The percent composition of a compound is its mass percent with respect to the sample total mass.
We already know tha sample mass, 200.0 g, to find the mass of the carbon dioxide, CO₂, we can use its molar mass and the measured number of moles:
[tex]\begin{gathered} M_{CO_{2}}=\frac{m_{CO_2}}{n_{CO_{2}}}_{} \\ m_{CO_2}=n_{CO_2}\cdot M_{CO_{2}} \end{gathered}[/tex]
We know that the molar mass is 44.01 g/mol, and the number of moles is 0.30 mol, so:
[tex]\begin{gathered} n_{CO_{2}}=0.30mol_{} \\ M_{CO_{2}}=44.01g\/mol \\ m_{CO_2}=0.30mol\cdot44.01g\/mol=13.203g \end{gathered}[/tex]
Now, we check the percent of 13.203g with respect to the total 200.0 g:
200.0g ---- 100%
13.203g ---- P
[tex]\begin{gathered} \frac{200.0g}{13.203g}=\frac{100\%}{P} \\ P\cdot200.0=13.203\cdot100\% \\ P=\frac{13.203}{200.0}\cdot100\%=6.6015\%\approx6.6\% \end{gathered}[/tex]
By comparison, we can see that this corresponds to alternative D.
