Question

The source of bitterness in dark chocolate is the compound theobromine, an alkaloid present in cocoa beans. In a sample of dark chocolate there are 6.62 × 10^20 molecules of theobromine present. How many milligrams of theobromine are present in the sample? The molecular formula for theobromine is C7H8N4O2.

Answer 1

Answer:

$198.01~mg~of ~theobromine$

Explanation:

For this case we have to go through molecules to moles, from moles to grams and finally from grams to mg. The ratios that we have to take into account are:

-) 1 mol = 6.023x10^23 molecules

-) The molar mass of $C_7H_8N_4O_2$ 180.16 g = 1 mol

-) 1 g= 1000 mg

With this in mind we can do the calculation:

$6.62x10^2^0~molecules\frac{1~mol}{6.023x10^2^3~molecules}\frac{180.16~g}{1~mol}\frac{1000~mg}{1~g}=198.01~mg~of ~theobromine$