The answers for the following sums is given below.
1.![Pd_{2}H_{2}](/tpl/images/0561/2103/f4821.png)
2.![C_{2} H_{6}](/tpl/images/0561/2103/45ca7.png)
3.![C_{2} H_{2} O_{2} Cl_{2}](/tpl/images/0561/2103/22ffb.png)
4.![C_{3}Cl_{3} N_{3}](/tpl/images/0561/2103/7199c.png)
5.![Tl_{2 } C_{4} H_{4}O_{6}](/tpl/images/0561/2103/1dba1.png)
6.![C_{8}H_{8}](/tpl/images/0561/2103/46b68.png)
7. ![N_{2}O_{5}](/tpl/images/0561/2103/4724a.png)
8.![P_{4}O_{6}](/tpl/images/0561/2103/ab63d.png)
9.![C_{4}H_{8} O_{2}](/tpl/images/0561/2103/69ea9.png)
Explanation:
1.Given:
    Molar mass=216.8g
Molecular formula=Pd![H_{2}](/tpl/images/0561/2103/afcdb.png)
    we know;
Molecular formula=n(Empirical formula)
molecular weight of palladium(Pd)=106.4u
molecular weight of hydrogen(H)=1u
Molar mass of Pd
:
Pd=106.4×1=106.4u
H=1×2=2
molar mass of Pd
=106.4+2=108.4
n=![\frac{216.8}{106.4}](/tpl/images/0561/2103/783d9.png)
n=2
Molecular formula=2(Pd
)
Molecular formula=![Pd_{2}H_{2}](/tpl/images/0561/2103/f4821.png)
Therefore the molecular formula of the compound is ![Pd_{2}H_{2}](/tpl/images/0561/2103/f4821.png)
2. Given:
    Molar mass=30.0g
Molecular formula=![CH_{3}](/tpl/images/0561/2103/be026.png)
    we know;
Molecular formula = n (Empirical formula)
molecular weight of Carbon(C)=12.01u
molecular weight of hydrogen(H)=1u
Molar mass of
:
C=12.01 × 1 = 12.01u
H=1 × 3 = 3u
molar mass of
=12.01 + 3 =15.01u
n=![\frac{30.0}{15.01}](/tpl/images/0561/2103/df483.png)
n=2
Molecular formula=2(
)
Molecular formula=![C_{2} H_{6}](/tpl/images/0561/2103/45ca7.png)
Therefore the molecular formula of the compound is ![C_{2} H_{6}](/tpl/images/0561/2103/45ca7.png)
3. Given:
    Molar mass=129g
Molecular formula=CHOCl
    we know;
Molecular formula = n (Empirical formula)
molecular weight of Carbon(C)=12.01u
molecular weight of hydrogen(H)=1u
molecular weight of oxygen(O)=16.00u
molecular weight of chlorie(Cl)=35.5u
Molar mass of CHOCl:
C=12.01 × 1 = 12.01u
H=1 × 1 = 1u
O=16.00×1=16.00u
Cl=35.5×1=35.5u
molar mass of CHOCl=12.01+1+16.00+35.5=64.5u
n=![\frac{129}{64.5}](/tpl/images/0561/2103/c8c77.png)
n=2
Molecular formula=2(CHOCl)
Molecular formula=![C_{2} H_{2} O_{2} Cl_{2}](/tpl/images/0561/2103/22ffb.png)
Therefore the molecular formula of the compound is ![C_{2} H_{2} O_{2} Cl_{2}](/tpl/images/0561/2103/22ffb.png)
5. Given:
    Molar mass=577g
Molecular formula=![TlC_{2} H_{2}O_{3}](/tpl/images/0561/2103/db491.png)
    we know;
Molecular formula = n (Empirical formula)
molecular weight of Carbon(C)=12.01u
molecular weight of Thallium(Tl)=204.3u
molecular weight of hydrogen(H)=1u
molecular weight of oxygen(O)=16.00u
Molar mass of
:
C=12.01 × 2= 24.02u
Tl=204.3×1=204.3u
H=1×2=2u
O=16.00×3=48.00
molar mass of
=204.3+24.02+1+48.00=278.32u
n=![\frac{577}{278.32}](/tpl/images/0561/2103/ea473.png)
n=2
Molecular formula=2 (
)
Molecular formula=![Tl_{2 } C_{4} H_{4}O_{6}](/tpl/images/0561/2103/1dba1.png)
Therefore the molecular formula of the compound is ![Tl_{2 } C_{4} H_{4}O_{6}](/tpl/images/0561/2103/1dba1.png)
4. Molar mass=184.5g
Molecular formula=CClN
    we know;
Molecular formula = n (Empirical formula)
molecular weight of Carbon(C)=12.01u
molecular weight of Nitrogen(N)=14u
molecular weight of chlorine(Cl)=35.5u
Molar mass of CClN:
C=12.01 × 1 = 12.01u
N=1×14=14U
Cl=35.5×1=35.5u
molar mass of CClN=12.01+14+35.5=61.5u
n=![\frac{184.5}{61.5}](/tpl/images/0561/2103/a61f8.png)
n=3
Molecular formula=3 (CClN)
Molecular formula=![C_{3}Cl_{3} N_{3}](/tpl/images/0561/2103/7199c.png)
Therefore the molecular formula of the compound is ![C_{3}Cl_{3} N_{3}](/tpl/images/0561/2103/7199c.png)
6. For the table refer the attached file.
Simplest ratio of elements:
Carbon=8
Hydrogen=8
Empirical formula=![C_{8}H_{8}](/tpl/images/0561/2103/46b68.png)
Molecular formula =![C_{8}H_{8}](/tpl/images/0561/2103/46b68.png)
Molar mass of
:
molecular weight of carbon=12.04u
molecular weight of hydrogen=1u
C=8×12.01=96.08u
H=1×8=8u
molar mass of
=96.08+8=104.08u
n=104.08÷78.0
n=1
Molecular formula = n(Empirical formula)
Molecular formula = 1(
)
Molecular formula =![C_{8}H_{8}](/tpl/images/0561/2103/46b68.png)
Therefore the molecular formula of a compound is ![C_{8}H_{8}](/tpl/images/0561/2103/46b68.png)
7. Given:
mass of oxide of nitrogen=108g
mass of nitrogen=4.02g
mass of oxygen=11.48g
moles of nitrogen=
= 0.289 moles
moles of oxygen=
=0.716 moles
We divide through by the lowest molar quantity to give an empirical formula  of
.
Now the molecular formula is multiple of the empirical formula.
So,
108 = n × (2×14.01 + 5×15.999)
Clearly,n=1, and the molecular formula is
.
8.For the table refer the attached file.
Simplest ratio of elements:
Phosphorus=2
Oxygen=3
We know;
Empirical formula=![P_{2} O_{3}](/tpl/images/0561/2103/d8be6.png)
molecular formula= 2(Empirical formula)
Molecular formula =2(
)
Molecular formula =![P_{4}O_{6}](/tpl/images/0561/2103/ab63d.png)
Therefore the molecular formula of the compound is ![P_{4}O_{6}](/tpl/images/0561/2103/ab63d.png)
9. For the table refer the attached file.
Simplest ratio of elements:
Carbon=2
Hydrogen=9
Oxygen=2
We know;
Empirical formula =![C_{2} H_{4} O](/tpl/images/0561/2103/d606b.png)
Molecular formula = 2(Empirical formula)
Molecular formula =2(
)
Molecular formula =![C_{4}H_{8} O_{2}](/tpl/images/0561/2103/69ea9.png)
Therefore the molecular formula of the compound is ![C_{4}H_{8} O_{2}](/tpl/images/0561/2103/69ea9.png)
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