Answer:
[tex]m_C=98.5mgC[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us obtain the mass of carbon in the sample by applying the following stoichiometric factor, considering that the only source of this atom in the sample is contained in the produced 361 mg of CO2 and one mole of this compound contain one mole of carbon atoms whose mass is 12.01 g/mol or mg/mmol as shown below:
[tex]m_C=361mgCO_2*\frac{1mmolCO_2}{44.01mgCO_2}* \frac{1mmolC}{1mmolCO_2}*\frac{12.01mgC}{1mmolC} \\\\m_C=98.5mgC[/tex]
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