Question

use the weight of the molecule to determine the molecular formula.
the weight of the empirical formula $(c_{4}h_{5}n_{2}o)$ doesnt match the weight of the actual substance (194 amu). the empirical formula weight (97 amu) is smaller than the molecular formula weight (194 amu). this means that actual molecule that were looking for is larger but has the same whole number ratio of elements.
keeping the ratio in mind, which of the following keeps that ratio while also having a molecular weight of 194 amu?
$c_{9}h_{10}n_{2}o_{5}=194$ amu
$c_{12}h_{15}n_{6}o_{3}=291$ amu
$c_{8}h_{10}n_{4}o_{2}=194$ amu
$c_{4}h_{5}n_{2}o = 97$ amu
$c_{9}h_{9}n_{2}=194$ amu
choose the correct molecular formula from the options below before moving on.

Explanation:

Step1: Calculate the weight of the empirical formula \(C_4H_5N_2O\)

The atomic weights are: \(C = 12\) amu, \(H=1\) amu, \(N = 14\) amu, \(O=16\) amu.

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Step2: Find the ratio of molecular formula weight to empirical formula weight

$$n=\frac{\text{Molecular formula weight}}{\text{Empirical formula weight}}=\frac{194}{97} = 2$$

Step3: Multiply the sub - scripts in the empirical formula by \(n = 2\)

For \(C_4H_5N_2O\), when multiplied by \(2\), we get \(C_{4\times2}H_{5\times2}N_{2\times2}O_{1\times2}=C_8H_{10}N_4O_2\)

Answer:

\(C_8H_{10}N_4O_2 = 194\) amu