≈ 16 a.m.u (g/mol)

• We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100g solution)

C: 49.2 % → 49.2 g

H: 9.68 % → 9.68 g

N: 19.19 % → 19.19 g

O:  (100% - 49.2% - 9.68% - 19.19%) = 21.93 % → 21.93 g

• The values ​​(in g) will be converted into quantity of substance (number of mols), dividing by molecular weight (g / mol) each of the values, we will see:

$C: \dfrac{49.2\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} = 4.1\:mol$

$H: \dfrac{9.68\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 9.68\:mol$

$N: \dfrac{19.19\:\diagup\!\!\!\!\!g}{14\:\diagup\!\!\!\!\!g/mol} \approx 1.37\:mol$

$O: \dfrac{21.93\:\diagup\!\!\!\!\!g}{16\:\diagup\!\!\!\!\!g/mol} \approx 1.37\:mol$

• We realize that the values ​​found above, some are not integers, so we divide these values ​​by the smallest of them, so that the proportion does not change, let's see:

$C: \dfrac{4.1}{1.37}\to\:\:\boxed{C \approx 3}$

$H: \dfrac{9.68}{1.37}\to\:\:\boxed{H \approx 7}$

$N: \dfrac{1.37}{1.37}\to\:\:\boxed{N = 1}$

$O: \dfrac{1.37}{1.37}\to\:\:\boxed{O = 1}$

= 3 : 7 : 1 : 1  ← whole number of atomic radio

• Thus, the minimum or Empirical Formula (E.F) found for the compound will be:

$\boxed{C_{3}H_{7}NO}\Longleftarrow(Empirical\:Formula)$

• Let's find the Molecular Weight (MW) of the Empirical Formula (EF), let's see:

if: C3H7NO  

C = 3*(12 a.m.u) = 36 a.m.u  

H = 7*(1 a.m.u) = 7 a.m.u  

N = 1*(14 a.m.u) = 14 a.m.u

O = 1*(16 a.m.u) = 16 a.m.u  

-------------------------------------  

$MW(E.F) = 36 + 7 + 14 + 16\to \boxed{MW\:(E.F) = 73\:a.m.u}$

• Knowing that the Molecular Weight of the Molecular Formula is 146 units of atomic mass (or 146 g/mol) and that the Molecular Weight of the Empirical Formula is 73 units of atomic mass (or 73 g/mol), then we will find the number of terms (n) for the molecular formula of the compound, let us see:

$n = \dfrac{MW_{M.F}}{MW_{E.F}}$

$n = \dfrac{146}{73}$

$\boxed{n = 3}$

• Therefore, the Molecular Formula (M.F) is the Empirical Formula (E.F) times the number of terms (n), then, we have:

$M.F = (E.F)*n$

$M.F = (C_3H_7NO)*2$

$\boxed{\boxed{M.F = C_6H_{14}N_2O_2}}\Longleftarrow(molecular\:formula\:of\:the\:compound)\:\:\:\:\:\:\bf\green{\checkmark}$

• Answer:

The molecular formula of compound (Lysine) is C6H14N2O2

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