13) calculate the percentage - by - mass of nitrogen in pb(no₃)₂
a)4.67 b)9.34 c)9.90 d)4.95 e)12.67
14) calculate the percentage - by - mass of oxygen in pb(no₃)₂
a)9.7 b)14.5 c)19.3 d)29.0 e)33.4
15) calculate the percentage - by - mass of chlorine in pbcl₂(no₃)₂
a)23.63 b)11.82 c)25.05 d)12.53 e)11.09
16) what is the empirical formula of a compound that contains 27.0% s, 13.4% o, and 59.6% cl by mass?
a)socl b)socl₂ c)s₂ocl d)so₂cl e)osocl₄
17) what is the empirical formula of a compound that contains 29% na, 41% s, and 30% o by mass?
a)na₂s₂o₃ b)naso₂ c)naso d)naso₃ e)na₂s₂o₆
18) a compound is composed of only c, h, and o. the combustion of a 0.519 - g sample of the compound yields 1.24 g of co₂ and 0.255 g of h₂o. what is the empirical formula of the compound?
a)c₆h₆o b)c₃h₃o c)ch₃o d)c₂h₆o₅ e)c₂h₆o₂
19) combustion of a 1.031 - g sample of a compound containing only carbon, hydrogen, and oxygen produced 2.265 g of co₂ and 1.236 g of h₂o. what is the empirical formula of the compound?
a)c₃h₆o b)c₃h₅o c)c₆h₁₆o₂ d)c₃h₉o₃ e)c₃h₆o₃
20) a compound that is composed of carbon, hydrogen, and oxygen contains 70.6% c, 5.9% h, and 23.5% o by mass. the molecular weight of the compound is 136 amu. what is the molecular formula?
a)c₈h₈o₂ b)c₈h₄o c)c₄h₄o d)c₉h₁₂o e)c₉h₆o₂

Question

Accepted Answer

13.
# Explanation:
## Step1: Find molar mass of $PCl_2(NH_2)_2$
The molar mass of $P = 30.97\ g/mol$, $Cl= 35.45\ g/mol$, $N = 14.01\ g/mol$, $H=1.01\ g/mol$.
$M = 30.97+2\times35.45 + 2\times14.01+4\times1.01=30.97 + 70.9+28.02 + 4.04=133.93\ g/mol$
## Step2: Find mass of nitrogen
Mass of $N$ in the compound: $m_N=2\times14.01 = 28.02\ g/mol$
## Step3: Calculate mass - percentage of nitrogen
$\text{Mass}\%=\frac{m_N}{M}\times100=\frac{28.02}{133.93}\times100\approx20.92\%$ (There seems to be an error in the problem - statement or options as the calculated value doesn't match the given options)

14.
# Explanation:
## Step1: Find molar mass of $Pb(NO_3)_2$
The molar mass of $Pb = 207.2\ g/mol$, $N = 14.01\ g/mol$, $O = 16.00\ g/mol$.
$M=207.2 + 2\times14.01+6\times16.00=207.2+28.02 + 96.00=331.22\ g/mol$
## Step2: Find mass of oxygen
Mass of $O$ in the compound: $m_O=6\times16.00 = 96.00\ g/mol$
## Step3: Calculate mass - percentage of oxygen
$\text{Mass}\%=\frac{m_O}{M}\times100=\frac{96.00}{331.22}\times100\approx29.0\%$
# Answer: D

15.
# Explanation:
## Step1: Find molar mass of $PCl_2(NH_2)_2$
$M = 30.97+2\times35.45+2\times14.01 + 4\times1.01=133.93\ g/mol$
## Step2: Find mass of chlorine
Mass of $Cl$ in the compound: $m_{Cl}=2\times35.45 = 70.9\ g/mol$
## Step3: Calculate mass - percentage of chlorine
$\text{Mass}\%=\frac{m_{Cl}}{M}\times100=\frac{70.9}{133.93}\times100\approx52.9\%$ (There seems to be an error in the problem - statement or options as the calculated value doesn't match the given options)

16.
# Explanation:
## Step1: Assume 100 g of the compound
So, we have $27.05\ g\ S$, $13.45\ g\ O$, and $59.50\ g\ Cl$
## Step2: Calculate moles of each element
$n_S=\frac{27.05\ g}{32.07\ g/mol}\approx0.8435\ mol$, $n_O=\frac{13.45\ g}{16.00\ g/mol}\approx0.8406\ mol$, $n_{Cl}=\frac{59.50\ g}{35.45\ g/mol}\approx1.68\ mol$
## Step3: Divide by the smallest number of moles
$n_S:n_O:n_{Cl}=\frac{0.8435}{0.8406}:\frac{0.8406}{0.8406}:\frac{1.68}{0.8406}\approx1:1:2$
The empirical formula is $SOCl_2$
# Answer: B

17.
# Explanation:
## Step1: Assume 100 g of the compound
We have $29\ g\ Na$, $41\ g\ S$, and $30\ g\ O$
## Step2: Calculate moles of each element
$n_{Na}=\frac{29\ g}{22.99\ g/mol}\approx1.26\ mol$, $n_S=\frac{41\ g}{32.07\ g/mol}\approx1.28\ mol$, $n_O=\frac{30\ g}{16.00\ g/mol}=1.875\ mol$
## Step3: Divide by the smallest number of moles
$n_{Na}:n_S:n_O=\frac{1.26}{1.26}:\frac{1.28}{1.26}:\frac{1.875}{1.26}\approx1:1:1.5$
Multiply by 2 to get whole - numbers: $n_{Na}:n_S:n_O = 2:2:3$
The empirical formula is $Na_2S_2O_3$
# Answer: A

18.
# Explanation:
## Step1: Calculate moles of $C$ from $CO_2$
$n_{CO_2}=\frac{1.34\ g}{44.01\ g/mol}\approx0.0304\ mol$, so $n_C = 0.0304\ mol$
## Step2: Calculate moles of $H$ from $H_2O$
$n_{H_2O}=\frac{0.255\ g}{18.02\ g/mol}\approx0.0141\ mol$, so $n_H=2\times0.0141 = 0.0282\ mol$
## Step3: Calculate mass of $C$ and $H$
Mass of $C$: $m_C=0.0304\ mol\times12.01\ g/mol = 0.365\ g$
Mass of $H$: $m_H=0.0282\ mol\times1.01\ g/mol=0.0285\ g$
## Step4: Calculate mass of $O$
$m_O=0.519-(0.365 + 0.0285)=0.1255\ g$
$n_O=\frac{0.1255\ g}{16.00\ g/mol}\approx0.00784\ mol$
## Step5: Divide by the smallest number of moles
$n_C:n_H:n_O=\frac{0.0304}{0.00784}:\frac{0.0282}{0.00784}:\frac{0.00784}{0.00784}\approx4:3.6:1\approx4:4:1$
The empirical formula is $C_4H_4O$ (There seems to be an error in the problem - statement or options as the closest option is not an exact match)

19.
# Explanation:
## Step1: Calculate moles of $C$ from $CO_2$
$n_{CO_2}=\frac{2.265\ g}{44.01\ g/mol}\approx0.0515\ mol$, so $n_C = 0.0515\ mol$
## Step2: Calculate moles of $H$ from $H_2O$
$n_{H_2O}=\frac{1.236\ g}{18.02\ g/mol}\approx0.0686\ mol$, so $n_H=2\times0.0686 = 0.1372\ mol$
## Step3: Calculate mass of $C$ and $H$
Mass of $C$: $m_C=0.0515\ mol\times12.01\ g/mol = 0.6185\ g$
Mass of $H$: $m_H=0.1372\ mol\times1.01\ g/mol = 0.1386\ g$
## Step4: Calculate mass of $O$
$m_O=1.031-(0.6185 + 0.1386)=0.2739\ g$
$n_O=\frac{0.2739\ g}{16.00\ g/mol}\approx0.0171\ mol$
## Step5: Divide by the smallest number of moles
$n_C:n_H:n_O=\frac{0.0515}{0.0171}:\frac{0.1372}{0.0171}:\frac{0.0171}{0.0171}\approx3:8:1$
The empirical formula is $C_3H_8O$
# Answer: A

20.
# Explanation:
## Step1: Assume 100 g of the compound
We have $70.6\ g\ C$, $5.9\ g\ H$, and $23.5\ g\ O$
## Step2: Calculate moles of each element
$n_C=\frac{70.6\ g}{12.01\ g/mol}\approx5.88\ mol$, $n_H=\frac{5.9\ g}{1.01\ g/mol}\approx5.84\ mol$, $n_O=\frac{23.5\ g}{16.00\ g/mol}\approx1.47\ mol$
## Step3: Divide by the smallest number of moles
$n_C:n_H:n_O=\frac{5.88}{1.47}:\frac{5.84}{1.47}:\frac{1.47}{1.47}\approx4:4:1$
The empirical formula is $C_4H_4O$
Empirical - formula mass $=4\times12.01+4\times1.01 + 16.00=68.08\ g/mol$
## Step4: Find the multiple $n$
$n=\frac{\text{Molecular weight}}{\text{Empirical - formula mass}}=\frac{136}{68.08}\approx2$
The molecular formula is $C_8H_8O_2$
# Answer: A

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QUESTION IMAGE

Question

Explanation:

Step1: Find molar mass of \(PCl_2(NH_2)_2\)

Step2: Find mass of nitrogen

Step3: Calculate mass - percentage of nitrogen

Step1: Find molar mass of \(Pb(NO_3)_2\)

Step2: Find mass of oxygen

Step3: Calculate mass - percentage of oxygen

Step1: Find molar mass of \(PCl_2(NH_2)_2\)

Step2: Find mass of chlorine

Step3: Calculate mass - percentage of chlorine

Answer: