Question

1. a compound is analyzed and found to contain 68.54% carbon, 8.63% hydrogen, and 22.83% oxygen. the molecular weight of this compound is known to be approximately 140 g/mol. what is the empirical formula? what is the molecular formula? (assume 100 grams of the substance is present.) 3. a compound is found to have (by mass) 48.38% carbon, 8.12% hydrogen and the rest oxygen. what is its empirical formula? 4. a compound is found to have 46.67% nitrogen, 6.70% hydrogen, 19.98% carbon and 26.65% oxygen. what is its empirical formula? element mass (g) atomic mass mole mole ratio empirical formula: element mass (g) atomic mass mole mole ratio empirical formula: element mass (g) atomic mass mole mole ratio empirical formula:

Explanation:

Step1: Assume 100 - g sample for problem - 2

Since we assume 100 g of the compound, the masses of elements are:
Carbon: $m_{C}=68.54\ g$, Hydrogen: $m_{H}=8.63\ g$, Oxygen: $m_{O}=22.83\ g$

Step2: Calculate moles of each element

The molar mass of $C$ is $M_{C}=12.01\ g/mol$, of $H$ is $M_{H}=1.01\ g/mol$, and of $O$ is $M_{O}=16.00\ g/mol$.
The number of moles of $C$: $n_{C}=\frac{m_{C}}{M_{C}}=\frac{68.54\ g}{12.01\ g/mol}\approx5.71\ mol$
The number of moles of $H$: $n_{H}=\frac{m_{H}}{M_{H}}=\frac{8.63\ g}{1.01\ g/mol}\approx8.54\ mol$
The number of moles of $O$: $n_{O}=\frac{m_{O}}{M_{O}}=\frac{22.83\ g}{16.00\ g/mol}\approx1.43\ mol$

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles ($n_{O} = 1.43\ mol$)
For $C$: $\frac{n_{C}}{n_{O}}=\frac{5.71\ mol}{1.43\ mol}\approx4$
For $H$: $\frac{n_{H}}{n_{O}}=\frac{8.54\ mol}{1.43\ mol}\approx6$
For $O$: $\frac{n_{O}}{n_{O}} = 1$
The empirical formula is $C_{4}H_{6}O$.

Step4: Calculate the empirical - formula mass

$M_{empirical}=4\times12.01\ g/mol+6\times1.01\ g/mol + 1\times16.00\ g/mol=48.04\ g/mol+6.06\ g/mol+16.00\ g/mol = 70.1\ g/mol$

Step5: Find the multiple $n$

The molecular weight is approximately $140\ g/mol$. $n=\frac{M_{molecular}}{M_{empirical}}=\frac{140\ g/mol}{70.1\ g/mol}\approx2$

Step6: Determine the molecular formula

The molecular formula is $(C_{4}H_{6}O)_{2}=C_{8}H_{12}O_{2}$

For problem - 3:

Step1: Assume 100 - g sample

Carbon: $m_{C}=48.38\ g$, Hydrogen: $m_{H}=8.12\ g$, Oxygen: $m_{O}=100-(48.38 + 8.12)=43.5\ g$

Step2: Calculate moles of each element

$n_{C}=\frac{m_{C}}{M_{C}}=\frac{48.38\ g}{12.01\ g/mol}\approx4.03\ mol$
$n_{H}=\frac{m_{H}}{M_{H}}=\frac{8.12\ g}{1.01\ g/mol}\approx8.04\ mol$
$n_{O}=\frac{m_{O}}{M_{O}}=\frac{43.5\ g}{16.00\ g/mol}\approx2.72\ mol$

Step3: Find the mole - ratio

Divide by the smallest number of moles ($n_{O}\approx2.72\ mol$)
For $C$: $\frac{n_{C}}{n_{O}}=\frac{4.03\ mol}{2.72\ mol}\approx1.5$
For $H$: $\frac{n_{H}}{n_{O}}=\frac{8.04\ mol}{2.72\ mol}\approx3$
For $O$: $\frac{n_{O}}{n_{O}} = 1$
Multiply each ratio by 2 to get whole - numbers. The empirical formula is $C_{3}H_{6}O_{2}$

For problem - 4:

Step1: Assume 100 - g sample

Nitrogen: $m_{N}=46.67\ g$, Hydrogen: $m_{H}=6.70\ g$, Carbon: $m_{C}=19.98\ g$, Oxygen: $m_{O}=26.65\ g$

Step2: Calculate moles of each element

$n_{N}=\frac{m_{N}}{M_{N}}=\frac{46.67\ g}{14.01\ g/mol}\approx3.33\ mol$
$n_{H}=\frac{m_{H}}{M_{H}}=\frac{6.70\ g}{1.01\ g/mol}\approx6.63\ mol$
$n_{C}=\frac{m_{C}}{M_{C}}=\frac{19.98\ g}{12.01\ g/mol}\approx1.66\ mol$
$n_{O}=\frac{m_{O}}{M_{O}}=\frac{26.65\ g}{16.00\ g/mol}\approx1.66\ mol$

Step3: Find the mole - ratio

Divide by the smallest number of moles ($n_{C}\approx1.66\ mol$)
For $N$: $\frac{n_{N}}{n_{C}}=\frac{3.33\ mol}{1.66\ mol}\approx2$
For $H$: $\frac{n_{H}}{n_{C}}=\frac{6.63\ mol}{1.66\ mol}\approx4$
For $C$: $\frac{n_{C}}{n_{C}} = 1$
For $O$: $\frac{n_{O}}{n_{C}}=\frac{1.66\ mol}{1.66\ mol}=1$
The empirical formula is $C_{1}H_{4}N_{2}O$

Answer:

1. Molecular formula: $C_{8}H_{12}O_{2}$, Empirical formula: $C_{4}H_{6}O$
2. Empirical formula: $C_{3}H_{6}O_{2}$
3. Empirical formula: $CH_{4}N_{2}O$