Question

The empirical formula of the anticancer drug altretamine is C3H6N2. A chemist experimentally determines its
molar mass to be 214 g/mol. What is the molecular formula for this drug?

Answer 1

Answer: The molecular formula will be $C_9H_{18}N_6$

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

To calculate the molecular formula, we need to find the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

empirical mass of $C_3H_6N_2=3\times 12+6\times 1+2\times 14=70g$

Molecular mass= 214 g/mol

$n=\frac{214}{70}=3$

Thus molecular formula will be $3\times C_3H_6N_2=C_9H_{18}N_6$

Answer 2

Altretamine, whose empirical formula is C₃H₆N₂, has the molecular formula C₉H₁₈N₆.

Differences between the empirical formula and the molecular formula

The empirical formula of a compound gives the simplest ratio of the number of different atoms present, whereas the molecular formula gives the actual number of each different atom present in a molecule.

• Step 1: Calculate the molar mass of the empirical formula.

M(C₃H₆N₂) = 3 × M(C) + 6 × M(H) + 2 × M(N)

M(C₃H₆N₂) = 3 × 12.01 g/mol + 6 × 1.01 g/mol + 2 × 14.01 g/mol

M(C₃H₆N₂) = 70.11 g/mol

• Step 2: Calculate "n".

"n" is the quotient between the molar mass of the molecular formula and the molar mass of the empirical formula.

n = (214 g/mol)/(70.11 g/mol) ≈ 3

• Step 3: Determine the molecular formula.

We will multiply the empirical formula by "n".

C₃H₆N₂ × 3 = C₉H₁₈N₆

Altretamine, whose empirical formula is C₃H₆N₂, has the molecular formula C₉H₁₈N₆.

Learn more about empirical and molecular formulas here: https://brainly.com/question/9255428