Answer
$8.0 \times 10^{-8} mol$ and $4.8 \times 10^{16}$ molecules.
Work Step by Step
We are given both the molecular formula $C_{21}H_{30}O_{2}$ and the sample size $25µ$. First, we obtain the molar mass of THC, by obtaining the atomic mass of each of its substances from the periodic table and adding them.
Carbon: $21 atoms \times 12 g/mol = 252 g/mol$
Hydrogen: $30 atoms \times 1 g/mol = 30 g/mol$
Oxygen: $2 atoms \times 16 g/mol = 32 g/mol$
Therefore, the molar mass of THC is $314 g/mol$ or $3.14 \times 10^{2} g/mol$
We can use the conversion factor $(10^{6}µ / 1g)$ to convert g/mol to µ/mol, multiplying to get $3.14 \times 10^{8} u/mol$
Then, just divide our sample size by the molecular mass in µ/mol: $2.5 \times 10^{1} µ \times 3.14 \times 10^{8} µ/mol = (0.796 \times 10^{-7})$ which when recorded properly is $8.0 \times 10^{-8} mol$
To get molecules, just use Avogadro's constant to determine $1 mol = 6.02 \times 10^{23} molecules$, multiplying to get $48 \times 10^{15}$ or $4.8 \times 10^{16}$ molecules
