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 6.72×1020 molecules of theobromine present. How many milligrams of theobromine are present in the sample? The molecular formula for theobromine is C7H8N4O2.

Respuesta :

Answer: [tex]1.98\times 10^{-4}[/tex] mg

Explanation:-

According to avogadro's law, 1 mole of every substance occupies 22.4 Liters at STP , contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles and weigh equal to the molecular mass of the substance.

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given molecules}}{\text {Avogadro's number}}[/tex]

[tex]\text{Number of moles}=\frac{6.72\times 10^{20}}{6.023\times 10^{23}}=1.1\times 10^{-3}moles[/tex]  

Now 1 mole of [tex]C_7H_8N_4O_2[/tex] molecule weigh = 180g

[tex]1.1\times 10^{-3}moles[/tex] of [tex]C_7H_8N_4O_2[/tex] molecule weigh = [tex]\frac{180}{1}\times 1.1\times 10^{-3}=0.198g=1.98\times 10^{-4}mg[/tex] (1g=1000mg)

Thus [tex]1.98\times 10^{-4}[/tex] mg of theobromine are present in the sample.

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