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 6.99×10²⁰ molecules of theobromine present. how many milligrams of theobromine are present in the sample? the molecular formula for theobromine is c₇h₈n₄o₂.

Explanation:

Step1: Calculate molar mass of theobromine

The molar mass of $C_7H_8N_4O_2$:
$M=(7\times12.01 + 8\times1.01+4\times14.01 + 2\times16.00)\text{ g/mol}$
$M=(84.07+8.08 + 56.04+32.00)\text{ g/mol}=180.19\text{ g/mol}$

Step2: Calculate number of moles

Use Avogadro's number $N_A = 6.022\times10^{23}\text{ molecules/mol}$.
The number of moles $n=\frac{N}{N_A}$, where $N = 6.99\times10^{20}$ molecules.
$n=\frac{6.99\times 10^{20}}{6.022\times10^{23}\text{ molecules/mol}}= 1.16\times10^{-3}\text{ mol}$

Step3: Calculate mass in grams

$m=n\times M$.
$m=(1.16\times10^{-3}\text{ mol})\times180.19\text{ g/mol}=0.21\text{ g}$

Step4: Convert to milligrams

Since $1\text{ g}=1000\text{ mg}$, then $m = 0.21\text{ g}\times1000\text{ mg/g}=112\text{ mg}$ (rounded to three - significant figures)

Answer:

$112$ mg