Question

1. calculate the percentage by mass of nitrogen in ptcl2(nh3)2.

a) 4.67 b) 9.34 c) 9.90 d) 4.95 e) 12.67

1. 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

1. calculate the percentage by mass of chlorine in ptcl2(nh3)2.

a) 23.63 b) 11.82 c) 25.05 d) 12.53 e) 18.09

1. 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) clso4

1. 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

1. 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) c3h3o d) c2h6o5 e) c2h6o2

1. 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) c3h8o b) c3h5o c) ch3o d) c3h9o3 e) c3h6o3

1. 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) c5h6o2

Explanation:

Step1: Recall molar - mass formula

The percentage by mass of an element $X$ in a compound $A_xB_yC_z\cdots$ is given by $\text{Mass}\%_X=\frac{n_X\times M_X}{M_{\text{compound}}}\times100\%$, where $n_X$ is the number of atoms of element $X$ in the formula, $M_X$ is the molar mass of element $X$, and $M_{\text{compound}}$ is the molar mass of the compound.

Step2: Calculate molar - mass of $PtCl_2(NH_3)_2$

The molar mass of $Pt = 195.08\ g/mol$, $Cl=35.45\ g/mol$, $N = 14.01\ g/mol$, $H = 1.01\ g/mol$.
$M_{PtCl_2(NH_3)_2}=195.08 + 2\times35.45+2\times(14.01 + 3\times1.01)$
$=195.08+70.9+2\times(14.01 + 3.03)$
$=195.08+70.9+2\times17.04$
$=195.08+70.9 + 34.08$
$=299.06\ g/mol$.
The number of $N$ atoms $n_N = 2$. The molar mass of $N$ is $M_N=14.01\ g/mol$.
The mass percentage of $N$ is $\text{Mass}\%_N=\frac{2\times14.01}{299.06}\times100\%=\frac{28.02}{299.06}\times100\%\approx9.37\%\approx9.34\%$ (answer to 13).

Step3: Calculate molar - mass of $Pb(NO_3)_2$

The molar mass of $Pb = 207.2\ g/mol$, $N = 14.01\ g/mol$, $O = 16.00\ g/mol$.
$M_{Pb(NO_3)_2}=207.2+2\times(14.01 + 3\times16.00)$
$=207.2+2\times(14.01+48.00)$
$=207.2+2\times62.01$
$=207.2 + 124.02$
$=331.22\ g/mol$.
The number of $O$ atoms $n_O = 6$. The molar mass of $O$ is $M_O = 16.00\ g/mol$.
The mass percentage of $O$ is $\text{Mass}\%_O=\frac{6\times16.00}{331.22}\times100\%=\frac{96.00}{331.22}\times100\%\approx29.0\%$ (answer to 14).

Step4: Calculate molar - mass of $PtCl_2(NH_3)_2$ again for $Cl$ percentage

We already know $M_{PtCl_2(NH_3)_2}=299.06\ g/mol$. The number of $Cl$ atoms $n_{Cl}=2$. The molar mass of $Cl$ is $M_{Cl}=35.45\ g/mol$.
The mass percentage of $Cl$ is $\text{Mass}\%_{Cl}=\frac{2\times35.45}{299.06}\times100\%=\frac{70.9}{299.06}\times100\%\approx23.7\%$ (closest to 23.63% in 15).

Step5: Determine empirical formula for 16

Assume a 100 - g sample. So, $m_S = 27.0\ g$, $m_O=13.4\ g$, $m_{Cl}=59.6\ g$.
$n_S=\frac{27.0\ g}{32.07\ g/mol}\approx0.842\ mol$, $n_O=\frac{13.4\ g}{16.00\ g/mol}=0.8375\ mol$, $n_{Cl}=\frac{59.6\ g}{35.45\ g/mol}\approx1.68\ mol$.
Divide by the smallest number ($n_O\approx0.8375$):
$\frac{n_S}{n_O}\approx1$, $\frac{n_O}{n_O} = 1$, $\frac{n_{Cl}}{n_O}\approx2$. The empirical formula is $SOCl_2$.

Step6: Determine empirical formula for 17

Assume a 100 - g sample. So, $m_{Na}=29\ g$, $m_S = 41\ g$, $m_O=30\ g$.
$n_{Na}=\frac{29\ g}{22.99\ g/mol}\approx1.26\ mol$, $n_S=\frac{41\ g}{32.07\ g/mol}\approx1.28\ mol$, $n_O=\frac{30\ g}{16.00\ g/mol}=1.875\ mol$.
Divide by the smallest number ($n_{Na}\approx1.26$):
$\frac{n_{Na}}{n_{Na}} = 1$, $\frac{n_S}{n_{Na}}\approx1$, $\frac{n_O}{n_{Na}}\approx1.5$. Multiply by 2 to get whole - numbers. The empirical formula is $Na_2S_2O_3$.

Step7: Determine empirical formula for 18

From the combustion data:
$n_{CO_2}=\frac{1.24\ g}{44.01\ g/mol}\approx0.0282\ mol$, so $n_C = 0.0282\ mol$, $m_C=0.0282\ mol\times12.01\ g/mol\approx0.339\ g$.
$n_{H_2O}=\frac{0.255\ g}{18.02\ g/mol}\approx0.0142\ mol$, so $n_H=0.0284\ mol$, $m_H=0.0284\ mol\times1.01\ g/mol\approx0.0287\ g$.
$m_O=0.519-(0.339 + 0.0287)=0.1513\ g$, $n_O=\frac{0.1513\ g}{16.00\ g/mol}\approx0.00946\ mol$.
Divide by the smallest number ($n_O\approx0.00946$):
$\frac{n_C}{n_O}\approx3$, $\frac{n_H}{n_O}\approx3$, $\frac{n_O}{n_O}=1$. The empirical formula is $C_3H_3O$.

Step8: Determine empirical formula for 19

$n_{CO_2}=\frac{2.265\ g}{44.01\ g/mol}\approx0.0515\ mol$, so $n_C = 0.0515\ mol$, $m_C=0.0515\ mol\times12.01\ g/mol\approx0.619\ g$.
$n_{H_2O}=\frac{1.236\ g}{18.02\ g/mol}\approx0.0686\ mol$, so $n_H=0.1372\ mol$, $m_H=0.1372\ mol\times1.01\ g/mol\approx0.1386\ g$.
$m_O=1.031-(0.619 + 0.1386)=0.2734\ g$, $n_O=\frac{0.2734\ g}{16.00\ g/mol}\approx0.0171\ mol$.
Divide by the smallest number ($n_O\approx0.0171$):
$\frac{n_C}{n_O}\approx3$, $\frac{n_H}{n_O}\approx8$, $\frac{n_O}{n_O}=1$. The empirical formula is $C_3H_8O$.

Step9: Determine molecular formula for 20

Assume a 100 - g sample. $m_C = 70.6\ g$, $m_H=5.9\ g$, $m_O=(100-(70.6 + 5.9))=23.5\ g$.
$n_C=\frac{70.6\ g}{12.01\ g/mol}\approx5.88\ mol$, $n_H=\frac{5.9\ g}{1.01\ g/mol}\approx5.84\ mol$, $n_O=\frac{23.5\ g}{16.00\ g/mol}\approx1.47\ mol$.
Divide by the smallest number ($n_O\approx1.47$):
$\frac{n_C}{n_O}\approx4$, $\frac{n_H}{n_O}\approx4$, $\frac{n_O}{n_O}=1$. The empirical formula is $C_4H_4O$.
The empirical - formula mass $M_{C_4H_4O}=4\times12.01+4\times1.01 + 16.00=68.08\ g/mol$.
The molecular weight is 136 amu. $\frac{M_{\text{molecular}}}{M_{\text{empirical}}}=\frac{136}{68.08}\approx2$. The molecular formula is $C_8H_8O_2$.

Answer:

1. B. 9.34
2. D. 29.0
3. A. 23.63
4. B. $SOCl_2$
5. A. $Na_2S_2O_3$
6. B. $C_3H_3O$
7. A. $C_3H_8O$
8. A. $C_8H_8O_2$