Question

Find the total number of atoms in a sample of cocaine hydrochloride, c17h22clno4, of mass 21.0 mg .

Answer 1

Answer: The number of atoms in given amount of cocaine hydrochloride is
`1.674* 10^(21)`

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Given mass of cocaine hydrochloride = 21.0 mg = 0.021 g (Conversion factor: 1 g = 1000 mg)

Molar mass of cocaine hydrochloride = 339.8 g/mol

Putting values in above equation, we get:

`\text{Moles of cocaine hydrochloride}=(0.021g)/(339.8g/mol)=6.18* 10^(-5)mol`

In 1 mole of cocaine hydrochloride, 17 moles of carbon atom, 22 moles of hydrogen atom, 1 mole of chlorine atom, 1 mole of nitrogen atom and 4 moles of oxygen atoms are present

Total number of atoms in 1 mole of cocaine hydrochloride = (17 + 22 + 1 + 1 + 4) = 45 atoms

According to mole concept:

1 mole of a substance contains
`6.022\time 10^(23)` number of atoms.

So,
`6.18* 10^(-5)mol` of cocaine hydrochloride will contain =
`(6.18* 10^(-5)* 45* 6.022* 10^(23))=1.674* 10^(21)` number of atoms

Hence, the number of atoms in given amount of cocaine hydrochloride is
`1.674* 10^(21)`

Answer 2

Answer is: the total number of atoms in a sample of cocaine hydrochloride is 1.65·10²¹.

m(C₁₇H₂₂ClNO₄) = 21.0 mg ÷ 1000 mg/g.

m(C₁₇H₂₂ClNO₄) = 0.021 g; mass of cacaine hydrochloride.

n(C₁₇H₂₂ClNO₄) = m(C₁₇H₂₂ClNO₄) ÷ M(C₁₇H₂₂ClNO₄).

n(C₁₇H₂₂ClNO₄) = 0.021 g ÷ 339.8 g/mol.

n(C₁₇H₂₂ClNO₄) = 0.000061 mol = 6.1·10⁻⁵ mol.

N(C₁₇H₂₂ClNO₄) = n(C₁₇H₂₂ClNO₄) · Na (Avogadro constant).

N(C₁₇H₂₂ClNO₄) = 6.1·10⁻⁵ mol · 6.022·10²³ 1/mol.

N(C₁₇H₂₂ClNO₄) = 3.67·10¹⁹; number of molecules.

In one molecule of cocaine hydrochloride there are 45 atoms (17 carbons, 22 hydrogens, 1 chlorine, 1 nitrogen and 4 oxygens):

N(atoms) = N(C₁₇H₂₂ClNO₄) · 45.

N(atoms) = 3.67·10¹⁹ · 45.

N(atoms) = 1.65·10²¹.