the source of bitterness in dark chocolate is the compound theobromine, an alkaloid present in cocoa beans…

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.

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.

Answer

Answer:

$112$ mg

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)