Question

what is the molecular formula for a compound with an empirical formula of ch and a molecular mass of 78.12 g/mol. ch c12h6 c6h6 c6h12 question 25 what is the molecular formula of a compound with an empirical formula of ch2o and a molecular mass of 180.18 g/mol? c18h36o18 c12h24o6 c6h12o6 c2h4o2

Explanation:

Step1: Calculate the empirical formula mass

The empirical formula is \(CH\). The atomic mass of \(C = 12.01\space g/mol\) and \(H= 1.01\space g/mol\).
The empirical formula mass \(M_{empirical}=(12.01 + 1.01)\space g/mol=13.02\space g/mol\)

Step2: Find the ratio \(n\)

We know that the molecular formula \(M_{molecular}=n\times M_{empirical}\). Given \(M_{molecular} = 78.12\space g/mol\)
\(n=\frac{M_{molecular}}{M_{empirical}}=\frac{78.12}{13.02}=6\)

Step3: Determine the molecular formula

Multiply the sub - scripts in the empirical formula \(CH\) by \(n = 6\).
The molecular formula is \(C_{6}H_{6}\)

Step1: Calculate the empirical formula mass

The empirical formula is \(CH_{2}O\). The atomic mass of \(C = 12.01\space g/mol\), \(H = 1.01\space g/mol\) and \(O=16.00\space g/mol\)
The empirical formula mass \(M_{empirical}=(12.01+2\times1.01 + 16.00)\space g/mol=(12.01 + 2.02+16.00)\space g/mol = 30.03\space g/mol\)

Step2: Find the ratio \(n\)

We know that \(M_{molecular}=n\times M_{empirical}\). Given \(M_{molecular}=180.18\space g/mol\)
\(n=\frac{M_{molecular}}{M_{empirical}}=\frac{180.18}{30.03}=6\)

Step3: Determine the molecular formula

Multiply the sub - scripts in the empirical formula \(CH_{2}O\) by \(n = 6\)
The molecular formula is \(C_{6}H_{12}O_{6}\)

Answer:

\(C_{6}H_{6}\)

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