chemistry i\nd. determine the molecular formula for each compound described.\n9. a compound has an empirical…

chemistry i\nd. determine the molecular formula for each compound described.\n9. a compound has an empirical formula of no₂ and a molar mass of 92.02 g/r\n10. a compound has an empirical formula of c₂h₃o and a molar mass of 172 g/\n11. ibuprofen, a common headache remedy, has an empirical formula of c₇h₉o and a molar mass of approximately 215 g/mol.\n12. nicotine is 74.1% carbon, 8.6% hydrogen, and 17.3% nitrogen by mass. its molar mass is about 160 g/mol.\n13. epinephrine (adrenaline) is a hormone secreted into the bloodstream in time of danger and stress. it is 59.0% carbon, 7.1% hydrogen, 26.2% oxygen, and nitrogen by mass. its molar mass is about 180 g/mol.

chemistry i\nd. determine the molecular formula for each compound described.\n9. a compound has an empirical formula of no₂ and a molar mass of 92.02 g/r\n10. a compound has an empirical formula of c₂h₃o and a molar mass of 172 g/\n11. ibuprofen, a common headache remedy, has an empirical formula of c₇h₉o and a molar mass of approximately 215 g/mol.\n12. nicotine is 74.1% carbon, 8.6% hydrogen, and 17.3% nitrogen by mass. its molar mass is about 160 g/mol.\n13. epinephrine (adrenaline) is a hormone secreted into the bloodstream in time of danger and stress. it is 59.0% carbon, 7.1% hydrogen, 26.2% oxygen, and nitrogen by mass. its molar mass is about 180 g/mol.

Answer

9.

Explanation:

Step1: Calculate empirical - formula mass

The empirical formula is $NO_2$. The atomic mass of $N = 14.01\ g/mol$ and $O=16.00\ g/mol$. So, the empirical - formula mass $M_{empirical}=14.01 + 2\times16.00=14.01 + 32.00 = 46.01\ g/mol$.

Step2: Find the ratio $n$

$n=\frac{M_{molecular}}{M_{empirical}}=\frac{92.02}{46.01}=2$.

Step3: Determine molecular formula

Multiply the sub - scripts in the empirical formula by $n$. The molecular formula is $N_2O_4$.

10.

Explanation:

Step1: Calculate empirical - formula mass

The empirical formula is $C_2H_3O$. The atomic mass of $C = 12.01\ g/mol$, $H = 1.01\ g/mol$, and $O = 16.00\ g/mol$. So, $M_{empirical}=2\times12.01+3\times1.01 + 16.00=24.02+3.03 + 16.00=43.05\ g/mol$.

Step2: Find the ratio $n$

$n=\frac{M_{molecular}}{M_{empirical}}=\frac{172}{43.05}\approx4$.

Step3: Determine molecular formula

Multiply the sub - scripts in the empirical formula by $n$. The molecular formula is $C_8H_{12}O_4$.

11.

Explanation:

Step1: Calculate empirical - formula mass

The empirical formula is $C_7H_9O$. The atomic mass of $C = 12.01\ g/mol$, $H = 1.01\ g/mol$, and $O = 16.00\ g/mol$. So, $M_{empirical}=7\times12.01+9\times1.01+16.00=84.07 + 9.09+16.00 = 109.16\ g/mol$.

Step2: Find the ratio $n$

$n=\frac{M_{molecular}}{M_{empirical}}=\frac{215}{109.16}\approx2$.

Step3: Determine molecular formula

Multiply the sub - scripts in the empirical formula by $n$. The molecular formula is $C_{14}H_{18}O_2$.

12.

Explanation:

Step1: Assume 100 g of the compound

So, we have $74.1\ g$ of $C$, $8.6\ g$ of $H$, and $17.3\ g$ of $N$.

Step2: Calculate moles of each element

$n_C=\frac{74.1\ g}{12.01\ g/mol}\approx6.17\ mol$, $n_H=\frac{8.6\ g}{1.01\ g/mol}\approx8.51\ mol$, $n_N=\frac{17.3\ g}{14.01\ g/mol}\approx1.23\ mol$.

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles ($n_N = 1.23\ mol$). $C:H:N=\frac{6.17}{1.23}:\frac{8.51}{1.23}:\frac{1.23}{1.23}\approx5:7:1$. The empirical formula is $C_5H_7N$.

Step4: Calculate empirical - formula mass

$M_{empirical}=5\times12.01+7\times1.01 + 14.01=60.05+7.07+14.01 = 81.13\ g/mol$.

Step5: Find the ratio $n$

$n=\frac{M_{molecular}}{M_{empirical}}=\frac{160}{81.13}\approx2$.

Step6: Determine molecular formula

Multiply the sub - scripts in the empirical formula by $n$. The molecular formula is $C_{10}H_{14}N_2$.

13.

Explanation:

Step1: Assume 100 g of the compound

So, we have $59.0\ g$ of $C$, $7.1\ g$ of $H$, $26.2\ g$ of $O$, and $(100-(59.0 + 7.1+26.2))\ g=7.7\ g$ of $N$.

Step2: Calculate moles of each element

$n_C=\frac{59.0\ g}{12.01\ g/mol}\approx4.91\ mol$, $n_H=\frac{7.1\ g}{1.01\ g/mol}\approx7.03\ mol$, $n_O=\frac{26.2\ g}{16.00\ g/mol}\approx1.64\ mol$, $n_N=\frac{7.7\ g}{14.01\ g/mol}\approx0.55\ mol$.

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles ($n_N = 0.55\ mol$). $C:H:O:N=\frac{4.91}{0.55}:\frac{7.03}{0.55}:\frac{1.64}{0.55}:\frac{0.55}{0.55}\approx9:13:3:1$. The empirical formula is $C_9H_{13}O_3N$.

Step4: Calculate empirical - formula mass

$M_{empirical}=9\times12.01+13\times1.01+3\times16.00 + 14.01=108.09+13.13+48.00+14.01 = 183.23\ g/mol$.

Step5: Find the ratio $n$

$n=\frac{M_{molecular}}{M_{empirical}}=\frac{180}{183.23}\approx1$.

Step6: Determine molecular formula

The molecular formula is $C_9H_{13}O_3N$.

Answer:

  1. $N_2O_4$
  2. $C_8H_{12}O_4$
  3. $C_{14}H_{18}O_2$
  4. $C_{10}H_{14}N_2$
  5. $C_9H_{13}O_3N$