question 20 (essay worth 8 points)\n(05.03 hc)\ndetermine the empirical formula of a compound containing…

Explanation:

Step1: Calculate moles of each element

The molar mass of carbon ($C$) is approximately $12.01\ g/mol$, hydrogen ($H$) is approximately $1.01\ g/mol$, and oxygen ($O$) is approximately $16.00\ g/mol$. Moles of $C=\frac{47.37\ g}{12.01\ g/mol}\approx3.944\ mol$ Moles of $H=\frac{10.59\ g}{1.01\ g/mol}\approx10.49\ mol$ Moles of $O=\frac{42.04\ g}{16.00\ g/mol}\approx2.628\ mol$

Step2: Find the mole - ratio

Divide each number of moles by the smallest number of moles (in this case, $2.628\ mol$). Ratio of $C=\frac{3.944\ mol}{2.628\ mol}\approx1.5$ Ratio of $H=\frac{10.49\ mol}{2.628\ mol}\approx4$ Ratio of $O=\frac{2.628\ mol}{2.628\ mol}=1$ Since we need whole - number ratios, multiply each ratio by 2 to get $C_3H_8O_2$. So the empirical formula is $C_3H_8O_2$.

Step3: Calculate the empirical formula mass

Empirical formula mass of $C_3H_8O_2=(3\times12.01)+(8\times1.01)+(2\times16.00)=36.03 + 8.08+32.00 = 76.11\ g/mol$

Step4: Determine the multiple ($n$)

$n=\frac{\text{Molar mass of compound}}{\text{Empirical formula mass}}=\frac{228.276\ g/mol}{76.11\ g/mol}\approx3$

Step5: Find the molecular formula

Multiply the sub - scripts in the empirical formula by $n$. Molecular formula $=(C_3H_8O_2)3 = C_9H{24}O_6$

Answer:

Empirical formula: $C_3H_8O_2$ Molecular formula: $C_9H_{24}O_6$