Given data

• Amount of organic compound burnt $= 0.01\ \text{mol}$
• Carbon percentage in compound $= 60\%$
• Mass of $\text{CO}_2$ produced $= 4.4\ \text{g}$

Find moles of $\text{CO}_2$ formed $n(\text{CO}_2)=\frac{4.4}{44}=0.1\ \text{mol}$

Find moles of carbon in the compound sample Each mole of $\text{CO}_2$ contains 1 mole of C, so: $n(\text{C})=0.1\ \text{mol}$

Find mass of carbon in the sample $m(\text{C})=0.1\times 12=1.2\ \text{g}$

Use carbon percentage to find total mass of compound burnt Since carbon is $60\%$ of the compound by mass, $\frac{1.2}{m_{\text{sample}}}=0.60$ $m_{\text{sample}}=\frac{1.2}{0.60}=2.0\ \text{g}$

Find molar mass of the compound We are told that $2.0\ \text{g}$ corresponds to $0.01\ \text{mol}$. Therefore, $M=\frac{2.0}{0.01}=200\ \text{g mol}^{-1}$

Final answer $\boxed{200}$

Comparison with stored answer Stored correct answer = $200$

My derived answer matches the stored answer.