chemistry i
d. determine the molecular formula for each compound described.
9. a compound has an empirical formula of no₂ and a molar mass of 92.02 g/r
10. a compound has an empirical formula of c₂h₃o and a molar mass of 172 g/
11. ibuprofen, a common headache remedy, has an empirical formula of c₇h₉o and a molar mass of approximately 215 g/mol.
12. nicotine is 74.1% carbon, 8.6% hydrogen, and 17.3% nitrogen by mass. its molar mass is about 160 g/mol.
13. 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.

Question

Accepted 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 $M_{empirical}=14.01 + 2	imes16.00=14.01 + 32.00 = 46.01\ g/mol$.
## 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, $M_{empirical}=2	imes12.01+3	imes1.01 + 16.00=24.02+3.03 + 16.00=43.05\ g/mol$.
## 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, $M_{empirical}=7	imes12.01+9	imes1.01+16.00=84.07 + 9.09+16.00 = 109.16\ g/mol$.
## 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
$M_{empirical}=5	imes12.01+7	imes1.01 + 14.01=60.05+7.07+14.01 = 81.13\ g/mol$.
## 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
$M_{empirical}=9	imes12.01+13	imes1.01+3	imes16.00 + 14.01=108.09+13.13+48.00+14.01 = 183.23\ g/mol$.
## 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:
9. $N_2O_4$
10. $C_8H_{12}O_4$
11. $C_{14}H_{18}O_2$
12. $C_{10}H_{14}N_2$
13. $C_9H_{13}O_3N$

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Question

Explanation:

9.

Step1: Calculate empirical - formula mass

Step2: Find the ratio $n$

Step3: Determine molecular formula

10.

Step1: Calculate empirical - formula mass

Step2: Find the ratio $n$

Step3: Determine molecular formula

11.

Step1: Calculate empirical - formula mass

Step2: Find the ratio $n$

Step3: Determine molecular formula

12.

Answer: