question 20 (essay worth 8 points)\n(05.03 hc...
question 20 (essay worth 8 points)\n(05.03 hc)\ndetermine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxygen.\nin an experiment, the molar mass of the compound was determined to be 228.276 g/mol. what is the molecular formula of the compound?\nfor both questions, show your work or explain how you determined the formulas by giving specific values used in calculations.
Answer
# 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: $C_9H_{24}O_6$