Answer
The mass is $0.1023 mg$
Work Step by Step
Using the molar mass of the compound provided, we can simply convert from molecules to moles and then from moles to grams and then finally to milligrams.
$1.8\times10^{17} molecules C_{12}H_{22}O_{11}\times\frac{1 mole C_{12}H_{22}O_{11}}{6.023\times10^{23} molecules} \times\frac{342.3 g/mole}{1 mole C_{12}H_{22}O_{11}}
= 1.023\times10^{-4} grams C_{12}H_{22}O_{11}$. To convert from grams to miligrams, we simply multiply by 1000;
$1.023\times10^{-4} grams C_{12}H_{22}O_{11} \times 1000 = 0.1023 mg$
