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Marysya12 [62]
3 years ago
14

A 6.175 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 13.30 g CO2 and 5.

447 g H2O are produced. In a separate experiment, the molar mass is found to be 102.1 g/mol. Determine the empirical formula and the molecular formula of the organic compound.
Chemistry
1 answer:
Bingel [31]3 years ago
6 0

Answer:

Empirical and molecular formulas are the same, C₅H₁₀O₂.

Explanation:

Hello!

In this case, when determining the empirical and molecular formulas of organic compounds via combustion analysis, we first need to compute the moles of carbon and hydrogen via the yielded mass of carbon dioxide and water:

n_C=13.30gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.30molC\\\\n_H=5.447gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}=0.60molH

Next, we need to compute the mass of oxygen by subtracting the mass of carbon and hydrogen to the mass of the sample of the compound:

m_O =6.175g-0.3molC*12.01gC/molC-0.6molH*1.01gH/molH =1.966gO

And consequently the moles:

n_O=0.12molO

Now, we need to divide the moles of each atom by the fewest moles, it in this case, those of oxygen to obtain the subscripts in the empirical formula:

C=\frac{0.30}{0.12} =2.5\\\\H=\frac{0.60}{0.12} =5\\\\O=\frac{0.12}{0.12} =1

Thus, the empirical formula, taken the nearest whole number is:

C_5H_{10}O_2

Now, if we divide the molar mass of the molecular formula (102.1 g/mol) by that of the empirical formula (102.1 g/mol) we infer they are both the same.

Best regards!

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