# An organic compound contains 20.0 % Carbon, 6.66% Hydrogen, 47.33% Nitrogen and the rest was Oxygen. Its molar mass is 60 gm. The molecular formula of the compound is? (A) CH_4 N_2 O (B) CH_2 NO_2 (C) C_2H_6 NO (D) CH_18 NO

Mar 16, 2017

$\text{Option A}$

#### Explanation:

With all these empirical formula problems, we assume that there are $100 \cdot g$ of unknown compound, and work out the empirical formula appropriately:

$\text{Moles of carbon:} = \frac{20.0 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 1.67 \cdot m o l .$

$\text{Moles of hydrogen:} = \frac{6.66 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 6.61 \cdot m o l .$

$\text{Moles of nitrogen:} = \frac{47.33 \cdot g}{14.01 \cdot g \cdot m o {l}^{-} 1} = 3.38 \cdot m o l .$

And the balance of mass was oxygen:

$\text{Moles of oxygen:} = \frac{\left(100 - 20.0 - 6.66 - 47.33\right) \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 1.63 \cdot m o l .$

And now we divide thru by the SMALLEST molar quantity, that of oxygen, to get the empirical formula:

$C : \frac{1.67 \cdot m o l}{1.63 \cdot m o l} \cong 1$

$H : \frac{6.61 \cdot m o l}{1.63 \cdot m o l} \cong 4$

$N : \frac{3.38 \cdot m o l}{1.63 \cdot m o l} \cong 2$

$O : \frac{1.63 \cdot m o l}{1.63 \cdot m o l} = 1$

We thus get an empirical formula of $C {H}_{4} {N}_{2} O$ (and possibly, this is $\text{urea}$, $O = C {\left(N {H}_{2}\right)}_{2}$).

(Note that I approve of this problem (for what that's worth!), because an analyst WOULD never quote the %O content. He/she would give you %C, %H, %N, and sometimes %X, but oxygen analysis is not routinely performed.)

Mar 16, 2017

A.

#### Explanation:

1. Find the percentage of Oxygen by adding all of the given percentages then subtracting from 100

$20.0 + 6.66 + 47.33 = 73. 999 = 74$
$100 - 74 = 26$

1. Change the percentages to grams. If there were 100 grams
C= 20 grams H = 6.66 grams N = 47.33 grams O = 26 grams.

2. Change the grams to moles by dividing by the molecular mass of the elements

Moles Carbon = $\frac{20}{12} = 1.66$ moles C

Moles Hydrogen = $\frac{6.66}{1} = \frac{6}{66}$ moles H

Moles Nitrogen = $\frac{47.33}{14} = 3.38$ moles N

Moles Oxygen = $\frac{26}{16} = 1.625$ moles O

1. Find the simplest mole ratios.
Since Oxygen is the smallest divide all the other nu=mber of moles by the moles of Oxygen

Carbon ratio = $\frac{C}{O} = \frac{1.66}{1.625} = 1.02$ or 1:1

Nitrogen ratio = $\frac{N}{O} = \frac{3.38}{1.625} = 2.04$ or 2:1

Hydrogen ratio = $\frac{H}{O} = \frac{6.66}{1.625} = 4.09$ or 4:1

So the compound has a ratio of 1 C: 2 N : 4 H : 1O for the empirical formula

The mass of one empirical formula is 62 grams per mole. This is slightly higher than the experimental molecular mass of 60 grams but is within experimental error.

So the compound most likely has a formula of $C {N}_{2} {H}_{4} O$

The Compound would look something like this

$$  H
I


H-N-C = O
I
N-H
I
H

The valance bonding of each atom is complete
C = 4 bonds
N = 3 bonds
O = 2 bonds
H = 1 bond