Question 21N.2.sl.TZ0.1a

«$\frac{8.802 \mathrm{g}}{44.01 \mathrm{g} {\mathrm{mol}}^{-1}}=$» 0.2000 «mol of C/CO₂»

AND «$\frac{3.604 \mathrm{g}}{18.02 \mathrm{g} {\mathrm{mol}}^{-1}}=$» 0.2000 «mol of H₂O»/0.4000 «mol of H»

OR

 «$\frac{8.802 \mathrm{g}}{44.01 \mathrm{g} {\mathrm{mol}}^{-1}}\times 12.01 \mathrm{g} {\mathrm{mol}}^{-1}$» 2.402 «g of C»

OR

«$\frac{3.604 \mathrm{g}}{18.02 \mathrm{g} {\mathrm{mol}}^{-1}}\times 2\times 1.01 \mathrm{g} {\mathrm{mol}}^{-1}=$» 0.404 «g of H» ✔

«4.406 g − 2.806 g» = 1.600 «g of O» ✔

«$\frac{2.402 \mathrm{g}}{12.01 \mathrm{g} {\mathrm{mol}}^{-1}}=0.2000 \mathrm{mol} \mathrm{C}; \frac{0.404 \mathrm{g}}{1.01 \mathrm{g} {\mathrm{mol}}^{-1}}=0.400 \mathrm{mol} \mathrm{H}; \frac{1.600 \mathrm{g}}{16.00 \mathrm{g} {\mathrm{mol}}^{-1}}=0.1000 \mathrm{mol} \mathrm{O}$»

C₂H₄O ✔

Award [3] for correct final answer.